4.9.3: Relative Strengths of Acids and Bases
By the end of this section, you will be able to:
- Assess the relative strengths of acids and bases according to their ionization constants
- Rationalize trends in acid–base strength in relation to molecular structure
- Carry out equilibrium calculations for weak acid–base systems
Acid and Base Ionization Constants
The relative strength of an acid or base is the extent to which it ionizes when dissolved in water. If the ionization reaction is essentially complete, the acid or base is termed strong ; if relatively little ionization occurs, the acid or base is weak. As will be evident throughout the remainder of this chapter, there are many more weak acids and bases than strong ones. The most common strong acids and bases are listed in Figure \(\PageIndex{1}\).
The relative strengths of acids may be quantified 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 consta nt, \(K_a\). For the reaction of an acid \(\ce{HA}\):
\[\ce{HA(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + A^{-}(aq)} \nonumber \]
the acid ionization constant is written
\[K_{ a }=\frac{[\ce{H3O^{+}}] [\ce{A^{-}}]}{[\ce{HA}]} \nonumber \]
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 \([\ce{H2O}]\) in the equation.
Equilibrium constant expressions are actually ratios of activities , and the value of K is determined at the limit of infinite dilution of the solutes. In these very dilute solutions, the activity of the solvent has a value of unity (1) and the activity of each solute can be approximated by its molar concentration.
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, \(\ce{HA}\), in an equilibrium mixture, and the stronger the acid. An acid is classified as “strong” when it undergoes complete ionization, in which case the concentration of HA is zero and the acid ionization constant is immeasurably large (\(K_a ≈ ∞\)). Acids that are partially ionized are called “weak,” and their acid ionization constants may be experimentally measured. A table of ionization constants for weak acids is provided in Appendix H.
To illustrate this idea, three acid ionization equations and K a values are shown below. The ionization constants increase from first to last of the listed equations, indicating the relative acid strength increases in the order:
\[\ce{CH3CO2H < HNO2 < HSO4^{−}} \nonumber \]
as demonstrated with the equations below:
\[\ce{CH3CO2H(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + CH3CO2^{-}(aq)} \quad \quad K_a=1.8 \times 10^{−5} \nonumber \]
\[\ce{HNO2(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + NO2^{-}(aq)} \quad \quad K_a=4.6 \times 10^{-4} \nonumber \]
\[\ce{HSO4^{-}(aq) + H2O(aq) \rightleftharpoons H3O^{+}(aq) + SO4^{2-}(aq) } \quad\quad K_a=1.2 \times 10^{-2} \nonumber \]
Another measure of the strength of an acid is its percent ionization. The percent ionization of a weak acid is defined in terms of the composition of an equilibrium mixture:
\[\% \text { ionization }=\frac{\left[ \ce{H3O^{+}} \right]_{ eq }}{[ \ce{HA} ]_0} \times 100 \nonumber \]
where the numerator is equivalent to the concentration of the acid's conjugate base (per stoichiometry, [A − ] = [H 3 O + ]). Unlike the K a value, the percent ionization of a weak acid varies with the initial concentration of acid, typically decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter can be used to confirm this behavior.
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:
\[\frac{\left[ \ce{H3O^{+}} \right]_{ eq }}{\left[ \ce{HNO2} \right]_0} \times 100 \nonumber \]
Converting the provided pH to hydronium ion molarity yields
\[\left[ \ce{H3O^{+}} \right]=10^{-2.09}=0.0081 M \nonumber \]
Substituting this value and the provided initial acid concentration into the percent ionization equation gives
\[\frac{8.1 \times 10^{-3}}{0.125} \times 100=6.5 \% \nonumber \]
(Recall the provided pH value of 2.09 is logarithmic, and so it contains just two significant digits, limiting the certainty of the computed percent ionization.)
Calculate the percent ionization of a 0.10- M solution of acetic acid with a pH of 2.89.
