Question 16.14: The solubility of CO2 in water at 25 °C and 10.13 kPa is 0.0...
The solubility of CO_{2} in water at 25 °C and 10.13 kPa is 0.0037 M. The common practice is to assume that all the dissolved CO_{2} is in the form of carbonic acid (H_{2}CO_{3}), which is produced in the reaction
CO_{2} (aq) + H_{2}O (l) ⇌ H_{2}CO_{3} (aq)What is the pH of a 0.0037 M solution of H_{2}CO_{3}?
Analyze We are asked to determine the pH of a 0.0037 M solution of a polyprotic acid.
Plan H_{2}CO_{3} is a diprotic acid; the two acid-dissociation constants, K_{a1} and K_{a2} (Table 16.3), differ by more
than a factor of 10³. Consequently, the pH can be determined by considering only K_{a1}, thereby treating the acid as if it were a monoprotic acid.
| TABLE 16.3 Acid-Dissociation Constants of Some Common Polyprotic Acids | ||||
| Name | Formula | K_{a1} | K_{a2} | K_{a3} |
| Ascorbic | H_{2}C_{6}H_{6}O_{6} | 8.0 × 10^{-5} | 1.6 × 10^{-12} | |
| Carbonic | H_{2}CO_{3} | 4.3 × 10^{-7} | 5.6 × 10^{-11} | |
| Citric | H_{3}C_{6}H_{5}O_{7} | 7.4 × 10^{-4} | 1.7 × 10^{-5} | 4.0 × 10^{-7} |
| Oxalic | HOOC—COOH | 5.9 × 10^{-2} | 6.4 × 10^{-5} | |
| Phosphoric | H_{3}PO_{4} | 7.5 × 10^{-3} | 6.2 × 10^{-8} | 4.2 × 10^{-13} |
| Sulfurous | H_{2}SO_{3} | 1.7 × 10^{-2} | 6.4 × 10^{-8} | |
| Sulfuric | H_{2}SO_{4} | Large | 1.2 × 10^{-2} | |
| Tartaric | C_{2}H_{2}O_{2}(COOH)_{2} | 1.0 × 10^{-3} | 4.6 × 10^{-5} | |
Solve
Proceeding as in Sample Exercises 16.12 and 16.13, we can write the equilibrium reaction and equilibrium concentrations as:
The equilibrium expression is:
K_{a1} =\frac{[H^{+}][HCO^{ -}_{3}]}{[H_{2}CO_{3}]}=\frac{(x)(x)}{0.0037 – x}= 4.3 × 10^{-7}Solving this quadratic equation, we get: x = 4.0 × 10^{-5} M
Alternatively, because K_{a1} is small, we can make the simplifying approximation that x is small, so that:
0.0037 – x \simeq 0.0037
Thus, \frac{(x)(x)}{0.0037}= 4.3 × 10^{-7}
Solving for x, we have:
x² = (0.0037)(4.3 × 10^{-7}) = 1.6 × 10^{-9}
x = [H^{+}] = [HCO^{-}_{3}] = \sqrt{1.6 × 10^{-9}} = 4.0 × 10^{-5} M
Because we get the same value (to two significant figures) our simplifying assumption was justified. The pH is therefore:
pH = – log [H^{+}] = – log(4.0 × 10^{-5}) = 4.40
Comment If we were asked for [CO^{2-}_{3}] we would need to use K_{a2}. Let’s illustrate that calculation. Using our calculated values of [HCO^{-}_{3}] and [H^{+}] and setting [CO^{2-}_{3}] = y, we have:
Assuming that y is small relative to 4.0 × 10^{-5}, we have:
K_{a2} =\frac{[H^{+}][CO^{2-}_{3}]}{[HCO^{ -}_{3}]}=\frac{(4.0 × 10^{-5})(y)}{4.0 × 10^{-5}}= 5.6 × 10^{-11}y = 5.6 × 10^{-11} M = [CO^{2-}_{3}]
We see that the value for y is indeed very small compared with 4.0 ×10^{-5}, showing that our assumption was justified. It also shows that the ionization of HCO^{-}_{3} is negligible relative to that of H_{2}CO_{3}, as far as production of H^{+} is concerned. However, it is the only source of CO^{2-}_{3}, which has a very low concentration in the solution. Our calculations thus tell us that in a solution of carbon dioxide in water, most of the CO_{2} is in the form of CO_{2} or H_{2}CO_{3}, only a small fraction ionizes to form H^{+} and HCO^{-}_{3}, and an even smaller fraction ionizes to give CO^{2-}_{3}. Notice also that [CO^{2-}_{3}] is numerically equal to K_{a2}.