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6.4: Equilibria Involving Weak Acids and Bases

  • Page ID
    43433
  • Learning Objectives

    • Rationalize trends in acid–base strength in relation to molecular structure
    • Carry out equilibrium calculations for weak acid–base systems

    Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid).

    Acetic acid (\(\ce{CH3CO2H}\)) is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation:

    \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber\]

    giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure \(\PageIndex{1}\)). The remaining weak acid is present in the nonionized form.

    For acetic acid, at equilibrium:

    \[K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=1.8 \times 10^{−5} \nonumber\]

    This image shows two bottles containing clear colorless solutions. Each bottle contains a single p H indicator strip. The strip in the bottle on the left is red, and a similar red strip is placed on a filter paper circle in front of the bottle on surface on which the bottles are resting. Similarly, the second bottle on the right contains and orange strip and an orange strip is placed in front of it on a filter paper circle. Between the two bottles is a pack of p Hydrion papers with a p H color scale on its cover.
    Figure \(\PageIndex{1}\): pH paper indicates that a 0.l-M solution of \(\ce{HCl}\) (beaker on left) has a pH of 1. The acid is fully ionized and \(\ce{[H3O+]}\) = 0.1 M. A 0.1-M solution of CH3CO2H (beaker on right) has a pH of 3 ( \(\ce{[H3O+]}\) = 0.001 M) because the weak acid CH3CO2H is only partially ionized. In this solution, \(\ce{[H3O+]} < [\ce{CH3CO2H}]\). (credit: modification of work by Sahar Atwa)
    Table \(\PageIndex{1}\): Ionization Constants of Some Weak Acids
    Ionization Reaction Ka at 25 °C
    \(\ce{HSO4- + H2O ⇌ H3O+ + SO4^2-}\) 1.2 × 10−2
    \(\ce{HF + H2O ⇌ H3O+ + F-}\) 3.5 × 10−4
    \(\ce{HNO2 + H2O ⇌ H3O+ + NO2-}\) 4.6 × 10−4
    \(\ce{HNCO + H2O ⇌ H3O+ + NCO-}\) 2 × 10−4
    \(\ce{HCO2H + H2O ⇌ H3O+ + HCO2-}\) 1.8 × 10−4
    \(\ce{CH3CO2H + H2O ⇌ H3O+ + CH3CO2-}\) 1.8 × 10−5
    \(\ce{HCIO + H2O ⇌ H3O+ + CIO-}\) 2.9 × 10−8
    \(\ce{HBrO + H2O ⇌ H3O+ + BrO-}\) 2.8 × 10−9
    \(\ce{HCN + H2O ⇌ H3O+ + CN-}\) 4.9 × 10−10

    Table \(\PageIndex{1}\) gives the ionization constants for several weak acids; additional ionization constants can be found in Table E1.

    At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base).

    For example, a solution of the weak base trimethylamine, (CH3)3N, in water reacts according to the equation:

    \[\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \nonumber\]

    This gives an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids.

    We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure \(\PageIndex{2}\)). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, \(K_b\), is called the ionization constant of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium:

    \[K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}} \nonumber\]

    alt
    Figure \(\PageIndex{2}\): pH paper indicates that a 0.1-M solution of NH3 (left) is weakly basic. The solution has a pOH of 3 ([OH] = 0.001 M) because the weak base NH3 only partially reacts with water. A 0.1-M solution of NaOH (right) has a pOH of 1 because NaOH is a strong base (credit: modification of work by Sahar Atwa).

    The ionization constants of several weak bases are given in Table \(\PageIndex{2}\) and Table E2.

    Table \(\PageIndex{2}\): Ionization Constants of Some Weak Bases
    Ionization Reaction Kb at 25 °C
    \(\ce{(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-}\) 5.9 × 10−4
    \(\ce{CH3NH2 + H2O ⇌ CH3NH3+ + OH-}\) 4.4 × 10−4
    \(\ce{(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-}\) 6.3 × 10−5
    \(\ce{NH3 + H2O ⇌ NH4+ + OH-}\) 1.8 × 10−5
    \(\ce{C6H5NH2 + H2O ⇌ C6N5NH3+ + OH-}\) 4.3 × 10−10

    Example \(\PageIndex{3}\): Determination of Ka from Equilibrium Concentrations

    Acetic acid is the principal ingredient in vinegar; that's why it tastes sour. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and \(\ce{[H3O+]}=\ce{[CH3CO2- ]}=0.00118\:M\). What is the value of \(K_a\) for acetic acid?

    alt
    Vinegar is a solution of acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)

    Solution

    We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:

    \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber \]

    \[\begin{align*} K_\ce{a} &=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}} \\[4pt] &=\dfrac{(0.00118)(0.00118)}{0.0787} \\[4pt] &=1.77×10^{−5} \end{align*}\]

    Exercise \(\PageIndex{3}\)

    What is the equilibrium constant for the ionization of the \(\ce{HSO4-}\) ion, the weak acid used in some household cleansers:

    \[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \nonumber\]

    In one mixture of NaHSO4 and Na2SO4 at equilibrium, \(\ce{[H3O+]}\) = 0.027 M; \(\ce{[HSO4- ]}=0.29\:M\); and \(\ce{[SO4^2- ]}=0.13\:M\).

    Answer

    \(K_a\) for \(\ce{HSO_4^-}= 1.2 ×\times 10^{−2}\)

    Example \(\PageIndex{4}\): Determination of Kb from Equilibrium Concentrations

    Caffeine, C8H10N4O2 is a weak base. What is the value of Kb for caffeine if a solution at equilibrium has [C8H10N4O2] = 0.050 M, \(\ce{[C8H10N4O2H+]}\) = 5.0 × 10−3 M, and [OH] = 2.5 × 10−3 M?

    Solution

    At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:

    \[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \nonumber \]

    so

    \[K_\ce{b}=\ce{\dfrac{[C8H10N4O2H+][OH- ]}{[C8H10N4O2]}}=\dfrac{(5.0×10^{−3})(2.5×10^{−3})}{0.050}=2.5×10^{−4} \nonumber\]

    Exercise \(\PageIndex{4}\)

    What is the equilibrium constant for the ionization of the \(\ce{HPO4^2-}\) ion, a weak base:

    \[\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H2PO4-}(aq)+\ce{OH-}(aq) \nonumber \]

    In a solution containing a mixture of \(\ce{NaH2PO4}\) and \(\ce{Na2HPO4}\) at equilibrium with:

    • \([\ce{OH^{−}}] = 1.3 × 10^{−6} M\)
    • \(\ce{[H2PO4^{-}]=0.042\:M}\) and
    • \(\ce{[HPO4^{2-}]=0.341\:M}\).
    Answer

    Kb for \(\ce{HPO4^2-}=1.6×10^{−7} \)

    Example \(\PageIndex{5}\): Determination of Ka or Kb from pH

    The pH of a 0.0516-M solution of nitrous acid, \(\ce{HNO2}\), is 2.34. What is its \(K_a\)?

    \[\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) \nonumber\]

    Solution

    We determine an equilibrium constant starting with the initial concentrations of HNO2, \(\ce{H3O+}\), and \(\ce{NO2-}\) as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)

    We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction:

    Example 14.3.5.png

    We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its concentration):

    Example 14.3.5B.png

    To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate \(\ce{[H3O+]}\), the equilibrium concentration of \(\ce{H3O+}\), from the pH:

    \[\ce{[H3O+]}=10^{−2.34}=0.0046\:M \nonumber \]

    The change in concentration of \(\ce{H3O+}\), \(x_{\ce{[H3O+]}}\), is the difference between the equilibrium concentration of H3O+, which we determined from the pH, and the initial concentration, \(\mathrm{[H_3O^+]_i}\). The initial concentration of \(\ce{H3O+}\) is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0).

    The change in concentration of \(\ce{NO2-}\) is equal to the change in concentration of \(\ce{[H3O+]}\). For each 1 mol of \(\ce{H3O+}\) that forms, 1 mol of \(\ce{NO2-}\) forms. The equilibrium concentration of HNO2 is equal to its initial concentration plus the change in its concentration.

    Now we can fill in the ICE table with the concentrations at equilibrium, as shown here:

    150749131940304.png

    Finally, we calculate the value of the equilibrium constant using the data in the table:

    \[K_\ce{a}=\ce{\dfrac{[H3O+][NO2- ]}{[HNO2]}}=\dfrac{(0.0046)(0.0046)}{(0.0470)}=4.5×10^{−4} \nonumber\]

    Exercise \(\PageIndex{5}\)

    The pH of a solution of household ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb for NH3.

    Answer

    \(K_b = 1.8 × 10^{−5}\)

    Example \(\PageIndex{6}\): Equilibrium Concentrations in a Solution of a Weak Acid

    Formic acid, HCO2H, is the irritant that causes the body’s reaction to ant stings.

    A photograph is shown of a large black ant on the end of a human finger.
    The pain of an ant’s sting is caused by formic acid. (credit: John Tann)

    What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid?

    \[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−4} \nonumber \]

    Solution

    1. Determine x and equilibrium concentrations. The equilibrium expression is:

    \[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \nonumber \]

    The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table.

