How would you slove this qestion?

Question

Sodium acetate, CH3COONa, is used for developing photographs. Find the value of Kb for the acetate ion. Then calculate the pH of a solution that contains 12.5 g of sodium acetate dissolved in 1.00 L of water.

Answer 1

You can't calculate the Kb for acetate from the information given, but you can find it by looking it up in a book or on the internet. I found it to be 5.7x10^-10. If you have the Ka for acetic acid (1.76x10^-5) you can calculate the Kb for acetate as follows: KaKb = 1x10^-14. Kb = 1x10^-14 / 1.76x10^-5 = 5.6810^-10

Now, on to solving the problem.

molar mass sodium acetate (CH3COONa) = 80.0 g/mol

Reaction: CH3COO^- + H2O ==> CH3COOH + OH^-

Kb = 5.7x10^-10 = [CH3COOH][OH^-] / [CH3COO^-]

and [CH3COOH] and [OH-] = x and [CH3COO-] = 12.5 g/L and 12.5 g x 1 mol/80.0 g = 0.156 mol/L

5.7x10^-10 = (x)(x) / 0.156

x² = 8.91x10^-11

x = [OH-] = 9.44x10^-6 M

pOH = -log 9.44x10-6 = 5.03

pH = 14 - 5.03 = 8.97