- Answer
-
1.3% ionized
Just as for acids, the relative strength of a base is reflected in the magnitude of its base-ionization constant ( K b ) in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, \(\ce{B}\):
\[\ce{B(aq) + H2O(l) \rightleftharpoons HB^{+}(aq) + OH^{-}(aq)}, \nonumber \]
the ionization constant is written as
\[K_{ b }=\frac{\left[ \ce{HB^{+}} \right]\left[ \ce{OH^{-}} \right]}{[ \ce{B} ]} \nonumber \]
Inspection of the data for three weak bases presented below shows the base strength increases in the order: \(\ce{NO2^{−} < CH2CO2^{−} < NH3} \)
\[\begin{align*}
\ce{NO_2^{-}(aq) + H2O(l) & \rightleftharpoons HNO_2(aq) + OH^{-}(aq)} & K_{ b }=2.17 \times 10^{-11} \\[4pt]
\ce{CH_3 CO_2^{-}(aq) + H2O(l) & \rightleftharpoons CH_3 CO_2 H(aq) + OH^{-}(aq)} & K_{ b }=5.6 \times 10^{-10} \\[4pt]
\ce{NH_3(aq) + H2O(l) & \rightleftharpoons NH_4^{+}(aq) + OH^{-}(aq)} & K_{ b }=1.8 \times 10^{-5}
\end{align*} \nonumber \]
A table of ionization constants for weak bases appears in Appendix I. As for acids, the relative strength of a base is also reflected in its percent ionization, computed as
\[\% \text { ionization }= \left( \dfrac{[ \ce{OH^{-}}]_{eq}}{[ \ce{B} ]_0} \right) \times 100 \% \nonumber \]
but will vary depending on the base ionization constant and the initial concentration of the solution.
Relative Strengths of Conjugate Acid-Base Pairs
Brønsted-Lowry acid-base chemistry is the transfer of protons; thus, logic suggests a relation between the relative strengths of conjugate acid-base pairs. The strength of an acid or base is quantified in its ionization constant, \(K_a\) or \(K_b\), which represents the extent of the acid or base ionization reaction. For the conjugate acid-base pair \(\ce{HA / A^{-}}\), ionization equilibrium equations and ionization constant expressions are
\[\ce{HA(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + A^{-}(aq)} \nonumber \]
with
\[K_a=\ce{\dfrac{[H3O^{+}][A^{−}]}{[HA]}} \nonumber \]
and
\[\ce{A^{−}(aq) + H2O(l) \rightleftharpoons OH^{-}(aq) + HA(aq)} \nonumber \]
with
\[K_b=\ce{\dfrac{[HA][OH–]}{[A−]}} \nonumber \]
Adding these two chemical equations yields the equation for the autoionization for water:
\[\begin{align*} \ce{\cancel{HA(aq)} + H2O(l) + \cancel{A^{−}(aq)} + H2O(l) &\rightleftharpoons H3O^{+}(aq) + \cancel{A^{-}(aq)} + OH^{-}(aq) + \cancel{HA(aq)}} \\[4pt] \ce{2H2O(l) &<=> H3O+(aq) + OH^{−}(aq)} \end{align*} \nonumber \]
As discussed in another chapter on equilibrium, the equilibrium constant for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and so
\[K_a \times K_b= \dfrac{[\ce{H3O^{+}}][\ce{A^{−}}]}{[\ce{HA}]} × \dfrac{[\ce{HA}][\ce{OH^{-}}]}{[\ce{A^{-}}]}=[\ce{H3O^{+}}][\ce{OH^{-}}]=K_w \nonumber \]
This equation states the relation between ionization constants for any conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water, \(K_w\). By rearranging this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:
\[K_{ a }= \dfrac{K_{ w }}{ K_{ b }} \nonumber \]
or
\[K_{ b }=\dfrac{K_{ w }}{K_{ a }} \nonumber \]
The inverse proportional relation between \(K_a\) and \(K_b\) means the stronger the acid or base, the weaker its conjugate partner . A strong acid exhibits an immeasurably large K a , and so its conjugate base will exhibit a K b that is essentially zero:
- strong acid: \(\quad K_{ a } \approx \infty\)
- conjugate base: \(K_{ b }=K_{ w } / K_{ a }=K_{ w } / \infty \approx 0\)
A similar approach can be used to support the observation that conjugate acids of strong bases ( K b ≈ ∞) are of negligible strength ( K a ≈ 0).
Another way of writing this relationship between K a , K b and K w is using logs: -log(K a ) -log(K b ) = -log(K w ) = pK a + pK b = pK w .
Use the K b for the nitrite ion, \(\ce{NO2^{-}}\), to calculate the K a for its conjugate acid.
