Question 18.10: Calculating Equilibrium Concentrations for a Polyprotic Acid......

Plan We know the initial concentration (0.050 M) and both K_a’s for H_2Asc, and we have to calculate the equilibrium concentrations of all species and convert [H_3O^+] to pH. We first write the equations and K_a expressions. Because K_{a1}  >>  K_{a2}, we can assume that the first dissociation produces almost all the H_3O^+:  [H_3O^+]_{\text{from }H_2Asc}  >>  [H_3O^+]_{\text{from }HAsc^−}. Also, because K_{a1} is small, the amount of H_2Asc that dissociates can be neglected. We set up a reaction table for the first dissociation, with x  =  [H_2Asc]_{\text{dissoc}}, and then we solve for [H_3O^+] and [HAsc^−]. Because the second dissociation occurs to a much lesser extent, we can substitute values from the first dissociation directly to find [Asc^{2−}] from the second.

Solution Writing the equations and K_a expressions:
              H_2Asc(aq)  +  H_2O(l) \xrightleftharpoons[]  HAsc^−(aq)  +  H_3O^+(aq)
              K_{a1}  =  \frac{[HAsc^−][H_3O^+]}{[H_2Asc]}  =  1.0×10^{−5}
              HAsc^−(aq)  +  H_2O(l)  \xrightleftharpoons[]  Asc^{2−}(aq)  +  H_3O^+(aq)
              K_{a2}  =  \frac{[Asc^{2−}][H_3O^+]}{[HAsc^−]}  =  5×10^{−12}

Setting up a reaction table (Table 1) with x  =  [H_2Asc]_{\text{dissoc}}  =  [HAsc^−]  ≈  [H_3O^+]:

Making the assumptions:
1. Because K_{a2}  <<  K_{a1},  [H_3O^+]_{\text{from }HAsc^−}  <<  [H_3O^+]_{\text{from }H_2Asc}. Therefore,
             [H_3O^+]_{\text{from }H_2Asc}  ≈  [H_3O^+]
2. Because K_{a1} is small, [H_2Asc]_{\text{init}}  −  x  =  [H_2Asc]  ≈  [H_2Asc]_{\text{init}}. Thus,
             [H_2Asc]  =  0.050  M  −  x  ≈  0.050  M
Substituting into the expression for K_{a1} and solving for x:

             K_{a1}  =  \frac{[H_3O^+][HAsc^−]}{[H_2Asc]}  =  1.0×10^{−5}  = \frac{x^2}{0.050  −  x}  ≈  \frac{x^2}{0.050}
             x  =  [HAsc^−]  ≈  [H_3O^+]  ≈  7.1×10^{−4}  M
             pH  =  −\log  [H_3O^+]  =  −\log  (7.1×10^{−4})  =  3.15

Checking the assumptions:

1. [H_3O^+]_{\text{from }HAsc^−}  <<  [H_3O^+]_{\text{from }H_2Asc}: For any second dissociation that does occur,

            K_{a2}  =  \frac{[H_3O^+][Asc^{2−}]}{[HAsc^−]}  =  5×10^{−12}  =  \frac{(x)(x)}{7.1×10^{−4}}
            x  =  [H_3O^+]_{\text{from }HAsc^−}  =  6×10^{−8}  M

This is even less than [H_3O^+]_{\text{from }H_2O}, so the assumption is justified.

2. [H_2Asc]_{\text{dissoc}}  <<  [H_2Asc]_{\text{init}}:  \frac{7.1×10^{−4}  M}{0.050  M}  ×  100  =  1.4\% (<5%; assumption is justified).
Also, note that           \frac{[H_2Asc]_{\text{init}}}{K_{a1}}  =  \frac{0.050 }{1.0×10^{−5}}  =  5000  >  400

Using the equilibrium concentrations from the first dissociation to calculate [Asc^{2−}]:

            K_{a2}  =  \frac{[H_3O^+][Asc^{2−}]}{[HAsc^−]}      and      [Asc^{2−}]  =  \frac{(K_{a2})[HAsc^−]}{[H_3O^+]}
            [Asc^{2−}]  =  \frac{(5×10^{−12})(7.1×10^{−4})}{7.1×10^{−4}}  =  5×10^{−12}  M

Check K_{a1}  > >  K_{a2}, so it makes sense that [HAsc^−]  >>  [Asc ^{2−} ] because Asc^{2−} is produced only in the second (much weaker) dissociation. Both K_a’s are small, so all concentrations except [H_2Asc] should be much lower than the original 0.050 M.