Question 16.19: Oxalic acid (H2C2O4) is a poisonous substance used mainly as...
Oxalic acid (\text{H}_2\text{C}_2\text{O}_4) is a poisonous substance used mainly as a bleaching agent. Calculate the concentrations of all species present at equilibrium in a 0.10-M solution at 25°\text{C}.
Strategy Follow the same procedure for each ionization as for the determination of equilibrium concentrations for a monoprotic acid. The conjugate base resulting from the first ionization is the acid for the second ionization, and its starting concentration is the equilibrium concentration from the first ionization.
Setup The ionizations of oxalic acid and the corresponding ionization constants are
\text{H}_2\text{C}_2\text{O}_4(aq) \rightleftarrows \text{H}^+(aq) + \text{HC}_2\text{O}_4^–(aq) K_{a_1} = 6.5 × 10^{–2}
\text{HC}_2\text{O}_4^– (aq) \rightleftarrows \text{H}^+(aq) + \text{C}_2\text{O}_4^{2–} (aq) K_{a_2} = 6.1 × 10^{–5}
Construct an equilibrium table for each ionization, using x as the unknown in the first ionization and \text{y} as the unknown in the second ionization.
\text{H}_2\text{C}_2\text{O}_4(aq) \rightleftarrows \text{H}^+(aq) + \text{HC}_2\text{O}_4^–(aq)
| Initial concentration (M): | 0.10 | 0 | 0 |
| Change in concentration (M): | –x | +x | +x |
| Equilibrium concentration (M): | 0.10 − x | x | x |
The equilibrium concentration of the hydrogen oxalate ion (\text{HC}_2\text{O}_4^–) after the first ionization becomes the starting concentration for the second ionization. Additionally, the equilibrium concentration of \text{H}^+ is the starting concentration for the second ionization.
\text{HC}_2\text{O}_4^– (aq) \rightleftarrows \text{H}^+(aq) + \text{C}_2\text{O}_4^{2–} (aq)
| Initial concentration (M): | x | x | 0 |
| Change in concentration (M): | –\text{y} | +\text{y} | +\text{y} |
| Equilibrium concentration (M): | x − \text{y} | x + \text{y} | \text{y} |
K_{a_1}=\frac{[\text{H}^+]\text{[HC}_2\text{O}_4^–]}{[\text{H}_2\text{C}_2\text{O}_4]}\\6.5\times 10^{-2}=\frac{x^2}{0.10-x}
Applying the approximation and neglecting x in the denominator of the expression gives
6.5 × 10^{–2} ≈ \frac{x^2}{0.10}
x^2 = 6.5 × 10^{–3}
x = 8.1 × 10^{–2} M
Testing the approximation,
\frac{8.1 × 10^{–2} M}{0.10 M} × 100\% = 81\%
Clearly the approximation is not valid, so we must solve the following quadratic equation:
x^2 + 6.5 × 10^{–2} x − 6.5 × 10^{–3} = 0
The result is x = 0.054 M. Thus, after the first ionization, the concentrations of species in solution are
[\text{H}^+] = 0.054 M
[\text{HC}_2\text{O}_4^–] = 0.054 M
[\text{H}_2\text{C}_2\text{O}_4] = (0.10 − 0.054) M = 0.046 M
Rewriting the equilibrium table for the second ionization, using the calculated value of x, gives the following:
\text{HC}_2\text{O}_4^– (aq) \rightleftarrows \text{H}^+(aq) + \text{C}_2\text{O}_4^{2–} (aq)
| Initial concentration (M): | 0.054 | 0.054 | 0 |
| Change in concentration (M): | –\text{y} | +\text{y} | +\text{y} |
| Equilibrium concentration (M): | 0.054 − \text{y} | 0.054 + \text{y} | \text{y} |
K_{a_2}= \frac{[\text{H}^+][\text{C}_2\text{O}_4^{2–}]}{[\text{HC}_2\text{O}_4^–] }
6.1 × 10^{–5} = \frac{(0.054 + \text{y})(\text{y})}{0.054 − \text{y} }
Assuming that \text{y} is very small and applying the approximations 0.054 + \text{y} ≈ 0.054 and 0.054 −\text{y} ≈ 0.054 gives
\frac{(0.054)(\text{y})}{0.054} = \text{y} = 6.1 × 10^{–5}
We must test the approximation as follows to see if it is valid:
\frac{6.1 × 10^{–5} M}{0.054 M} × 100\% = 0.11\%
This time, because the ionization constant is much smaller, the approximation is valid. At equilibrium, the concentrations of all species are
[\text{H}_2\text{C}_2\text{O}_4] = 0.046 M
[\text{HC}_2\text{O}_4^–] = (0.054 − 6.1 × 10^{–5}) M = 0.054 M
[\text{H}^+] = (0.054 + 6.1 × 10^{–5}) M = 0.054 M
[\text{C}_2\text{O}_4^{2–}] = 6.1 × 10^{–5} M