Solution
\(K_b\) for \(\ce{NO2^{-}}\) is given in this section as \(2.17 \times 10^{−11}\). The conjugate acid of \(\ce{NO2^{-}}\) is \(\ce{HNO2}\); \(K_a\) for \(\ce{HNO2}\) can be calculated using the relationship:
\[K_{ a } \times K_{ b }=1.0 \times 10^{-14}=K_{ w } \nonumber \]
Solving for K a yields
\[K_{ a }=\frac{K_{ w }}{K_{ b }}=\frac{1.0 \times 10^{-14}}{2.17 \times 10^{-11}}=4.6 \times 10^{-4} \nonumber \]
This answer can be verified by finding the Ka for \(\ce{HNO2}\) in Appendix H.
Determine the relative acid strengths of \(\ce{ NH 4^{ +}}\) and \(\ce{HCN}\) by comparing their ionization constants. The ionization constant of \(\ce{HCN}\) is given in Appendix H as \(4.9 \times 10^{−10}\). The ionization constant of \(\ce{ NH 4^{ +}}\) is not listed, but the ionization constant of its conjugate base, \(\ce{NH3}\), is listed as \(1.8 \times 10^{−5}\).
- Answer
-
\(\ce{NH4^{+}}\) is the slightly stronger acid (\(K_a\) for \(\ce{NH4^{+}}\) is \(5.6 \times 10^{−10}\)).
Acid-Base Equilibrium Calculations
The chapter on chemical equilibria introduced several types of equilibrium calculations and the various mathematical strategies that are helpful in performing them. These strategies are generally useful for equilibrium systems regardless of chemical reaction class, and so they may be effectively applied to acid-base equilibrium problems. This section presents several example exercises involving equilibrium calculations for acid-base systems.
Acetic acid is the principal ingredient in vinegar (Figure \(\PageIndex{4}\)) that provides its sour taste. At equilibrium, a solution contains \(\ce{[CH3CO2H]} = 0.0787\, M\): and \(\ce{H3O^+}]=[\ce{CH3CO2^{-}}]=0.00118\,M\). What is the value of \(K_a\) for acetic acid?
Solution
The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the Ka for acetic acid.
\[\ce{CH3CO2H(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + CH3CO2^{-}(aq)} \nonumber \]
\[K_{ a }=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{CH3CO2^{-}} \right]}{\left[ \ce{CH3CO2H} \right]}=\frac{(0.00118)(0.00118)}{0.0787}=1.77 \times 10^{-5} \nonumber \]
The \(\ce{HSO4^{−}}\) ion, weak acid used in some household cleansers:
\[\ce{HSO4^{-}(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + SO4^{2-}(aq)} \nonumber \]
What is the acid ionization constant for this weak acid if an equilibrium mixture has the following composition: \([\ce{H3O^{+}}] = 0.027\, M\); \([\ce{HSO4^{-}}]=0.29\,M\); \([\ce{HSO4^{-}}] =0.29\,M\); and \([\ce{SO4^{2-}}]=0.13\,M\)?
- Answer
-
K a for \(\ce{HSO4^{-}} = 1.2 \times 10^{−2}\)
Caffeine, \(\ce{C8H10N4O2}\) is a weak base. What is the value of \(K_b\) for caffeine if a solution at equilibrium has \([\ce{C8H10N4O2}] = 0.050~\text{M}\), \([\ce{C8H10N4O2H^{+}}] = 5.0 \times 10^{−3} ~\text{M}\), and \([\ce{OH^{-}}] = 2.5 \times 10^{−3} ~\text{M}\)?
Solution
The relevant equilibrium equation and its equilibrium constant expression are shown below. Substitution of the provided equilibrium concentrations permits a straightforward calculation of the \(K_b\) for caffeine.
\[\ce{C8H10N4O2(aq) + H2O(l) \rightleftharpoons C8H10N4O2H^{+}(aq) + OH^{−}(aq)} \nonumber \]
\[\begin{align*} K_b &=\ce{[C8H10N4O2H^{+}][OH^{-}][C8H10N4O2]} \\[4pt] &=(5.0 \times 10^{-3})(2.5 \times 10^{-3}) (0.050) \\[4pt]&=2.5 \times 10^{-4} \end{align*} \nonumber \]
What is the equilibrium constant for the ionization of the \(\ce{HPO4^{2−}}\) ion, a weak base
\[\ce{HPO4^{2-}(aq) + H2O(l) \rightleftharpoons H2PO4^{−}(aq) + OH^{-}(aq)} \nonumber \]
if the composition of an equilibrium mixture is as follows: \(\ce{[OH^{-}}] = 1.3 \times 10^{−6} M}\); \(\ce{H2PO4^{-}}]=0.042\,M\); and \([\ce{HPO4^{2-}}]=0.341\,M\)?
