Calculations with Polyprotic Acids

If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This approach is demonstrated in the following example exercise.

Ionization of a Diprotic Acid
“Carbonated water” contains a palatable amount of dissolved carbon dioxide. The solution is acidic because CO2 reacts with water to form carbonic acid, H2CO3. What are $[H_3O^+]$, $[HCO_3^-]$, and $[CO_3^{2-}]$ in a saturated solution of CO2 with an initial [H2CO3] = 0.033 M?

$$H_2CO_3(aq)+H_2O(l)⇌H_3O^+(aq)+HCO_3^-(aq)\qquad K_{H_2CO_3}=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}=4.3×10^{-7}$$ $$HCO_3^-(aq)+H_2O(l)⇌H_3O^+(aq)+CO_3^{2-}(aq)\qquad K_{HCO_3^-}=\frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]}=4.7×10^{-11}$$

Solution
As indicated by the ionization constants, H2CO3 is a much stronger acid than $HCO_3^-$, so the stepwise ionization reactions may be treated separately.

The first ionization reaction is $$H_2CO_3(aq)+H_2O(l)⇌H_3O^+(aq)+HCO_3^-(aq)\qquad K_{H_2CO_3}=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}=4.3×10^{-7}$$



Using provided information, an ICE table for this first step is prepared:



Substituting the equilibrium concentrations into the equilibrium equation gives $$K_{H_2CO_3}=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}=\frac{(x)(x)}{0.033-x}=4.3×10^{-7}$$



Assuming x << 0.033 and solving the simplified equation yields $$x=1.2×10^{-4}$$



The ICE table defined x as equal to the bicarbonate ion molarity and the hydronium ion molarity:

$$[H_2CO_3]=0.033\;M$$ $$[H_3O^+]=[HCO_3^-]=1.2×10^{-4}\;M$$

Using the bicarbonate ion concentration computed above, the second ionization is subjected to a similar equilibrium calculation:

$$HCO_3^-(aq)+H_2O(l)⇌H_3O^+(aq)+CO_3^{2-}(aq)$$ $$K_{HCO_3^-}=\frac{[H_3O^+][CO_3^{2-}]}{[HCO_3^-]}=\frac{(1.2×10^{-4})[CO_3^{2-}]}{1.2×10^{-4}}$$ $$[CO_3^{2-}]=\frac{(4.7×10^{-11})(1.2×10^{-4})}{1.2×10^{-4}}=4.7×10^{-11}\;M$$

To summarize: at equilibrium [H2CO3] = 0.033 M; $[H_3O^+]=1.2×10^{-4}\;M$; $[HCO_3^-]=1.2×10^{-4}\;M$; $[CO_3^{2-}]=4.7×10^{-11}\;M$.

 

Check Your Learning

The concentration of H2S in a saturated aqueous solution at room temperature is approximately 0.1 M. Calculate $[H_3O^+]$, [HS−], and [S2−] in the solution:

$$H_2S(aq)+H_2O(l)⇌H_3O^+(aq)+HS^-(aq)\qquad K_{a1}=8.9×10^{-8}$$ $$HS^-(aq)+H_2O(l)⇌H_3O^+(aq)+S^{2-}(aq)\qquad K_{a2}=1.0×10^{-19}$$
Answer:

[H2S] = 0.1 M;$[H_3O^+]$= [HS−] = 0.000094 M; [S2−] = $1×10^{-19}\;M$