Question 18.p.9: The hypochlorite ion, ClO^-, acts as a weak base in water. W......
The hypochlorite ion, \text{ClO}^-, acts as a weak base in water. Write a balanced equation and equilibrium expression for this reaction. The K_\text{b}\text{ of ClO}^- is calculated from the K_\text{a} of its conjugate acid, hypochlorous acid, HClO (from Appendix C, K_\text{a} = 2.9 \times 10^{-8}). Make simplifying assumptions (if valid), solve for [\text{OH}^-] , \text{convert to [H}_3\text{O}^+] and calculate pH.
Assumptions:
1) Since K_\text{b} >> K_\text{w}, the dissociation of water provides a negligible amount of [\text{OH}^-] .
2) Since K_\text{b}\text{ is very small, [ClO}^-]_\text{eq} = 0.20 – \text{x} ≈ 0.2. \\ K_\text{b}=\frac{[\text{HClO}][\text{OH}^-]}{[\text{ClO}^-]}=\frac{\text{x}^2}{(0.20)}=3.448276\times 10^{-7} \\ \text{x}=2.6261\times 10^{-4} \\ \text{Therefore, [HClO] = [OH}^-]=2.6\times 10^{-4} \ M \\ \text{Since }\frac{[\text{OH}^-]}{[\text{ClO}^-]} (100)=\frac{2.6261\times 10^{-4}}{0.20}(100)=0.1313\text{ which <5\%, the assumption that the dissociation of ClO}^-\text{ is small is valid.} \\ [\text{H}_3\text{O}^+]=\frac{1.0\times 10^{-14}}{2.6261\times 10^{-4}} =3.8079\times 10^{-11} \ M\text{ and pH }=-\log(3.8079\times 10^{-11})=10.4193=10.42
Check: Since hypochlorite ion is a weak base, a pH > 7 is expected.