What is the pH of a 0.15-M Ca(CH3CO2)2 solution? Note: CH3COOH ; Ka = 1.8 x 10 -5

Question

Please solve this question! The answer that I got was 9.11 and I want to ensure I am correct. Some of my work is below:

Ca 2+ 2 CH3COO- + H2O <––> CH3COOH + OH-

Kw = 1.0 x 10^-14

Kw = (Kb) (Ka)

Kb = (1.0x10^-14)/ (1.8x10^-5) = 5.56 x10^-10

5.56 x 10^-10 = (x^2) / (0.30)

x = 1.29 x 10^-5M

pOH = -log[OH] = -log(1.29 x10^-5) = 4.89

pH = 14 - 4.89 = 9.11

Answer 1

I also calculated pH = 9.11. Well done.