Question 14.11: CALCULATING THE pH AND THE EQUILIBRIUM CONCENTRATIONS IN A S...
CALCULATING THE pH AND THE EQUILIBRIUM CONCENTRATIONS IN A SOLUTION OF A DIPROTIC ACID
Calculate the pH and the concentrations of all species present (H_{2}CO_{3}, HCO_{3}^{-}, CO_{3}^{2-}, H_{3}O^{+}, and OH^{-}) in a 0.020 M carbonic acid solution.
STRATEGY
Use the eight-step procedure summarized in Figure 14.6. The values of K_{a1} and K_{a2} may be found in Table 14.3.
TABLE 14.3 Stepwise Dissociation Constants for Polyprotic Acids at 25 °C
| Name Formula K_{a1} K_{a2} K_{a3} |
| Carbonic acid H_{2}CO_{3} 4.3 × 10^{-7} 5.6 × 10^{-11}
Hydrogen sulfide* H_{2}S 1.0 × 10^{-7} ∼10^{-19} Oxalic acid H_{2}C_{2}O_{4} 5.9 × 10^{-2} 6.4 × 10^{-5} Phosphoric acid H_{3}PO_{4} 7.5 × 10^{-3} 6.2 × 10^{-8} 4.8 × 10^{-13} Sulfuric acid H_{2}SO_{4} Very large 1.2× 10^{-2} Sulfurous acid H_{2}SO_{3} 1.5 × 10^{-2} 6.3 × 10^{-8} |
^{*}Because of its very small size, K_{a2} for H_{2}S is difficult to measure and its value is uncertain.
Steps 1–3. The species present initially are H_{2}CO_{3} (acid) and H_{2}O (acid or base).
Because K_{a1} >> K_{w}, the principal reaction is the dissociation of H_{2}CO_{3}.
Step 4.
Step 5. Substituting the equilibrium concentrations into the equilibrium equation for the principal reaction gives
K_{a1} = 4.3 × 10^{-7} = \frac{[H_{3}O^{+}][HCO_{3}^{-}]}{[H_{2}CO_{3}]} = \frac{(x)(x)}{(0.020 – x)}Assuming that (0.020 – x) ≈ 0.020,
x^{2} = (4.3 × 10^{-7})(0.020)x = 9.3 × 10^{-5} Approximation (0.020 – x) ≈ 0.020 is justified.
Step 6. The big concentrations are
[H_{3}O^{+}] = [HCO_{3}^{-}] = x = 9.3 × 10^{-5} M
[H_{2}CO_{3}] = 0.020 – x = 0.020 – 0.000 093 = 0.020 M
Step 7. The small concentrations are obtained from the subsidiary equilibria—(1) dissociation of HCO_{3}^{-} and (2) dissociation of water—and from the big concentrations already determined:
(1) HCO_{3}^{-}(aq) + H_{2}O(l) \rightleftharpoons H_{3}O^{+}(aq) + CO_{3}^{2-}(aq)
K_{a2} = 5.6 × 10^{-11} = \frac{[H_{3}O^{+}][CO_{3}^{2-}]}{[HCO_{3}^{-}]} = \frac{(9.3 × 10^{-5})[CO_{3}^{2-}]}{9.3 × 10^{-5}}[CO_{3}^{2-}] = K_{a2} = 5.6 × 10^{-11} M
(In general, for a solution of a weak diprotic acid that has a very small value of K_{a2},
[A^{2-}] = K_{a2}.)
(2) [OH^{-}] = \frac{K_{w}}{[H_{3}O^{+}]} = \frac{1.0 × 10^{-14}}{9.3 × 10^{-5}} = 1.1 × 10^{-10} M
The second dissociation of H_{2}CO_{3} produces a negligible amount of H_{3}O^{+} compared with the H_{3}O^{+} obtained from the first dissociation. Of the 9.3 × 10^{-5} mol/L of HCO_{3}^{-} produced by the first dissociation, only 5.6 × 10^{-11} mol/L dissociates to form H_{3}O^{+} and CO_{3}^{2-}.
Step 8. pH = -log [H_{3}O^{+}] = -log (9.3 × 10^{-5}) = 4.03
BALLPARK CHECK
When the value of x can be neglected compared with the initial concentration of the acid (Step 5), the [H_{3}O^{+}] equals the square root of the product of K_{a1} and the initial concentration of the acid. In this problem, [H_{3}O^{+}] is the square root of approximately (4 × 10^{-7})(2 × 10^{-2}) , or about 10^{-4}. Therefore, the pH ≈ 4, in agreement with the solution.
| Principal reaction H_{2}CO_{3}(aq) + H_{2}O(l) \rightleftharpoons H_{3}O^{+}(aq) + HCO_{3}^{-}(aq) |
| Initial concentration (M) 0.020 ∼0 0
Change (M) -x +x +x Equilibrium concentration (M) 0.020 – x x x |