    The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):

    Example 14.3.6.png

    2. Solve for \(x\) and the equilibrium concentrations. At equilibrium:

    \[\begin{align*} K_\ce{a} &=1.8×10^{−4}=\ce{\dfrac{[H3O+][HCO2- ]}{[HCO2H]}} \\[4pt] &=\dfrac{(x)(x)}{0.534−x}=1.8×10^{−4} \end{align*}\]

    Now solve for \(x\). Because the initial concentration of acid is reasonably large and \(K_a\) is very small, we assume that \(x << 0.534\), which permits us to simplify the denominator term as \((0.534 − x) = 0.534\). This gives:

    \[K_\ce{a}=1.8×10^{−4}=\dfrac{x^{2}}{0.534} \nonumber\]

    Solve for \(x\) as follows:

    \[\begin{align*} x^2 &=0.534×(1.8×10^{−4}) \\[4pt] &=9.6×10^{−5} \\[4pt] x &=\sqrt{9.6×10^{−5}} \\[4pt] &=9.8×10^{−3} \end{align*}\]

    To check the assumption that \(x\) is small compared to 0.534, we calculate:

    \[\begin{align*} \dfrac{x}{0.534} &=\dfrac{9.8×10^{−3}}{0.534} \\[4pt] &=1.8×10^{−2} \, \textrm{(1.8% of 0.534)} \end{align*}\]

    \(x\) is less than 5% of the initial concentration; the assumption is valid.

    We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:

    \[\begin{align*} \ce{[H3O+]} &=~0+x=0+9.8×10^{−3}\:M. \\[4pt] &=9.8×10^{−3}\:M \end{align*}\]

    The pH of the solution can be found by taking the negative log of the \(\ce{[H3O+]}\), so:

    \[pH = −\log(9.8×10^{−3})=2.01 \nonumber\]

    Exercise \(\PageIndex{6}\): acetic acid

    Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H?

    \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5} \nonumber\]

    Hint

    Determine \(\ce{[CH3CO2- ]}\) at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or \(\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]_{initial}}}×100\).

    Answer

    percent ionization = 1.3%

    The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.

    Example \(\PageIndex{7}\): Equilibrium Concentrations in a Solution of a Weak Base

    Find the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base:

    \[\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=6.3×10^{−5} \nonumber\]

    Solution This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example \(\PageIndex{6}\). The reactants and products will be different and the numbers will be different, but the logic will be the same:

    Example 14.3.7A.png

    1. Determine x and equilibrium concentrations. The table shows the changes and concentrations:

    Example 14.3.7B.png

    2. Solve for \(x\) and the equilibrium concentrations. At equilibrium:

    \[K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{0.25−x=}6.3×10^{−5} \nonumber\]

    If we assume that x is small relative to 0.25, then we can replace (0.25 − x) in the preceding equation with 0.25. Solving the simplified equation gives:

    \[x=4.0×10^{−3} \nonumber \]

    This change is less than 5% of the initial concentration (0.25), so the assumption is justified.

    Recall that, for this computation, \(x\) is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation):

    \[\begin{align*} (\ce{[OH- ]}=~0+x=x=4.0×10^{−3}\:M \\[4pt] &=4.0×10^{−3}\:M \end{align*}\]

    Then calculate pOH as follows:

    \[\ce{pOH}=−\log(4.3×10^{−3})=2.40 \nonumber\]

    Using the relation introduced in the previous section of this chapter:

    \[\mathrm{pH + pOH=p\mathit{K}_w=14.00}\nonumber \]

    permits the computation of pH:

    \[\mathrm{pH=14.00−pOH=14.00−2.37=11.60} \nonumber\]

    Check the work. A check of our arithmetic shows that \(K_b = 6.3 \times 10^{−5}\).

    Exercise \(\PageIndex{7}\)

    1. Show that the calculation in Step 2 of this example gives an x of 4.3 × 10−3 and the calculation in Step 3 shows Kb = 6.3 × 10−5.
    2. Find the concentration of hydroxide ion in a 0.0325-M solution of ammonia, a weak base with a Kb of 1.76 × 10−5. Calculate the percent ionization of ammonia, the fraction ionized × 100, or \(\ce{\dfrac{[NH4+]}{[NH3]}}×100 \%\)
    Answer a

    \(7.56 × 10^{−4}\, M\), 2.33%

    Answer b

    2.33%

    Some weak acids and weak bases ionize to such an extent that the simplifying assumption that x is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation.

    Example \(\PageIndex{8}\): Equilibrium Concentrations in a Solution of a Weak Acid

    Sodium bisulfate, NaHSO4, is used in some household cleansers because it contains the \(\ce{HSO4-}\) ion, a weak acid. What is the pH of a 0.50-M solution of \(\ce{HSO4-}\)?

    \[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \nonumber \]

    Solution

    We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of \(\ce{HSO4-}\) so that we can use \(\ce{[H3O+]}\) to determine the pH. As in the previous examples, we can approach the solution by the following steps:

    Example 14.3.8A.png

    1. Determine \(x\) and equilibrium concentrations. This table shows the changes and concentrations:

    Example 14.3.8B.png

    2. Solve for \(x\) and the concentrations.

    As we begin solving for \(x\), we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of \(x\).

    At equilibrium:

    \[K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+][SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x} \nonumber\]

    If we assume that x is small and approximate (0.50 − x) as 0.50, we find:

    \[x=7.7×10^{−2} \nonumber \]

    When we check the assumption, we confirm:

    \[\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \nonumber\]

    which for this system is

    \[\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \nonumber\]

    The value of \(x\) is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find \(x\).

    The equation:

    \[K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x}\nonumber \]

    gives

    \[6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \nonumber\]

    or

    \[x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0 \nonumber \]

    This equation can be solved using the quadratic formula. For an equation of the form

    \[ax^{2+} + bx + c=0, \nonumber\]

    \(x\) is given by the quadratic equation:

    \[x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \nonumber\]

    In this problem, \(a = 1\), \(b = 1.2 × 10^{−3}\), and \(c = −6.0 × 10^{−3}\).

    Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:

    \[x=7.2×10^{−2} \nonumber\]

    Now determine the hydronium ion concentration and the pH:

    \[\begin{align*} \ce{[H3O+]} &=~0+x=0+7.2×10^{−2}\:M \\[4pt] &=7.2×10^{−2}\:M \end{align*}\]

    The pH of this solution is:

    \[\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \nonumber\]

    Exercise \(\PageIndex{8}\)

    1. Show that the quadratic formula gives \(x = 7.2 × 10^{−2}\).
    2. Calculate the pH in a 0.010-M solution of caffeine, a weak base:

    \[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=2.5×10^{−4} \nonumber\]

    Hint

    It will be necessary to convert [OH] to \(\ce{[H3O+]}\) or pOH to pH toward the end of the calculation.

    Answer

    pH 11.16

    The Relative Strengths of Strong Acids and Bases

    Strong acids, such as \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\), all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl, Br, and I that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\) differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order \(\ce{HCl < HBr < HI}\), and so \(\ce{HI}\) is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the leveling effect of water.

    Water also exerts a leveling effect on the strengths of strong bases. For example, the oxide ion, O2−, and the amide ion, \(\ce{NH2-}\), are such strong bases that they react completely with water:

    \[\ce{O^2-}(aq)+\ce{H2O}(l)⟶\ce{OH-}(aq)+\ce{OH-}(aq)\]

    \[\ce{NH2-}(aq)+\ce{H2O}(l)⟶\ce{NH3}(aq)+\ce{OH-}(aq)\]

    Thus, O2− and \(\ce{NH2-}\) appear to have the same base strength in water; they both give a 100% yield of hydroxide ion.

    In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is \(\ce{HF < HCl < HBr < HI}\). Likewise, for group 16, the order of increasing acid strength is H2O < H2S < H2Se < H2Te. Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is \(\ce{CH4 < NH3 < H2O < HF}\); across the third row, it is \(\ce{SiH4 < PH3 < H2S < HCl}\) (see Figure \(\PageIndex{3}\)).

    This diagram has two rows and four columns. Red arrows point left across the bottom of the figure and down at the right side and are labeled “Increasing acid strength.” Blue arrows point left across the bottom and up at the right side of the figure and are labeled “Increasing base strength.” The first column is labeled 14 at the top and two white squares are beneath it. The first has the number 6 in the upper left corner and the formula C H subscript 4 in the center along with designation Neither acid nor base. The second square contains the number 14 in the upper left corner, the formula C H subscript 4 at the center and the designation Neither acid nor base. The second column is labeled 15 at the top and two blue squares are beneath it. The first has the number 7 in the upper left corner and the formula N H subscript 3 in the center along with the designation Weak base and K subscript b equals 1.8 times 10 superscript negative 5. The second square contains the number 15 in the upper left corner, the formula P H subscript 3 at the center and the designation Very weak base and K subscript b equals 4 times 10 superscript negative 28. The third column is labeled 16 at the top and two squares are beneath it. The first is shaded tan and has the number 8 in the upper left corner and the formula H subscript 2 O in the center along with the designation neutral. The second square is shaded pink, contains the number 16 in the upper left corner, the formula H subscript 2 S at the center and the designation Weak acid and K subscript a equals 9.5 times 10 superscript negative 8. The fourth column is labeled 17 at the top and two squares are beneath it. The first is shaded pink, has the number 9 in the upper left corner and the formula H F in the center along with the designation Weak acid and K subscript a equals 6.8 times 10 superscript negative 4. The second square is shaded a deeper pink, contains the number 17 in the upper left corner, the formula H C l at the center, and the designation Strong acid.
    Figure \(\PageIndex{3}\): As you move from left to right and down the periodic table, the acid strength increases. As you move from right to left and up, the base strength increases.

    Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula OnE(OH)m, and include sulfuric acid, \(\ce{O2S(OH)2}\), sulfurous acid, \(\ce{OS(OH)2}\), nitric acid, \(\ce{O2NOH}\), perchloric acid, \(\ce{O3ClOH}\), aluminum hydroxide, \(\ce{Al(OH)3}\), calcium hydroxide, \(\ce{Ca(OH)2}\), and potassium hydroxide, \(\ce{KOH}\):

    A diagram is shown that includes a central atom designated with the letter E. Single bonds extend above, below, left, and right of the E. An O atom is bonded to the right of the E, and an arrow points to the bond labeling it, “Bond a.” An H atom is single bonded to the right of the O atom. An arrow pointing to this bond connects it to the label, “Bond b.”

    If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond a between the element and oxygen is more readily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH)2 and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.

    If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond a relatively strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called oxyacids.

    Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H2SO3, or OS(OH)2 (with a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure \(\PageIndex{4}\)).

    A diagram is shown that includes four structural formulas for acids. A red, right pointing arrow is placed beneath the structures which is labeled “Increasing acid strength.” At the top left, the structure of Nitrous acid is provided. It includes an H atom to which an O atom with two unshared electron pairs is connected with a single bond to the right. A single bond extends to the right and slightly below to a N atom with one unshared electron pair. A double bond extends up and to the right from this N atom to an O atom which has two unshared electron pairs. To the upper right is a structure for Nitric acid. This structure differs from the previous structure in that the N atom is directly to the right of the first O atom and a second O atom with three unshared electron pairs is connected with a single bond below and to the right of the N atom which has no unshared electron pairs. At the lower left, an O atom with two unshared electron pairs is double bonded to its right to an S atom with a single unshared electron pair. An O atom with two unshared electron pairs is bonded above and an H atom is single bonded to this O atom. To the right of the S atom is a single bond to another O atom with two unshared electron pairs to which an H atom is single bonded. This structure is labeled “Sulfurous acid.” A similar structure which is labeled “Sulfuric acid” is placed in the lower right region of the figure. This structure differs in that an H atom is single bonded to the left of the first O atom, leaving it with two unshared electron pairs and a fourth O atom with two unshared electron pairs is double bonded beneath the S atom, leaving it with no unshared electron pairs.
    Figure \(\PageIndex{4}\): As the oxidation number of the central atom E increases, the acidity also increases.

    Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate \(\ce{Al(H2O)3(OH)3}\), is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, \(\ce{Al(H2O)3(OH)3}\), is converted into the soluble ion, \(\ce{[Al(H2O)2(OH)4]-}\), by reaction with hydroxide ion:

    \[[\ce{Al(H2O)3(OH)3}](aq)+\ce{OH-}(aq)⇌\ce{H2O}(l)+\ce{[Al(H2O)2(OH)4]-}(aq) \nonumber\]

    In this reaction, a proton is transferred from one of the aluminum-bound H2O molecules to a hydroxide ion in solution. The \(\ce{Al(H2O)3(OH)3}\) compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion \(\ce{[Al(H2O)6]^3+}\) by reaction with hydronium ion:

    \[\ce{3H3O+}(aq)+\ce{Al(H2O)3(OH)3}(aq)⇌\ce{Al(H2O)6^3+}(aq)+\ce{3H2O}(l) \nonumber\]

    In this case, protons are transferred from hydronium ions in solution to \(\ce{Al(H2O)3(OH)3}\), and the compound functions as a base.

    Summary

    The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH4 < NH3 < H2O < HF), and they increase down a group (HF < HCl < HBr < HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H2SO3 < H2SO4). The strengths of oxyacids also increase as the electronegativity of the central element increases [H2SeO4 < H2SO4].

    Key Equations

    • \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\)
    • \(K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}\)
    • \(K_a \times K_b = 1.0 \times 10^{−14} = K_w \,(\text{at room temperature})\)
    • \(\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\)

    Glossary

    acid ionization constant (Ka)
    equilibrium constant for the ionization of a weak acid
    base ionization constant (Kb)
    equilibrium constant for the ionization of a weak base
    leveling effect of water
    any acid stronger than \(\ce{H3O+}\), or any base stronger than OH will react with water to form \(\ce{H3O+}\), or OH, respectively; water acts as a base to make all strong acids appear equally strong, and it acts as an acid to make all strong bases appear equally strong
    oxyacid
    compound containing a nonmetal and one or more hydroxyl groups
    percent ionization
    ratio of the concentration of the ionized acid to the initial acid concentration, times 100

    Contributors and Attributions

    <div class="comment">
    <div class="mt-comment-content">
    <p>Forker source[1] start-chem-38278</p>
    </div>
    </div>

    <div class="box-objectives">
    <p class="box-legend lt-chem-38278"><span>Learning Objectives</span></p>

    <ul>
        <li class="lt-chem-38278">Rationalize trends in acid–base strength in relation to molecular structure</li>
        <li class="lt-chem-38278">Carry out equilibrium calculations for weak acid–base systems</li>
    </ul>
    </div>

    <p class="lt-chem-38278">Many acids and bases are weak; that is, they do not ionize fully in aqueous solution. A solution of a weak acid in water is a mixture of the nonionized acid, hydronium ion, and the conjugate base of the acid, with the nonionized acid present in the greatest concentration. Thus, a weak acid increases the hydronium ion concentration in an aqueous solution (but not as much as the same amount of a strong acid).</p>

    <p class="lt-chem-38278">Acetic acid (\(\ce{CH3CO2H}\)) is a weak acid. When we add acetic acid to water, it ionizes to a small extent according to the equation:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber\]</p>

    <p class="lt-chem-38278">giving an equilibrium mixture with most of the acid present in the nonionized (molecular) form. This equilibrium, like other equilibria, is dynamic; acetic acid molecules donate hydrogen ions to water molecules and form hydronium ions and acetate ions at the same rate that hydronium ions donate hydrogen ions to acetate ions to reform acetic acid molecules and water molecules. We can tell by measuring the pH of an aqueous solution of known concentration that only a fraction of the weak acid is ionized at any moment (Figure \(\PageIndex{1}\)). The remaining weak acid is present in the nonionized form.</p>

    <p class="lt-chem-38278">For acetic acid, at equilibrium:</p>

    <p class="mt-align-center lt-chem-38278">\[K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=1.8 \times 10^{−5} \nonumber\]</p>

    <figure><img alt="This image shows two bottles containing clear colorless solutions. Each bottle contains a single p H indicator strip. The strip in the bottle on the left is red, and a similar red strip is placed on a filter paper circle in front of the bottle on surface on which the bottles are resting. Similarly, the second bottle on the right contains and orange strip and an orange strip is placed in front of it on a filter paper circle. Between the two bottles is a pack of p Hydrion papers with a p H color scale on its cover." height="306" src="/@api/deki/files/65686/CNX_Chem_14_03_acetate_img.jpg" width="402" />
    <figcaption>Figure \(\PageIndex{1}\): pH paper indicates that a 0.l-M solution of \(\ce{HCl}\) (beaker on left) has a pH of 1. The acid is fully ionized and \(\ce{[H3O+]}\) = 0.1 M. A 0.1-M solution of CH<sub>3</sub>CO<sub>2</sub>H (beaker on right) has a pH of 3 ( \(\ce{[H3O+]}\) = 0.001 M) because the weak acid CH<sub>3</sub>CO<sub>2</sub>H is only partially ionized. In this solution, \(\ce{[H3O+]} &lt; [\ce{CH3CO2H}]\). (credit: modification of work by Sahar Atwa)</figcaption>
    </figure>