- Answer
-
\(K_b\) for \(\ce{HPO4^{2-}}\) is \(1.6 \times 10^{−7}\)
The pH of a 0.0516-M solution of nitrous acid, \(\ce{HNO2}\), is 2.34. What is its \(K_a\)?
\[\ce{HNO2(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + NO2^{−}(aq)} \nonumber \]
Solution
The nitrous acid concentration provided is a formal concentration, one that does not account for any chemical equilibria that may be established in solution. Such concentrations are treated as “initial” values for equilibrium calculations using the ICE table approach. Notice the initial value of hydronium ion is listed as approximately zero because a small concentration of H 3 O + is present (\(1 × \times 10^{−7}~ \text{M}\)) due to the autoprotolysis of water. In many cases, such as all the ones presented in this chapter, this concentration is much less than that generated by ionization of the acid (or base) in question and may be neglected.
The pH provided is a logarithmic measure of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and so it represents an “equilibrium” value for the ICE table:
\[\left[ \ce{H3O^{+}} \right]=10^{-2.34}=0.0046 ~\text{M} \nonumber \]
The ICE table for this system is then
Finally, calculate the value of the equilibrium constant using the data in the table:
\[K_{ a }=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{NO2^{-}} \right]}{\left[ \ce{HNO2} \right]}=\frac{(0.0046)(0.0046)}{(0.0470)}=4.6 \times 10^{-4} \nonumber \]
The pH of a solution of household ammonia, a 0.950-M solution of \(\ce{NH3}\), is 11.612. What is \(K_b\) for \(\ce{NH3}\).
- Answer
-
\(K_b = 1.8 \times 10^{−5}\)
Formic acid, HCO 2 H, is one irritant that causes the body’s reaction to some ant bites and stings (Figure \(\PageIndex{5}\)).
What is the concentration of hydronium ion and the pH of a 0.534- M solution of formic acid?
\[\ce{HCO2H(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + HCO2^{-}(aq)} \quad K_{ a }=1.8 \times 10^{-4} \nonumber \]
Solution
The ICE table for this system is
Substituting the equilibrium concentration terms into the K a expression gives
\[\begin{align*}
K_{ a }&=1.8 \times 10^{-4}=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{HCO2^{-}} \right]}{\left[ \ce{HCO2H} \right]}\\[4pt]
&=\frac{(x)(x)}{0.534-x}=1.8 \times 10^{-4}
\end{align*} \nonumber \]
The relatively large initial concentration and small equilibrium constant permits the simplifying assumption that x will be much lesser than 0.534, and so the equation becomes
\[K_{ a }=1.8 \times 10^{-4}=\frac{x^2}{0.534} \nonumber \]
Solving the equation for \(x\) yields
\[\begin{align*}
x^2 &=0.534 \times\left(1.8 \times 10^{-4}\right)=9.6 \times 10^{-5}\\[4pt]
x&=\sqrt{9.6 \times 10^{-5}}\\[4pt]
&=9.8 \times 10^{-3} M
\end{align*} \nonumber \]
To check the assumption that x is small compared to 0.534, its relative magnitude can be estimated:
\[\frac{x}{0.534}=\frac{9.8 \times 10^{-3}}{0.534}=1.8 \times 10^{-2}(1.8 \% \text { of } 0.534) \nonumber \]
Because x is less than 5% of the initial concentration, the assumption is valid.
As defined in the ICE table, x is equal to the equilibrium concentration of hydronium ion:
\[x=\left[ \ce{H3O^{+}} \right]=0.0098 ~\text{M} \nonumber \]
Finally, the pH is calculated to be
\[\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=-\log (0.0098)=2.01 \nonumber \]
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100- M solution of acetic acid, CH 3 CO 2 H?
