6.7: Polyprotic Acids and Bases
- Page ID
- 93343
Whereas monoprotic acids (like HCl pictured below) only have one hydrogen ion (aka proton - pictured below in white) to donate, polyprotic acids can donate multiple protons (see the H2SO4 molecule pictured below). Following a similar logic, polyprotic bases are bases that can accept multiple protons. With the basics of polyprotic acids and bases covered, let's dive into some of the details of how these special acids and bases work.
In the case of polyprotic acids and bases we can write down an equilibrium constant for each proton lost or gained. These constants are subscripted 1, 2, etc., to distinguish them. For sulfurous acid, a diprotic acid, we can, for example, write
\[\text{H}_{2}\text{SO}_{3} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{H}\text{SO}_{3}^{-}\]
\[K_{a\text{1}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ HSO}_{\text{3}}^{-}\text{ }]\text{ }}{\text{ }[\text{ H}_{\text{2}}\text{SO}_{\text{3}}\text{ }]\text{ }}=\text{1}\text{.7 }\times \text{ 10}^{-\text{2}}\text{ mol L}^{-\text{1}}\]
and
\[\text{H}\text{SO}_{3}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{SO}_{3}^{2-}\]
\[K_{a\text{2}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ SO}_{\text{3}}^{\text{2}-}\text{ }]\text{ }}{\text{ }[\text{ HSO}_{\text{3}}\text{ }]\text{ }}=\text{5}\text{.6 }\times \text{ 10}^{-\text{8}}\text{ mol L}^{-\text{1}}\]
The carbonate ion, CO32–, is an example of a diprotic base for which the appropriate base constants are
\[\text{CO}_{3}^{2-} + \text{H}_{2}\text{O} \rightleftharpoons \text{HCO}_{3}^{-} + \text{OH}^{-}\]
\[K_{b\text{1}}=\frac{\text{ }[\text{ HCO}_{\text{3}}^{-}\text{ }]\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{\text{ }[\text{ CO}_{\text{3}}^{\text{2}-}\text{ }]\text{ }}=\text{2}\text{.1 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\]
and
\[\text{HCO}_{3}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{2}\text{CO}_{3}^{} + \text{OH}^{-}\]
\[K_{b\text{2}}=\frac{\text{ }[\text{ H}_{\text{2}}\text{CO}_{\text{3}}\text{ }]\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{\text{ }[\text{ CO}_{\text{3}}^{-}\text{ }]\text{ }}=\text{2}\text{.4 }\times \text{ 10}^{-\text{8}}\text{ mol L}^{-\text{1}}\]
A general treatment of the pH of solutions of polyprotic species is beyond our intended scope, but it is worth noting that in many cases we can treat polyprotic species as monoprotic. In the case of H2SO3, for example, Ka1 is very much larger than Ka2 indicating that H2SO3 is a very much stronger acid than HSO3–. This means that when H2SO3 is dissolved in water, we can treat it as a monoprotic acid and ignore the possible loss of a second proton. Solutions of salts containing the carbonate ion, such as N2CO3 or K2CO3 can be treated similarly.
Example \(\PageIndex{1}\): pH Calculation
Find the pH of a 0.100-M solution of sodium carbonate, N2CO3. Use the base constant Kb1 = 2.10 × 10–4 mol L–1.
Solution
We ignore the acceptance of a second proton and treat the carbonate ion as a monoprotic base. We then have
\(\begin{align}\text{ }[\text{ OH}^{-}\text{ }]\text{ }=\sqrt{K_{b}c_{b}}=\sqrt{\text{2}\text{.10 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\text{ }\times \text{ 0}\text{.100 mol L}^{-\text{1}}} \\\text{ }=\text{4}\text{.58 }\times \text{ 10}^{-\text{3}}\text{ mol L}^{-\text{1}} \\\end{align}\)
Checking, we find that
\[\frac{\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{c_{b}}=\frac{\text{4}\text{.58 }\times\text{ 10}^{-\text{3}}}{\text{0}\text{.1}}\approx \text{4}\text{.6 percent}\]
so that our approximation is only just valid.
We now find
\[\text{pOH}=-\text{log}\left(\text{4.58}\times\text{10}^{-3}\right)= \text{2.34}\]
while
\[\text{pH}={14}-\text{pOH}={14}-{2.34}={11.6}\]
Since the carbonate ion is a somewhat stronger base than NH3, we expect a 0.1-M solution to be somewhat more basic, as actually found.
A glance at the Ka and and Kb tables reveals that most acid and base constants involve numbers having negative powers of 10. As in the case of [H3O+] and [OH–], then, it is convenient to define
\[\text{p}K_{a}=-\text{log}\frac{K_{a}}{\text{mol L}^{-\text{1}}}\]]
\[{p}K_{b}=-\text{log}\frac{K_{b}}{\text{mol L}^{-\text{1}}}\]
Using these definitions, the larger Ka or Kb is (i.e., the stronger an acid or base, respectively), the smaller pKaor pKb will be. For a strong acid like HNO3, Ka = 20 mol L–1 and
\[\text{p}K_{a} = -\text{log}\text{ 20}=-\left(\text{1.30}\right)=-\text{1.30}\]
Thus for very strong acids or bases pK values can even be negative.
Contributors
Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.