    <table align="center" class="mt-responsive-table" summary="This table has two columns and ten rows. The first row is a header row and it labels each column, “Ionization Reaction,” and “K subscript a at 25 degrees C.” Under the “Ionization Reaction” column are the following equations: “H S O subscript 4 superscript negative sign plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus S O subscript 4 superscript 2 negative sign,” “H F plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus F superscript negative sign,” “H N O subscript 2 plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus N O subscript 2 superscript negative sign,” “H N C O plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sing plus H C O subscript 2 superscript negative sign,” “C H subscript 3 C O subscript 2 H plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus C I O superscript negative sign,” “ H B r O plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus B r O superscript negative sign,” and “H C N plus H subscript 2 O equilibrium arrow H subscript 3 O superscript plus sign plus C N superscript negative sign.” Under the “K subscript a at 25 degrees C” column are the following values: 1.2 times ten to the negative two, 7.2 times ten to the negative 4, 4.5 times ten to the negative 4, 3.46 times ten to the negative 4, 1.8 times ten to the negative 4, 1.8 times ten to the negative 5, 3.5 times ten to the negative 8, 2 times ten to the negative 9, and 4 times ten to the negative ten.">
        <caption>Table \(\PageIndex{1}\): Ionization Constants of Some Weak Acids</caption>
        <thead>
            <tr valign="top">
                <th scope="col" style="text-align:center;">Ionization Reaction</th>
                <th scope="col" style="text-align:center;"><em>K</em><sub>a</sub> at 25 °C</th>
            </tr>
        </thead>
        <tbody>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{HSO4- + H2O ⇌ H3O+ + SO4^2-}\)</td>
                <td class="lt-chem-38278" data-th="Ka at 25 °C" style="text-align:center;">1.2 × 10<sup>−2</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{HF + H2O ⇌ H3O+ + F-}\)</td>
                <td class="lt-chem-38278" data-th="Ka at 25 °C" style="text-align:center;">3.5 × 10<sup>−4</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{HNO2 + H2O ⇌ H3O+ + NO2-}\)</td>
                <td class="lt-chem-38278" data-th="Ka at 25 °C" style="text-align:center;">4.6 × 10<sup>−4</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{HNCO + H2O ⇌ H3O+ + NCO-}\)</td>
                <td class="lt-chem-38278" data-th="Ka at 25 °C" style="text-align:center;">2 × 10<sup>−4</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{HCO2H + H2O ⇌ H3O+ + HCO2-}\)</td>
                <td class="lt-chem-38278" data-th="Ka at 25 °C" style="text-align:center;">1.8 × 10<sup>−4</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{CH3CO2H + H2O ⇌ H3O+ + CH3CO2-}\)</td>
                <td class="lt-chem-38278" data-th="Ka at 25 °C" style="text-align:center;">1.8 × 10<sup>−5</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{HCIO + H2O ⇌ H3O+ + CIO-}\)<span> </span></td>
                <td class="lt-chem-38278" data-th="Ka at 25 °C" style="text-align:center;">2.9 × 10<sup>−8</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{HBrO + H2O ⇌ H3O+ + BrO-}\)</td>
                <td class="lt-chem-38278" data-th="Ka at 25 °C" style="text-align:center;">2.8 × 10<sup>−9</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{HCN + H2O ⇌ H3O+ + CN-}\)</td>
                <td class="lt-chem-38278" data-th="Ka at 25 °C" style="text-align:center;">4.9 × 10<sup>−10</sup></td>
            </tr>
        </tbody>
    </table>

    <p class="lt-chem-38278">Table \(\PageIndex{1}\) gives the ionization constants for several weak acids; additional ionization constants can be found in Table E1.</p>

    <p class="lt-chem-38278">At equilibrium, a solution of a weak base in water is a mixture of the nonionized base, the conjugate acid of the weak base, and hydroxide ion with the nonionized base present in the greatest concentration. Thus, a weak base increases the hydroxide ion concentration in an aqueous solution (but not as much as the same amount of a strong base).</p>

    <p class="lt-chem-38278">For example, a solution of the weak base trimethylamine, (CH<sub>3</sub>)<sub>3</sub>N, in water reacts according to the equation:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \nonumber\]</p>

    <p class="lt-chem-38278">This gives an equilibrium mixture with most of the base present as the nonionized amine. This equilibrium is analogous to that described for weak acids.</p>

    <p class="lt-chem-38278">We can confirm by measuring the pH of an aqueous solution of a weak base of known concentration that only a fraction of the base reacts with water (Figure \(\PageIndex{2}\)). The remaining weak base is present as the unreacted form. The equilibrium constant for the ionization of a weak base, \(K_b\), is called the <strong>ionization constant</strong> of the weak base, and is equal to the reaction quotient when the reaction is at equilibrium. For trimethylamine, at equilibrium:</p>

    <p class="mt-align-center lt-chem-38278">\[K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}} \nonumber\]</p>

    <figure><img alt="alt" height="313" src="/@api/deki/files/65962/CNX_Chem_14_03_ammonia.jpg" width="547" />
    <figcaption>Figure \(\PageIndex{2}\): pH paper indicates that a 0.1-M solution of NH<sub>3</sub> (left) is weakly basic. The solution has a pOH of 3 ([OH<sup>−</sup>] = 0.001 M) because the weak base NH<sub>3</sub> only partially reacts with water. A 0.1-M solution of NaOH (right) has a pOH of 1 because NaOH is a strong base (credit: modification of work by Sahar Atwa).</figcaption>
    </figure>

    <p class="lt-chem-38278">The ionization constants of several weak bases are given in Table \(\PageIndex{2}\) and Table E2.</p>

    <table align="center" class="mt-responsive-table" summary="This table has two columns and six rows. The first row is a header row and it labels each column: “Ionization Reaction,” and “K subscript b at 25 degrees C.” Under the “Ionization Reaction” column are the following reactions: “( C H subscript 3 ) subscript 2 N H plus H subscript 2 O equilibrium arrow ( C H subscript 3 ) subscript 2 N H subscript 2 superscript plus sign plus O H superscript negative sign,” “C H subscript 3 N H subscript 2 plus H subscript 2 O equilibrium arrow C H subscript 3 N H subscript 3 superscript plus sign plus O H superscript negative sign,” “( C H subscript 3 ) subscript 3 N plus H subscript 2 O equilibrium arrow ( C H subscript 3 ) subscript 3 N H superscript plus sign plus O H superscript negative sign,” “N H subscript 3 plus H subscript 2 O equilibrium arrow N H subscript 4 superscript plus sign plus O H superscript negative sign,” and “C subscript 6 H subscript 5 N H subscript 2 plus H subscript 2 O equilibrium arrow C subscript 6 N subscript 5 N H subscript 3 superscript plus sign plus O H superscript negative sign.” Under the “K subscript b at 25 degrees C” column are the following values: “7.4 times ten to the negative 4,” “4.4 times ten to the negative 4,” “7.4 times ten to the negative 5,” “1.8 times ten to the negative 5,” and “4.6 times ten to the negative 10.”">
        <caption>Table \(\PageIndex{2}\): Ionization Constants of Some Weak Bases</caption>
        <thead>
            <tr valign="top">
                <th scope="col" style="text-align:center;"><strong>Ionization Reaction</strong></th>
                <th scope="col" style="text-align:center;"><strong>K<sub>b</sub> at 25 °C</strong></th>
            </tr>
        </thead>
        <tbody>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{(CH3)2NH + H2O ⇌ (CH3)2NH2+ + OH-}\)</td>
                <td class="lt-chem-38278" data-th="Kb at 25 °C" style="text-align:center;">5.9 × 10<sup>−4</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{CH3NH2 + H2O ⇌ CH3NH3+ + OH-}\)</td>
                <td class="lt-chem-38278" data-th="Kb at 25 °C" style="text-align:center;">4.4 × 10<sup>−4</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{(CH3)3N + H2O ⇌ (CH3)3NH+ + OH-}\)</td>
                <td class="lt-chem-38278" data-th="Kb at 25 °C" style="text-align:center;">6.3 × 10<sup>−5</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{NH3 + H2O ⇌ NH4+ + OH-}\)</td>
                <td class="lt-chem-38278" data-th="Kb at 25 °C" style="text-align:center;">1.8 × 10<sup>−5</sup></td>
            </tr>
            <tr valign="top">
                <td class="lt-chem-38278" data-th="Ionization Reaction" style="text-align:center;">\(\ce{C6H5NH2 + H2O ⇌ C6N5NH3+ + OH-}\)</td>
                <td class="lt-chem-38278" data-th="Kb at 25 °C" style="text-align:center;">4.3 × 10<sup>−10</sup></td>
            </tr>
        </tbody>
    </table>

    <div class="box-example">
    <p class="box-legend lt-chem-38278"><span>Example \(\PageIndex{3}\): <span>Determination of <em>K</em><sub>a</sub> from Equilibrium Concentrations</span></span></p>

    <p class="lt-chem-38278">Acetic acid is the principal ingredient in vinegar; that's why it tastes sour. At equilibrium, a solution contains [CH<sub>3</sub>CO<sub>2</sub>H] = 0.0787 <em>M</em> and \(\ce{[H3O+]}=\ce{[CH3CO2- ]}=0.00118\:M\). What is the value of \(K_a\) for acetic acid?</p>

    <figure><img alt="alt" height="266" src="/@api/deki/files/65687/CNX_Chem_14_03_Vinegar.jpg" width="280" />
    <figcaption>Vinegar is a solution of acetic acid, a weak acid. (credit: modification of work by “HomeSpot HQ”/Flickr)</figcaption>
    </figure>

