# Polyprotic Acids and Bases

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:53

In the case of polyproticDescribes an acid that can donate two or more hydrogen ions (&quot;protons&quot;) to a base. Examples are sulfuric and phosphoric acids. acids and bases we can write down an equilibrium constantThe value of the equilibrium constant expression when equilibrium concentrations are substituted; a value greater than one indicates the position of equilibrium lies toward products (product-favored), and a value less than one indicates the position of equilibrium lies toward reactants (reactant-favored). for each protonThe positively charged particle in an atomic nucleus; its mass is similar to the mass of a hydrogen atom. lost or gained. These constants are subscripted 1, 2, etc., to distinguish them. For sulfurous acid, a diproticDescribes an acid that can donate two hydrogen ions (protons) to a base. acid, we can, for example, write

$\text{H}_{2}\text{SO}_{3} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{H}\text{SO}_{3}^{-}$

$K_{a\text{1}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ HSO}_{\text{3}}^{-}\text{ }]\text{ }}{\text{ }[\text{ H}_{\text{2}}\text{SO}_{\text{3}}\text{ }]\text{ }}=\text{1}\text{.7 }\times \text{ 10}^{-\text{2}}\text{ mol dm}^{-\text{3}}$

and

$\text{H}\text{SO}_{3}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{SO}_{3}^{2-}$

$K_{a\text{2}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ SO}_{\text{3}}^{\text{2}-}\text{ }]\text{ }}{\text{ }[\text{ HSO}_{\text{3}}\text{ }]\text{ }}=\text{5}\text{.6 }\times \text{ 10}^{-\text{8}}\text{ mol dm}^{-\text{3}}$

The carbonate ionAn atom or covalently bonded set of atoms that carries an overall net charge., CO32–, is an example of a diprotic base for which the appropriate base constants are

$\text{CO}_{3}^{2-} + \text{H}_{2}\text{O} \rightleftharpoons \text{HCO}_{3}^{-} + \text{OH}^{-}$

$K_{b\text{1}}=\frac{\text{ }[\text{ HCO}_{\text{3}}^{-}\text{ }]\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{\text{ }[\text{ CO}_{\text{3}}^{\text{2}-}\text{ }]\text{ }}=\text{2}\text{.1 }\times \text{ 10}^{-\text{4}}\text{ mol dm}^{-\text{3}}$

and

$\text{HCO}_{3}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{2}\text{CO}_{3}^{} + \text{OH}^{-}$

$K_{b\text{2}}=\frac{\text{ }[\text{ H}_{\text{2}}\text{CO}_{\text{3}}\text{ }]\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{\text{ }[\text{ CO}_{\text{3}}^{-}\text{ }]\text{ }}=\text{2}\text{.4 }\times \text{ 10}^{-\text{8}}\text{ mol dm}^{-\text{3}}$

A general treatment of the pHA logarithmic measure of the concentration of hydrogen (hydronium) ion; pH = -log10([H+]) or pH = -log10([H3O+]). 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 Find the pH of a 0.100-M solution of sodium carbonate, N2CO3. Use the base constant Kb1 = 2.10 × 10–4 mol dm–3.

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 dm}^{-\text{3}}\text{ }\times \text{ 0}\text{.100 mol dm}^{-\text{3}}} \ & \text{ }=\text{4}\text{.58 }\times \text{ 10}^{-\text{3}}\text{ mol dm}^{-\text{3}} \ \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

$\text{pH}=\text{14}-\text{pOH}= \text{14}-\text{2.34}=\text{11.6}\,$

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

$\text{p}K_{a}=-\text{log}\frac{K_{a}}{\text{mol dm}^{-\text{3}}}$...and...$\text{p}K_{b}=-\text{log}\frac{K_{b}}{\text{mol dm}^{-\text{3}}}$

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 acidAn acid that ionizes completely in a particular solvent. like HNO3, Ka = 20 mol dm–3 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.