Question 15.11: Calculate the pH and the concentrations of all species prese......
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 15.7.
| IDENTIFY | |
| Known | Unknown |
| Concentration of H_{2}CO_{3} (0.020 M) | pH |
| K_{a1}=4.3\times 10^{-7}, K_{a2}=5.6\times 10^{-11} (Table 15.3) | Equilibrium concentrations |
| Table 15.3 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 \times 10^{-7} | 5.6 \times 10^{-11} | 4.8 \times 10^{-13} |
| Hydrogen sulfidea^{a} | H_{2}S | 1.0 \times 10^{-7} | \sim 10^{-19} | |
| Oxalic acid | H_{2}C_{2}O_{4} | 5.9 \times 10^{-2} | 6.4 \times 10^{-5} | |
| Phosphoric acid | H_{3}PO_{4} | 7.5 \times 10^{-3} | 6.2 \times 10^{-8} | |
| Sulfuric acid | H_{2}SO_{4} | Very large | 1.2 \times 10^{-2} | |
| Sulfurous acid | H_{2}SO_{3} | 1.5 \times 10^{-2} | 6.3 \times 10^{-8} | |
| ^{a}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.
Table 1
Step 5. Substituting the equilibrium concentrations into the equilibrium equation for the principal reaction gives
K_{a1}=4.3\times 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\times 10^{-7})(0.020)x = 9.3 \times 10^{-5} Approximation (0.020 – x) ≈ 0.020 is justified.
Step 6. The big concentrations are
[H_{3}O^{+}]=[HCO_{3} ^{-}]= x = 9.3\times 10^{-5} M[H_{2}CO_{3}] = 0.020 – x = 0.020 – 0.000 093 = 0.020 M
Step 7. The small concentrations (CO_{3} ^{2-} and OH^{-}) 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)\xrightleftharpoons{}H_{3}O^{+}(aq)+CO_{3} ^{2-}(aq)
K_{a2}=5.6\times 10^{-11}=\frac{[H_{3}O^{+}][CO_{3} ^{2-}]}{[HCO_{3} ^{-}]}=\frac{(9.3\times 10^{-5})[CO_{3} ^{2-}]}{(9.3\times 10^{-5})}[CO_{3} ^{2-}]=K_{a2}=5.6\times 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 \times 10^{-14}}{9.3\times 10^{-5}}=1.1\times 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\times 10^{-5} M of HCO_{3} ^{-} produced by the first dissociation, only 5.6\times 10^{-11} M dissociates to form H_{3}O^{+} and CO_{3} ^{2-}.
Step 8. pH = -log [H_{3}O^{+}] = – log (9.3\times 10^{-5}) = 4.03
Table 1
| Principal reaction | H_{2}CO_{3}(aq) | + | H_{2}O(l) | \xrightleftharpoons{} | 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 |