    <p class="lt-chem-38278"><strong><span>Solution</span> </strong></p>

    <p class="lt-chem-38278">We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber \]</p>

    <p class="mt-align-center lt-chem-38278">\[\begin{align*} K_\ce{a} &amp;=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}} \\[4pt] &amp;=\dfrac{(0.00118)(0.00118)}{0.0787} \\[4pt] &amp;=1.77×10^{−5} \end{align*}\]</p>
    </div>

    <div class="box-exercise">
    <p class="box-legend lt-chem-38278"><span>Exercise \(\PageIndex{3}\)</span></p>

    <p class="lt-chem-38278">What is the equilibrium constant for the ionization of the \(\ce{HSO4-}\) ion, the weak acid used in some household cleansers:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \nonumber\]</p>

    <p class="lt-chem-38278">In one mixture of NaHSO<sub>4</sub> and Na<sub>2</sub>SO<sub>4</sub> at equilibrium, \(\ce{[H3O+]}\) = 0.027 <em>M</em>; \(\ce{[HSO4- ]}=0.29\:M\); and \(\ce{[SO4^2- ]}=0.13\:M\).</p>

    <dl>
        <dt><strong class="emphasis bold">Answer</strong></dt>
        <dd>
        <p class="lt-chem-38278">\(K_a\) for \(\ce{HSO_4^-}= 1.2 ×\times 10^{−2}\)</p>
        </dd>
    </dl>
    </div>

    <div class="box-example">
    <p class="box-legend lt-chem-38278"><span>Example \(\PageIndex{4}\): <span>Determination of <em>K</em><sub>b</sub> from Equilibrium Concentrations</span></span></p>

    <p class="lt-chem-38278">Caffeine, C<sub>8</sub>H<sub>10</sub>N<sub>4</sub>O<sub>2</sub> is a weak base. What is the value of <em>K</em><sub>b</sub> for caffeine if a solution at equilibrium has [C<sub>8</sub>H<sub>10</sub>N<sub>4</sub>O<sub>2</sub>] = 0.050 <em>M</em>, \(\ce{[C8H10N4O2H+]}\) = 5.0 × 10<sup>−3</sup> <em>M</em>, and [OH<sup>−</sup>] = 2.5 × 10<sup>−3</sup> <em>M</em>?</p>

    <p class="lt-chem-38278"><strong><span>Solution</span> </strong></p>

    <p class="lt-chem-38278">At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \nonumber \]</p>

    <p class="lt-chem-38278">so</p>

    <p class="mt-align-center lt-chem-38278">\[K_\ce{b}=\ce{\dfrac{[C8H10N4O2H+][OH- ]}{[C8H10N4O2]}}=\dfrac{(5.0×10^{−3})(2.5×10^{−3})}{0.050}=2.5×10^{−4} \nonumber\]</p>
    </div>

    <div class="box-exercise">
    <p class="box-legend lt-chem-38278"><span>Exercise \(\PageIndex{4}\)</span></p>

    <p class="lt-chem-38278">What is the equilibrium constant for the ionization of the \(\ce{HPO4^2-}\) ion, a weak base:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H2PO4-}(aq)+\ce{OH-}(aq) \nonumber \]</p>

    <p class="lt-chem-38278">In a solution containing a mixture of \(\ce{NaH2PO4}\) and&nbsp;\(\ce{Na2HPO4}\) at equilibrium with:</p>

    <ul>
        <li>\([\ce{OH^{−}}] = 1.3 × 10^{−6} M\)</li>
        <li>\(\ce{[H2PO4^{-}]=0.042\:M}\)&nbsp;and</li>
        <li>\(\ce{[HPO4^{2-}]=0.341\:M}\).</li>
    </ul>

    <dl>
        <dt><strong class="emphasis bold">Answer</strong></dt>
        <dd>
        <p class="lt-chem-38278"><em>K</em><sub>b</sub> for \(\ce{HPO4^2-}=1.6×10^{−7} \)</p>
        </dd>
    </dl>
    </div>

    <div class="box-example">
    <p class="box-legend lt-chem-38278"><span>Example \(\PageIndex{5}\): <span>Determination of <em>K</em><sub>a</sub> or <em>K</em><sub>b</sub> from pH</span></span></p>

    <p class="lt-chem-38278">The pH of a 0.0516-<em>M</em> solution of nitrous acid, \(\ce{HNO2}\), is 2.34. What is its \(K_a\)?</p>

    <p class="lt-chem-38278">\[\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) \nonumber\]</p>

    <p class="lt-chem-38278"><strong>Solution </strong></p>

    <p class="lt-chem-38278">We determine an equilibrium constant starting with the initial concentrations of HNO<sub>2</sub>, \(\ce{H3O+}\), and \(\ce{NO2-}\) as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)</p>

    <p class="lt-chem-38278">We can solve this problem with the following steps in which <em>x</em> is a change in concentration of a species in the reaction:</p>

    <p class="mt-align-center lt-chem-38278"><img alt="Example 14.3.5.png" height="386px" src="/@api/deki/files/123057/Example_14.3.5.png" style="width: 766px; height: 386px;" width="766px" /></p>

    <p class="lt-chem-38278">We can summarize the various concentrations and changes as shown here (the concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its concentration):</p>

    <p class="mt-align-center lt-chem-38278"><img alt="Example 14.3.5B.png" height="235px" src="/@api/deki/files/123058/Example_14.3.5B.png" style="width: 735px; height: 235px;" width="735px" /></p>

    <p class="lt-chem-38278">To get the various values in the ICE (Initial, Change, Equilibrium) table, we first calculate \(\ce{[H3O+]}\), the equilibrium concentration of \(\ce{H3O+}\), from the pH:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{[H3O+]}=10^{−2.34}=0.0046\:M \nonumber \]</p>

    <p class="lt-chem-38278">The change in concentration of \(\ce{H3O+}\), \(x_{\ce{[H3O+]}}\), is the difference between the equilibrium concentration of H<sub>3</sub>O<sup>+</sup>, which we determined from the pH, and the initial concentration, \(\mathrm{[H_3O^+]_i}\). The initial concentration of \(\ce{H3O+}\) is its concentration in pure water, which is so much less than the final concentration that we approximate it as zero (~0).</p>

    <p class="lt-chem-38278">The change in concentration of \(\ce{NO2-}\) is equal to the change in concentration of \(\ce{[H3O+]}\). For each 1 mol of \(\ce{H3O+}\) that forms, 1 mol of \(\ce{NO2-}\) forms. The equilibrium concentration of HNO<sub>2</sub> is equal to its initial concentration plus the change in its concentration.</p>

    <p class="lt-chem-38278">Now we can fill in the ICE table with the concentrations at equilibrium, as shown here:</p>

    <p class="mt-align-center lt-chem-38278"><img alt="150749131940304.png" height="198px" src="/@api/deki/files/123059/150749131940304.png" style="width: 755px; height: 198px;" width="755px" /></p>

    <p class="lt-chem-38278">Finally, we calculate the value of the equilibrium constant using the data in the table:</p>

    <p class="lt-chem-38278">\[K_\ce{a}=\ce{\dfrac{[H3O+][NO2- ]}{[HNO2]}}=\dfrac{(0.0046)(0.0046)}{(0.0470)}=4.5×10^{−4} \nonumber\]</p>
    </div>

    <div class="box-exercise">
    <p class="box-legend lt-chem-38278"><span>Exercise \(\PageIndex{5}\)</span></p>

    <p class="lt-chem-38278">The pH of a solution of household ammonia, a 0.950-<em>M</em> solution of NH<sub>3,</sub> is 11.612. What is <em>K</em><sub>b</sub> for NH<sub>3</sub>.</p>

    <dl>
        <dt><strong class="emphasis bold">Answer</strong></dt>
        <dd>
        <p class="lt-chem-38278">\(K_b = 1.8 × 10^{−5}\)</p>
        </dd>
    </dl>
    </div>

    <div class="box-example">
    <p class="box-legend lt-chem-38278"><span>Example \(\PageIndex{6}\): <span>Equilibrium Concentrations in a Solution of a Weak Acid</span></span></p>

    <p class="lt-chem-38278">Formic acid, HCO<sub>2</sub>H, is the irritant that causes the body’s reaction to ant stings.</p>

    <figure><img alt="A photograph is shown of a large black ant on the end of a human finger." height="205" src="/@api/deki/files/65966/CNX_Chem_14_03_AntSting.jpg" width="364" />
    <figcaption>The pain of an ant’s sting is caused by formic acid. (credit: John Tann)</figcaption>
    </figure>

    <p class="lt-chem-38278">What is the concentration of hydronium ion and the pH in a 0.534-<em>M</em> solution of formic acid?</p>

    <p>\[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−4} \nonumber \]</p>

    <div>
    <p class="lt-chem-38278"><strong>Soluti<span>on</span></strong></p>

    <p class="lt-chem-38278"><em>1. Determine x and equilibrium concentrations</em>. The equilibrium expression is:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \nonumber \]</p>

    <p class="lt-chem-38278">The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table.</p>

    <p class="lt-chem-38278">The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):</p>

