12.4: Acid Strength
- Page ID
- 195173
Learning Objectives
- Assess the relative strengths of acids according to their ionization constants
- Carry out equilibrium calculations for weak acid systems
We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The ionization of an acid in water is given by the general expression:
\[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq)\]
In the ionization process, water acts as the base accepting a proton from the acid \(\ce{HA}\), \(\ce{A^{−}}\) is the conjugate base of the acid \(\ce{HA}\), and the hydronium ion (\(\ce{H3O+}\)) is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of \(\ce{H3O+}\) and \(\ce{A^{−}}\) when the acid ionizes in water. Table \(\PageIndex{1}\) lists several strong acids. A weak acid does not completely ionize in water, yielding only small amounts of \(\ce{H3O+}\) and \(\ce{A^{−}}\).
| Common Strong Acids | Common Strong Bases | ||
|---|---|---|---|
| \(\ce{HClO4}\) | perchloric acid | ||
| \(\ce{HCl}\) | hydrochloric acid | \(\ce{NaOH}\) | sodium hydroxide |
| \(\ce{HBr}\) | hydrobromic acid | \(\ce{KOH}\) | potassium hydroxide |
| \(\ce{HI}\) | hydroiodic acid | \(\ce{Ba(OH)2}\) | barium hydroxide |
| \(\ce{HNO3}\) | nitric acid | ||
| \(\ce{H2SO4}\) | sulfuric acid | ||
The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the acid-ionization constant, Ka. For the reaction of an acid \(\ce{HA}\):
\[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq)\]
we write the equation for the ionization constant as:
\[K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\]
where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we do not include [H2O] in the equation. The larger the \(K_a\) of an acid, the larger the concentration of \(\ce{H3O+}\) and \(\ce{A^{−}}\) relative to the concentration of the nonionized acid, \(\ce{HA}\) at equilibrium. Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase.
The following data on acid-ionization constants indicate the order of acid strength: \(\ce{CH3CO2H} < \ce{HNO2} < \ce{HSO4-}\)
\[ \begin{aligned} \ce{CH3CO2H}(aq) + \ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \quad &K_\ce{a}=1.8×10^{−5} \\[4pt] \ce{HNO2}(aq)+\ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) &K_\ce{a}=4.6×10^{-4} \\[4pt] \ce{HSO4-}(aq)+\ce{H2O}(aq) &⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) & K_\ce{a}=1.2×10^{−2} \end{aligned}\]
Table \(\PageIndex{2}\) gives the ionization constants for several weak acids; additional ionization constants can be found in Reference Table E1.
| Acid | Ka at 25 °C |
|---|---|
| \(\ce{H2SO3}\) | 1.5 × 10−2 |
| \(\ce{HSO4-}\) | 1.2 × 10−2 |
| \(\ce{H3PO4}\) | 7.5 × 10−3 |
| \(\ce{HF}\) | 3.5 × 10−4 |
| \(\ce{HNO2}\) | 4.6 × 10−4 |
| \(\ce{HNCO}\) | 2 × 10−4 |
| \(\ce{HCO2H}\) | 1.8 × 10−4 |
| \(\ce{HC3H5O3}\) | 1.4 × 10−4 |
| \(\ce{HC2H3O2}\) | 1.8 × 10−5 |
| \(\ce{H2CO3}\) | 4.3 × 10−7 |
| \(\ce{HSO3-}\) | 1.0 × 10−7 |
| \(\ce{H2S}\) | 9.1 × 10−8 |
| \(\ce{HClO}\) | 3.5 × 10−8 |
| \(\ce{HBrO}\) | 2.0 × 10−9 |
| \(\ce{HCN}\) | 6.2 × 10−10 |
| \(\ce{NH4+}\) | 5.6 × 10−10 |
| \(\ce{HCO3-}\) | 4.8 × 10−11 |
| \(\ce{H2O2}\) | 2.4 × 10−12 |
| \(\ce{H2O}\) | 1.0 × 10−14 |
Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100:
\[\% \:\ce{ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\% \label{PercentIon} \]
Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration.
Example \(\PageIndex{1}\): Calculation of Percent Ionization from pH
Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.
Solution
The percent ionization for an acid is:
\[\ce{\dfrac{[H3O+]_{eq}}{[HNO2]_0}}×100 \]
The chemical equation for the dissociation of the nitrous acid is:
\[\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{NO2-}(aq)+\ce{H3O+}(aq). \]
Since 10−pH = \(\ce{[H3O+]}\), we find that \(10^{−2.09} = 8.1 \times 10^{−3}\, M\), so that percent ionization (Equation \ref{PercentIon}) is:
Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures.
Exercise \(\PageIndex{1}\)
Calculate the percent ionization of a 0.10 M solution of acetic acid with a pH of 2.89.
- Answer
-
1.3% ionized
We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by:
\[\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq)\]
Water is the acid that reacts with the base, \(\ce{HB^{+}}\) is the conjugate acid of the base \(\ce{B}\), and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH− and HB+ when it reacts with water; Table \(\PageIndex{1}\) lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water.
The extent to which an acid, HA, donates protons to water molecules is relative to the strength of the conjugate base, A−, of the acid. If A− is a strong base, any protons that are donated to water molecules are recaptured by A−. Thus there is relatively little A− and \(\ce{H3O+}\) in solution, and the acid, HA, is weak. However, if A− is a weak base, water binds the protons more strongly, and the solution contains primarily A− and H3O+—the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases.