\[\ce{CH3CO2H(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + CH3CO2^{-}(aq)} \quad K_{ a }=1.8 \times 10^{-5} \nonumber \]
- Answer
-
percent ionization = 1.3%
Find the concentration of hydroxide ion, the pOH, and the pH of a 0.25- M solution of trimethylamine, a weak base:
\[\ce{(CH3)3N(aq) + H2O(l) \rightleftharpoons (CH3)3NH^{+}(aq) + OH^{-}(aq)} \quad K_{ b }=6.3 \times 10^{-5} \nonumber \]
Solution
The ICE table for this system is
Substituting the equilibrium concentration terms into the K b expression gives
\[K_{ b }=\frac{\left[ \ce{(CH3)3NH^{+}} \right]\left[ \ce{OH^{-}} \right]}{\left[ \ce{(CH3)3N} \right]}=\frac{(x)(x)}{0.25-x}=6.3 \times 10^{-5} \nonumber \]
Assuming \(x \ll 0.25\) and solving for \(x\) yields
\[x=4.0 \times 10^{-3}~\text{M}\nonumber \]
This value is less than 5% of the initial concentration (0.25), so the assumption is justified. As defined in the ICE table, \(x\) is equal to the equilibrium concentration of hydroxide ion:
\[\begin{align*}
\left[ OH^{-}\right] &=\sim 0+x \\[4pt] &=x=4.0 \times 10^{-3} M\\[4pt]
\end{align*} \nonumber \]
The pOH is calculated to be
\[\text{pOH} =-\log \left(4.0 \times 10^{-3}\right)=2.40 \nonumber \]
Using the relation introduced in the previous section of this chapter:
\[\text{pH} + \text{pOH} = p K_{ w }=14.00 \nonumber \]
permits the computation of pH:
\[\text{pH} = 14.00 - \text{pOH} =14.00-2.40=11.60 \nonumber \]
Calculate the hydroxide ion concentration and the percent ionization of a 0.0325- M solution of ammonia, a weak base with a K b of \(1.76 \times 10^{−5}\).
- Answer
-
\(7.56 \times 10^{−4}~ \text{M}\), 2.33%
In some cases, the strength of the weak acid or base and its formal (initial) concentration result in an appreciable ionization. Though the ICE strategy remains effective for these systems, the algebra is a bit more involved because the simplifying assumption that x is negligible cannot be made. Calculations of this sort are demonstrated in Example \(\PageIndex{8}\) below.
Sodium bisulfate, NaHSO4, is used in some household cleansers as a source of the \(\ce{HSO4^{−}}\) ion, a weak acid. What is the pH of a 0.50-M solution of \(\ce{HSO4^{-}}\)?
\[\ce{HSO4^{-}(aq) + H2O(l) \rightleftharpoons H3O^{+}(aq) + SO4^{2-}(aq)} \quad K_{ a }=1.2 \times 10^{-2} \nonumber \]
Solution
The ICE table for this system is
Substituting the equilibrium concentration terms into the K a expression gives
\[K_{ a }=1.2 \times 10^{-2}=\frac{\left[ \ce{H3O^{+}} \right]\left[ \ce{SO4^{2-}} \right]}{\left[ \ce{HSO4^{-}} \right]}=\frac{(x)(x)}{0.50-x} \nonumber \]
If the assumption that \(x \ll 0.5\) is made, simplifying and solving the above equation yields
\[x=0.077~\text{M} \nonumber \]
This value of \(x\) is clearly not significantly less than 0.50
M
; rather, it is approximately 15% of the initial concentration:
When we check the assumption, we calculate:
\[\frac{x}{\left[ \ce{HSO4^{-}} \right]_{ i }} \nonumber \]
\[\frac{x}{0.50}=\frac{7.7 \times 10^{-2}}{0.50}=0.15(15 \%) \nonumber \]
Because the simplifying assumption is not valid for this system, the equilibrium constant expression is solved as follows:
\[K_{ a }=1.2 \times 10^{-2}=\frac{(x)(x)}{0.50-x} \nonumber \]
Rearranging this equation yields
\[6.0 \times 10^{-3}-1.2 \times 10^{-2} x=x^2 \nonumber \]
Writing the equation in quadratic form gives
\[x^2+1.2 \times 10^{-2} x-6.0 \times 10^{-3}=0 \nonumber \]
Solving for the two roots of this quadratic equation results in a negative value that may be discarded as physically irrelevant and a positive value equal to x . As defined in the ICE table, \(x\) is equal to the hydronium concentration.
\[x=\left[ H3O^{+}\right]=0.072~\text{M} \nonumber \]
\[\text{pH} =-\log \left[ \ce{H3O^{+}} \right]=-\log (0.072)=1.14 \nonumber \]
Calculate the pH in a 0.010- M solution of caffeine, a weak base:
\[\ce{C8H{10}N4O2(aq) + H2O(l) \rightleftharpoons C8H{10}N4O2H^{+}(aq) + OH^{-}(aq)} \quad K_{ b }=2.5 \times 10^{-4} \nonumber \]
- Answer
-
pH = 11.16