    <p class="mt-align-center lt-chem-38278"><img alt="Example 14.3.6.png" height="183px" src="/@api/deki/files/123064/Example_14.3.6.png" style="width: 677px; height: 183px;" width="677px" /></p>

    <p class="lt-chem-38278"><em>2. Solve for \(x\) and the equilibrium concentrations.</em> At equilibrium:</p>

    <p class="lt-chem-38278">\[\begin{align*} K_\ce{a} &amp;=1.8×10^{−4}=\ce{\dfrac{[H3O+][HCO2- ]}{[HCO2H]}} \\[4pt] &amp;=\dfrac{(x)(x)}{0.534−x}=1.8×10^{−4} \end{align*}\]</p>

    <p class="lt-chem-38278">Now solve for \(x\). Because the initial concentration of acid is reasonably large and \(K_a\) is very small, we assume that \(x &lt;&lt; 0.534\), which <em>permits </em>us to simplify the denominator term as \((0.534 − x) = 0.534\). This gives:</p>

    <p class="lt-chem-38278">\[K_\ce{a}=1.8×10^{−4}=\dfrac{x^{2}}{0.534} \nonumber\]</p>

    <p class="lt-chem-38278">Solve for \(x\) as follows:</p>

    <p class="lt-chem-38278">\[\begin{align*} x^2 &amp;=0.534×(1.8×10^{−4}) \\[4pt] &amp;=9.6×10^{−5} \\[4pt] x &amp;=\sqrt{9.6×10^{−5}} \\[4pt]&nbsp;&amp;=9.8×10^{−3} \end{align*}\]</p>
    </div>

    <p class="lt-chem-38278">To check the assumption that \(x\) is small compared to 0.534, we calculate:</p>

    <p class="mt-align-center lt-chem-38278">\[\begin{align*} \dfrac{x}{0.534} &amp;=\dfrac{9.8×10^{−3}}{0.534} \\[4pt] &amp;=1.8×10^{−2} \, \textrm{(1.8% of 0.534)} \end{align*}\]</p>

    <p class="lt-chem-38278">\(x\) is less than 5% of the initial concentration; the assumption is valid.</p>

    <p class="lt-chem-38278">We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:</p>

    <p class="lt-chem-38278">\[\begin{align*} \ce{[H3O+]} &amp;=~0+x=0+9.8×10^{−3}\:M. \\[4pt] &amp;=9.8×10^{−3}\:M \end{align*}\]</p>

    <p class="lt-chem-38278">The pH of the solution can be found by taking the negative log of the \(\ce{[H3O+]}\), so:</p>

    <p class="mt-align-center lt-chem-38278">\[pH = −\log(9.8×10^{−3})=2.01 \nonumber\]</p>
    </div>

    <div class="box-exercise">
    <p class="box-legend lt-chem-38278"><span>Exercise \(\PageIndex{6}\): acetic acid</span></p>

    <p class="lt-chem-38278">Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-<em>M</em> solution of acetic acid, CH<sub>3</sub>CO<sub>2</sub>H?</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5} \nonumber\]</p>

    <dl>
        <dt><strong class="emphasis bold">Hint</strong></dt>
        <dd>
        <p class="lt-chem-38278">Determine \(\ce{[CH3CO2- ]}\) at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or \(\ce{\dfrac{[CH3CO2- ]}{[CH3CO2H]_{initial}}}×100\).</p>
        </dd>
    </dl>

    <dl>
        <dt><strong class="emphasis bold">Answer</strong></dt>
        <dd>
        <p class="lt-chem-38278">percent ionization = 1.3%</p>
        </dd>
    </dl>
    </div>

    <p class="lt-chem-38278">The following example shows that the concentration of products produced by the ionization of a weak base can be determined by the same series of steps used with a weak acid.</p>

    <div class="box-example">
    <p class="box-legend lt-chem-38278"><span>Example \(\PageIndex{7}\): <span>Equilibrium Concentrations in a Solution of a Weak Base</span></span></p>

    <p class="lt-chem-38278">Find the concentration of hydroxide ion in a 0.25-<em>M</em> solution of trimethylamine, a weak base:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{(CH3)3N}(aq)+\ce{H2O}(l)⇌\ce{(CH3)3NH+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=6.3×10^{−5}&nbsp;\nonumber\]</p>

    <p class="lt-chem-38278">Solution This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example \(\PageIndex{6}\). The reactants and products will be different and the numbers will be different, but the logic will be the same:</p>

    <p class="mt-align-center lt-chem-38278"><img alt="Example 14.3.7A.png" height="129px" src="/@api/deki/files/123063/Example_14.3.7A.png" style="width: 727px; height: 129px;" width="727px" /></p>

    <p class="lt-chem-38278"><em>1. Determine x and equilibrium concentrations</em>. The table shows the changes and concentrations:</p>

    <p class="mt-align-center lt-chem-38278"><img alt="Example 14.3.7B.png" height="190px" src="/@api/deki/files/123062/Example_14.3.7B.png" style="width: 726px; height: 190px;" width="726px" /></p>

    <p class="lt-chem-38278"><em>2. Solve for \(x\) and the equilibrium concentrations</em>. At equilibrium:</p>

    <p class="mt-align-center lt-chem-38278">\[K_\ce{b}=\ce{\dfrac{[(CH3)3NH+][OH- ]}{[(CH3)3N]}}=\dfrac{(x)(x)}{0.25−x=}6.3×10^{−5} \nonumber\]</p>

    <p class="lt-chem-38278">If we assume that <em>x</em> is small relative to 0.25, then we can replace (0.25 − <em>x</em>) in the preceding equation with 0.25. Solving the simplified equation gives:</p>

    <p class="lt-chem-38278">\[x=4.0×10^{−3} \nonumber \]</p>

    <p class="lt-chem-38278">This change is less than 5% of the initial concentration (0.25), so the assumption is justified.</p>

    <p class="lt-chem-38278">Recall that, for this computation, \(x\) is equal to the equilibrium concentration of <em>hydroxide ion</em> in the solution (see earlier tabulation):</p>

    <p class="lt-chem-38278">\[\begin{align*} (\ce{[OH- ]}=~0+x=x=4.0×10^{−3}\:M \\[4pt] &amp;=4.0×10^{−3}\:M \end{align*}\]</p>

    <div>
    <p>Then calculate pOH as follows:</p>

    <p>\[\ce{pOH}=−\log(4.3×10^{−3})=2.40 \nonumber\]</p>

    <p>Using the relation introduced in the previous section of this chapter:</p>

    <p>\[\mathrm{pH + pOH=p\mathit{K}_w=14.00}\nonumber \]</p>

    <p>permits the computation of pH:</p>

    <p>\[\mathrm{pH=14.00−pOH=14.00−2.37=11.60} \nonumber\]</p>

    <p>Check the work. A check of our arithmetic shows that \(K_b = 6.3 \times&nbsp;10^{−5}\).</p>
    </div>
    </div>

    <ol class="stepwise">
    </ol>

    <div class="box-exercise">
    <p class="box-legend lt-chem-38278"><span>Exercise \(\PageIndex{7}\)</span></p>

    <ol start="1" style="list-style-type: lower-alpha;">
        <li class="lt-chem-38278">Show that the calculation in Step 2 of this example gives an <em>x</em> of 4.3 × 10<sup>−3</sup> and the calculation in Step 3 shows <em>K</em><sub>b</sub> = 6.3 × 10<sup>−5</sup>.</li>
        <li class="lt-chem-38278">Find the concentration of hydroxide ion in a 0.0325-<em>M</em> solution of ammonia, a weak base with a <em>K</em><sub>b</sub> of 1.76 × 10<sup>−5</sup>. Calculate the percent ionization of ammonia, the fraction ionized × 100, or \(\ce{\dfrac{[NH4+]}{[NH3]}}×100 \%\)</li>
    </ol>

    <dl>
        <dt><strong class="emphasis bold">Answer a</strong></dt>
        <dd>
        <p class="lt-chem-38278">\(7.56 × 10^{−4}\, M\), 2.33%</p>
        </dd>
        <dt><strong class="emphasis bold">Answer b</strong></dt>
        <dd>
        <p class="lt-chem-38278">2.33%</p>
        </dd>
    </dl>
    </div>

    <p class="lt-chem-38278">Some weak acids and weak bases ionize to such an extent that the simplifying assumption that <em>x</em> is small relative to the initial concentration of the acid or base is inappropriate. As we solve for the equilibrium concentrations in such cases, we will see that we cannot neglect the change in the initial concentration of the acid or base, and we must solve the equilibrium equations by using the quadratic equation.</p>

    <div class="box-example">
    <p class="box-legend lt-chem-38278"><span>Example \(\PageIndex{8}\): <span>Equilibrium Concentrations in a Solution of a Weak Acid</span></span></p>

    <p class="lt-chem-38278">Sodium bisulfate, NaHSO<sub>4</sub>, is used in some household cleansers because it contains the \(\ce{HSO4-}\) ion, a weak acid. What is the pH of a 0.50-<em>M</em> solution of \(\ce{HSO4-}\)?</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \nonumber \]</p>

