Question 13.9: The distilled water you use in the laboratory is slightly ac...
The distilled water you use in the laboratory is slightly acidic because of dissolved CO_{2}, which reacts to form carbonic acid, H_{2}CO_{3}. Calculate the pH of a 0.0010 M solution of H_{2}CO_{3} and [{CO_{3}}^ {2-}] at equilibrium.
ANALYSIS
Information given: [H_{2}CO_{3}]_{o} (0.0010 M)
Information implied: two-step ionization, K_{a} value for each ionization (Table 13.3)
Asked for: pH, [{CO_{3}}^ {2-}]_{eq}
STRATEGY
1. Write the ionization reaction for each ionization. Recall that ionization of polyprotic acids takes place one H^+ at a time.
2. Most of the H^+is obtained from the first ionization. Find [H^+]_{eq} for the first ionization and convert to pH.
3. Note [H^+] = [{HCO_{3}}^{-}] . Substitute into the expression for the second ionization to obtain [{CO_{3}}^ {2-}]_{eq}.
| Table 13.3 Equilibrium Constants for Some Weak Polyprotic Acids at 25°C |
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| Acid | Formula | K_{a1} | K_{a2} | K_{a3} |
| Carbonic acid* | H_{2}CO_{3} | 4.4 \times 10^{-7} | 4.7 \times 10^{-11} | |
| Oxalic acid | H_{2}C_{2}O_{4} | 5.9 \times 10^{-2} | 5.2 \times 10^{-5} | |
| Phosphoric acid | H_{3}PO_{4} | 7.1 \times 10^{-3} | 6.2 \times 10^{-8} | 4.5 \times 10^{-13} |
| Sulfurous acid | H_{2}SO_{3} | 1.7 \times 10^{-2} | 6.0 \times 10^{-8} | |
*Carbonic acid is a water solution of carbon dioxide:
CO_{2}(g) + H_{2}O \rightleftharpoons H_{2}CO_{3}(aq)
The ionization constants listed are calculated assuming that all the carbon dioxide that dissolves is in the form of H_{2}CO_{3}.