Figure \(\PageIndex{1}\) lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other.
The first six acids in Figure \(\PageIndex{1}\) are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base.
Those acids that lie between the hydronium ion and water in Figure \(\PageIndex{1}\) form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure \(\PageIndex{1}\) exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid.
The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure \(\PageIndex{1}\). A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution.
Exercise \(\PageIndex{2}\)
We can determine the relative acid strengths of \(\ce{NH4+}\) and HCN by comparing their ionization constants. The ionization constant of HCN is given in Table E1 as 4.9 × 10−10. The ionization constant of \(\ce{NH4+}\) is not listed, but the ionization constant of its conjugate base, NH3, is listed as 1.8 × 10−5. Determine the ionization constant of \(\ce{NH4+}\), and decide which is the stronger acid, HCN or \(\ce{NH4+}\).
- Answer
-
\(\ce{NH4+}\) is the slightly stronger acid (Ka for \(\ce{NH4+}\) = 5.6 × 10−10).
The Ionization of Weak Acids and Weak Bases
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:
\[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)\]
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 \(\PageIndex{2}\)). The remaining weak acid is present in the nonionized form.
For acetic acid, at equilibrium:
\[K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=1.8 \times 10^{−5}\]
Example \(\PageIndex{2}\): 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 Ka for acetic acid?
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:
\[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \]
\[K_\ce{a}=\ce{\dfrac{[H3O+][CH3CO2- ]}{[CH3CO2H]}}=\dfrac{(0.00118)(0.00118)}{0.0787}=1.77×10^{−5} \]
Exercise \(\PageIndex{3}\)
What is the equilibrium constant for the ionization of the \(\ce{HSO4-}\) ion, the weak acid used in some household cleansers:
\[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \]
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 \(\PageIndex{3}\): Determination of Ka from pH
The pH of a 0.0516-M solution of nitrous acid, \(\ce{HNO2}\), is 2.34. What is its Ka?
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:
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):
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:
\[\ce{[H3O+]}=10^{−2.34}=0.0046\:M \]
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:
Finally, we calculate the value of the equilibrium constant using the data in the table:
\[K_\ce{a}=\ce{\dfrac{[H3O+][NO2- ]}{[HNO2]}}=\dfrac{(0.0046)(0.0046)}{(0.0470)}=4.5×10^{−4} \]
Example \(\PageIndex{4}\): Equilibrium Concentrations in a Solution of a Weak Acid
Formic acid, HCO2H, is the irritant that causes the body’s reaction to ant stings.
What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid?
Solution
1. Determine x and equilibrium concentrations. The equilibrium expression is:
\[\ce{HCO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO2-}(aq) \]
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):
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 Ka 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:
\[K_\ce{a}=1.8×10^{−4}=\dfrac{x^{2}}{0.534} \]
Solve for x as follows:
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:
The pH of the solution can be found by taking the negative log of the \(\ce{[H3O+]}\), so:
\(pH = −\log(9.8×10^{−3})=2.01\)
Exercise \(\PageIndex{4}\): 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?
\[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \hspace{20px} K_\ce{a}=1.8×10^{−5} \]
- 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%
Some weak acids 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 \(\PageIndex{5}\): 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 \(\ce{HSO4-}\)?
\[\ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) \hspace{20px} K_\ce{a}=1.2×10^{−2} \]
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:
1. Determine x and equilibrium concentrations. This table shows the changes and concentrations:
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:
\[K_\ce{a}=1.2×10^{−2}=\ce{\dfrac{[H3O+][SO4^2- ]}{[HSO4- ]}}=\dfrac{(x)(x)}{0.50−x} \]
If we assume that x is small and approximate (0.50 − x) as 0.50, we find:
When we check the assumption, we confirm:
\[\dfrac{x}{\mathrm{[HSO_4^- ]_i}} \overset{?}{\le} 0.05 \]
which for this system is
\[\dfrac{x}{0.50}=\dfrac{7.7×10^{−2}}{0.50}=0.15(15\%) \]
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:
\[K_\ce{a}=1.2×10^{−2}=\dfrac{(x)(x)}{0.50−x} \]
gives
\[6.0×10^{−3}−1.2×10^{−2}x=x^{2+} \]
or
\[x^{2+}+1.2×10^{−2}x−6.0×10^{−3}=0\ \]
This equation can be solved using the quadratic formula. For an equation of the form
x is given by the equation:
\[x=\dfrac{−b±\sqrt{b^{2+}−4ac}}{2a} \]
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:
\[x=7.2×10^{−2} \]
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:
\[\mathrm{pH=−log[H_3O^+]=−log7.2×10^{−2}=1.14} \]
Summary
The strengths of Brønsted-Lowry acids in aqueous solutions can be determined by their acid 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.
Key Equations
- \(K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}}\)
- \(\textrm{Percent ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\)
Glossary
- acid ionization constant (Ka)
- equilibrium constant for the ionization of a weak acid
- percent ionization
- ratio of the concentration of the ionized acid to the initial acid concentration, times 100
Contributors and Attributions
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).