    <p class="lt-chem-38278"><strong>Soluti<span>on</span> </strong></p>

    <p class="lt-chem-38278">We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of \(\ce{HSO4-}\) so that we can use \(\ce{[H3O+]}\) to determine the pH. As in the previous examples, we can approach the solution by the following steps:</p>

    <p class="mt-align-center lt-chem-38278"><img alt="Example 14.3.8A.png" height="122px" src="/@api/deki/files/123061/Example_14.3.8A.png" style="width: 692px; height: 122px;" width="692px" /></p>

    <p class="lt-chem-38278"><em>1. Determine \(x\) and equilibrium concentrations</em>. This table shows the changes and concentrations:</p>

    <p class="mt-align-center lt-chem-38278"><img alt="Example 14.3.8B.png" height="241px" src="/@api/deki/files/123060/Example_14.3.8B.png" style="width: 752px; height: 241px;" width="752px" /></p>

    <p class="lt-chem-38278"><em>2. Solve for \(x\) and the concentrations</em>.</p>

    <p class="lt-chem-38278">As we begin solving for \(x\), we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of \(x\).</p>

    <p class="lt-chem-38278">At equilibrium:</p>

    <p class="mt-align-center lt-chem-38278">\[K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+][SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x}&nbsp;\nonumber\]</p>

    <p class="lt-chem-38278">If we assume that <em>x</em> is small and approximate (0.50 − <em>x</em>) as 0.50, we find:</p>

    <p class="lt-chem-38278">\[x=7.7×10^{−2} \nonumber \]</p>

    <p class="lt-chem-38278">When we check the assumption, we confirm:</p>

    <p class="mt-align-center lt-chem-38278">\[\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \nonumber\]</p>

    <p class="lt-chem-38278">which for this system is</p>

    <p class="mt-align-center lt-chem-38278">\[\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \nonumber\]</p>

    <p class="lt-chem-38278">The value of \(x\) is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find \(x\).</p>

    <p class="lt-chem-38278">The equation:</p>

    <p class="mt-align-center lt-chem-38278">\[K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x}\nonumber \]</p>

    <p class="lt-chem-38278">gives</p>

    <p class="mt-align-center lt-chem-38278">\[6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \nonumber\]</p>

    <p class="lt-chem-38278">or</p>

    <p class="mt-align-center lt-chem-38278">\[x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0 \nonumber \]</p>

    <p class="lt-chem-38278">This equation can be solved using the quadratic formula. For an equation of the form</p>

    <p>\[ax^{2+} + bx + c=0, \nonumber\]</p>

    <div>
    <p>\(x\) is given by the quadratic equation:</p>

    <p>\[x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \nonumber\]</p>

    <p>In this problem, \(a = 1\), \(b = 1.2 × 10^{−3}\), and \(c = −6.0 × 10^{−3}\).</p>

    <p>Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:</p>

    <p>\[x=7.2×10^{−2} \nonumber\]</p>

    <p>Now determine the hydronium ion concentration and the pH:</p>

    <p>\[\begin{align*} \ce{[H3O+]} &amp;=~0+x=0+7.2×10^{−2}\:M \\[4pt] &amp;=7.2×10^{−2}\:M \end{align*}\]</p>

    <p>The pH of this solution is:</p>

    <p>\[\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \nonumber\]</p>
    </div>
    </div>

    <div class="box-exercise">
    <p class="box-legend lt-chem-38278"><span>Exercise \(\PageIndex{8}\)</span></p>

    <ol start="1" style="list-style-type: lower-alpha;">
        <li class="lt-chem-38278">Show that the quadratic formula gives \(x = 7.2 × 10^{−2}\).</li>
        <li class="lt-chem-38278">Calculate the pH in a 0.010-<em>M</em> solution of caffeine, a weak base:</li>
    </ol>

    <p class="mt-align-center lt-chem-38278">\[\ce{C8H10N4O2}(aq)+\ce{H2O}(l)⇌\ce{C8H10N4O2H+}(aq)+\ce{OH-}(aq) \hspace{20px} K_\ce{b}=2.5×10^{−4} \nonumber\]</p>

    <dl>
        <dt>Hint</dt>
        <dd>
        <p class="lt-chem-38278">It will be necessary to convert [OH<sup>−</sup>] to \(\ce{[H3O+]}\) or pOH to pH toward the end of the calculation.</p>
        </dd>
    </dl>

    <dl>
        <dt><strong class="emphasis bold">Answer</strong></dt>
        <dd>
        <p class="lt-chem-38278">pH 11.16</p>
        </dd>
    </dl>
    </div>

    <h2 class="lt-chem-38278">The Relative Strengths of Strong Acids and Bases</h2>

    <p class="lt-chem-38278">Strong acids, such as \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\), all exhibit the same strength in water. The water molecule is such a strong base compared to the conjugate bases Cl<sup>−</sup>, Br<sup>−</sup>, and I<sup>−</sup> that ionization of these strong acids is essentially complete in aqueous solutions. In solvents less basic than water, we find \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\) differ markedly in their tendency to give up a proton to the solvent. For example, when dissolved in ethanol (a weaker base than water), the extent of ionization increases in the order \(\ce{HCl &lt; HBr &lt; HI}\), and so \(\ce{HI}\) is demonstrated to be the strongest of these acids. The inability to discern differences in strength among strong acids dissolved in water is known as the <span>leveling effect of water</span>.</p>

    <p class="lt-chem-38278">Water also exerts a leveling effect on the strengths of strong bases. For example, the oxide ion, O<sup>2−</sup>, and the amide ion, \(\ce{NH2-}\), are such strong bases that they react completely with water:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{O^2-}(aq)+\ce{H2O}(l)⟶\ce{OH-}(aq)+\ce{OH-}(aq)\]</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{NH2-}(aq)+\ce{H2O}(l)⟶\ce{NH3}(aq)+\ce{OH-}(aq)\]</p>

    <p class="lt-chem-38278">Thus, O<sup>2−</sup> and \(\ce{NH2-}\) appear to have the same base strength in water; they both give a 100% yield of hydroxide ion.</p>

    <p class="lt-chem-38278">In the absence of any leveling effect, the acid strength of binary compounds of hydrogen with nonmetals (A) increases as the H-A bond strength decreases down a group in the periodic table. For group 17, the order of increasing acidity is \(\ce{HF &lt; HCl &lt; HBr &lt; HI}\). Likewise, for group 16, the order of increasing acid strength is H<sub>2</sub>O &lt; H<sub>2</sub>S &lt; H<sub>2</sub>Se &lt; H<sub>2</sub>Te. Across a row in the periodic table, the acid strength of binary hydrogen compounds increases with increasing electronegativity of the nonmetal atom because the polarity of the H-A bond increases. Thus, the order of increasing acidity (for removal of one proton) across the second row is \(\ce{CH4 &lt; NH3 &lt; H2O &lt; HF}\); across the third row, it is \(\ce{SiH4 &lt; PH3 &lt; H2S &lt; HCl}\) (see Figure \(\PageIndex{3}\)).</p>

    <figure><img alt="This diagram has two rows and four columns. Red arrows point left across the bottom of the figure and down at the right side and are labeled “Increasing acid strength.” Blue arrows point left across the bottom and up at the right side of the figure and are labeled “Increasing base strength.” The first column is labeled 14 at the top and two white squares are beneath it. The first has the number 6 in the upper left corner and the formula C H subscript 4 in the center along with designation Neither acid nor base. The second square contains the number 14 in the upper left corner, the formula C H subscript 4 at the center and the designation Neither acid nor base. The second column is labeled 15 at the top and two blue squares are beneath it. The first has the number 7 in the upper left corner and the formula N H subscript 3 in the center along with the designation Weak base and K subscript b equals 1.8 times 10 superscript negative 5. The second square contains the number 15 in the upper left corner, the formula P H subscript 3 at the center and the designation Very weak base and K subscript b equals 4 times 10 superscript negative 28. The third column is labeled 16 at the top and two squares are beneath it. The first is shaded tan and has the number 8 in the upper left corner and the formula H subscript 2 O in the center along with the designation neutral. The second square is shaded pink, contains the number 16 in the upper left corner, the formula H subscript 2 S at the center and the designation Weak acid and K subscript a equals 9.5 times 10 superscript negative 8. The fourth column is labeled 17 at the top and two squares are beneath it. The first is shaded pink, has the number 9 in the upper left corner and the formula H F in the center along with the designation Weak acid and K subscript a equals 6.8 times 10 superscript negative 4. The second square is shaded a deeper pink, contains the number 17 in the upper left corner, the formula H C l at the center, and the designation Strong acid." height="386px" src="/@api/deki/files/65690/CNX_Chem_14_03_AcidpH.jpg" style="width: 608px; height: 386px;" width="608px" />
    <figcaption>Figure \(\PageIndex{3}\): As you move from left to right and down the periodic table, the acid strength increases. As you move from right to left and up, the base strength increases.</figcaption>
    </figure>

    <p class="lt-chem-38278">Compounds containing oxygen and one or more hydroxyl (OH) groups can be acidic, basic, or amphoteric, depending on the position in the periodic table of the central atom E, the atom bonded to the hydroxyl group. Such compounds have the general formula O<sub>n</sub>E(OH)<sub>m</sub>, and include sulfuric acid, \(\ce{O2S(OH)2}\), sulfurous acid,&nbsp;\(\ce{OS(OH)2}\), nitric acid, \(\ce{O2NOH}\), perchloric acid, \(\ce{O3ClOH}\), aluminum hydroxide, \(\ce{Al(OH)3}\), calcium hydroxide, \(\ce{Ca(OH)2}\), and potassium hydroxide, \(\ce{KOH}\):</p>

    <p class="mt-align-center lt-chem-38278"><img alt="A diagram is shown that includes a central atom designated with the letter E. Single bonds extend above, below, left, and right of the E. An O atom is bonded to the right of the E, and an arrow points to the bond labeling it, “Bond a.” An H atom is single bonded to the right of the O atom. An arrow pointing to this bond connects it to the label, “Bond b.”" height="101" src="/@api/deki/files/65689/CNX_Chem_14_03_OHbonds_img.jpg" width="242" /></p>

    <p class="lt-chem-38278">If the central atom, E, has a low electronegativity, its attraction for electrons is low. Little tendency exists for the central atom to form a strong covalent bond with the oxygen atom, and bond <em>a</em> between the element and oxygen is more readily broken than bond <em>b</em> between oxygen and hydrogen. Hence bond <em>a</em> is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case with Ca(OH)<sub>2</sub> and KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic elements form ionic hydroxides that are by definition basic compounds.</p>

    <p class="lt-chem-38278">If, on the other hand, the atom E has a relatively high electronegativity, it strongly attracts the electrons it shares with the oxygen atom, making bond <em>a</em> relatively strongly covalent. The oxygen-hydrogen bond, bond <em>b</em>, is thereby weakened because electrons are displaced toward E. Bond <em>b</em> is polar and readily releases hydrogen ions to the solution, so the material behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Thus, nonmetallic elements form covalent compounds containing acidic −OH groups that are called <span>oxyacids</span>.</p>

    <p class="lt-chem-38278">Increasing the oxidation number of the central atom E also increases the acidity of an oxyacid because this increases the attraction of E for the electrons it shares with oxygen and thereby weakens the O-H bond. Sulfuric acid, H<sub>2</sub>SO<sub>4</sub>, or O<sub>2</sub>S(OH)<sub>2</sub> (with a sulfur oxidation number of +6), is more acidic than sulfurous acid, H<sub>2</sub>SO<sub>3</sub>, or OS(OH)<sub>2</sub> (with a sulfur oxidation number of +4). Likewise nitric acid, HNO<sub>3</sub>, or O<sub>2</sub>NOH (N oxidation number = +5), is more acidic than nitrous acid, HNO<sub>2</sub>, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure \(\PageIndex{4}\)).</p>

    <figure><img alt="A diagram is shown that includes four structural formulas for acids. A red, right pointing arrow is placed beneath the structures which is labeled “Increasing acid strength.” At the top left, the structure of Nitrous acid is provided. It includes an H atom to which an O atom with two unshared electron pairs is connected with a single bond to the right. A single bond extends to the right and slightly below to a N atom with one unshared electron pair. A double bond extends up and to the right from this N atom to an O atom which has two unshared electron pairs. To the upper right is a structure for Nitric acid. This structure differs from the previous structure in that the N atom is directly to the right of the first O atom and a second O atom with three unshared electron pairs is connected with a single bond below and to the right of the N atom which has no unshared electron pairs. At the lower left, an O atom with two unshared electron pairs is double bonded to its right to an S atom with a single unshared electron pair. An O atom with two unshared electron pairs is bonded above and an H atom is single bonded to this O atom. To the right of the S atom is a single bond to another O atom with two unshared electron pairs to which an H atom is single bonded. This structure is labeled “Sulfurous acid.” A similar structure which is labeled “Sulfuric acid” is placed in the lower right region of the figure. This structure differs in that an H atom is single bonded to the left of the first O atom, leaving it with two unshared electron pairs and a fourth O atom with two unshared electron pairs is double bonded beneath the S atom, leaving it with no unshared electron pairs." height="350px" src="/@api/deki/files/65688/CNX_Chem_14_03_Oxyacid.jpg" style="width: 617px; height: 350px;" width="617px" />
    <figcaption>Figure \(\PageIndex{4}\): As the oxidation number of the central atom E increases, the acidity also increases.</figcaption>
    </figure>

    <p class="lt-chem-38278">Hydroxy compounds of elements with intermediate electronegativities and relatively high oxidation numbers (for example, elements near the diagonal line separating the metals from the nonmetals in the periodic table) are usually amphoteric. This means that the hydroxy compounds act as acids when they react with strong bases and as bases when they react with strong acids. The amphoterism of aluminum hydroxide, which commonly exists as the hydrate&nbsp;\(\ce{Al(H2O)3(OH)3}\), is reflected in its solubility in both strong acids and strong bases. In strong bases, the relatively insoluble hydrated aluminum hydroxide, \(\ce{Al(H2O)3(OH)3}\), is converted into the soluble ion, \(\ce{[Al(H2O)2(OH)4]-}\), by reaction with hydroxide ion:</p>

    <p class="lt-chem-38278">\[[\ce{Al(H2O)3(OH)3}](aq)+\ce{OH-}(aq)⇌\ce{H2O}(l)+\ce{[Al(H2O)2(OH)4]-}(aq) \nonumber\]</p>

    <p class="lt-chem-38278">In this reaction, a proton is transferred from one of the aluminum-bound H<sub>2</sub>O molecules to a hydroxide ion in solution. The \(\ce{Al(H2O)3(OH)3}\) compound thus acts as an acid under these conditions. On the other hand, when dissolved in strong acids, it is converted to the soluble ion \(\ce{[Al(H2O)6]^3+}\) by reaction with hydronium ion:</p>

    <p class="mt-align-center lt-chem-38278">\[\ce{3H3O+}(aq)+\ce{Al(H2O)3(OH)3}(aq)⇌\ce{Al(H2O)6^3+}(aq)+\ce{3H2O}(l) \nonumber\]</p>

    <p class="lt-chem-38278">In this case, protons are transferred from hydronium ions in solution to \(\ce{Al(H2O)3(OH)3}\), and the compound functions as a base.</p>

    <h2 class="lt-chem-38278">Summary</h2>

    <p class="lt-chem-38278">The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. The strengths of the binary acids increase from left to right across a period of the periodic table (CH<sub>4</sub> &lt; NH<sub>3</sub> &lt; H<sub>2</sub>O &lt; HF), and they increase down a group (HF &lt; HCl &lt; HBr &lt; HI). The strengths of oxyacids that contain the same central element increase as the oxidation number of the element increases (H<sub>2</sub>SO<sub>3</sub> &lt; H<sub>2</sub>SO<sub>4</sub>). The strengths of oxyacids also increase as the electronegativity of the central element increases [H<sub>2</sub>SeO<sub>4</sub> &lt; H<sub>2</sub>SO<sub>4</sub>].</p>

    <section class="key-equations">
    <h2 class="lt-chem-38278">Key Equations</h2>

    <ul>
        <li class="lt-chem-38278">\(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\)</li>
        <li class="lt-chem-38278">\(K_\ce{b}=\ce{\dfrac{[HB+][OH- ]}{[B]}}\)</li>
        <li class="lt-chem-38278">\(K_a \times K_b = 1.0 \times 10^{−14} = K_w \,(\text{at room temperature})\)</li>
        <li class="lt-chem-38278">\(\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\)</li>
    </ul>
    </section>

    <h2 class="lt-chem-38278">Glossary</h2>

    <dl>
        <dt>acid ionization constant (<em>K</em><sub>a</sub>)</dt>
        <dd>equilibrium constant for the ionization of a weak acid</dd>
    </dl>

    <dl>
        <dt>base ionization constant (<em>K</em><sub>b</sub>)</dt>
        <dd>equilibrium constant for the ionization of a weak base</dd>
    </dl>

    <dl>
        <dt>leveling effect of water</dt>
        <dd>any acid stronger than \(\ce{H3O+}\), or any base stronger than OH<sup>−</sup> will react with water to form \(\ce{H3O+}\), or OH<sup>−</sup>, respectively; water acts as a base to make all strong acids appear equally strong, and it acts as an acid to make all strong bases appear equally strong</dd>
    </dl>

    <dl>
        <dt>oxyacid</dt>
        <dd>compound containing a nonmetal and one or more hydroxyl groups</dd>
    </dl>

    <dl>
        <dt>percent ionization</dt>
        <dd>ratio of the concentration of the ionized acid to the initial acid concentration, times 100</dd>
    </dl>

    <h2 class="lt-chem-38278">Contributors and Attributions</h2>

    <ul>
        <li class="lt-chem-38278">

    Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).

    </li>
    </ul>

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    <p>Forker source[1] end-chem-38278</p>
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