Question

A 0.725 M solution of a weak acid has [H+] = 5.23 × 10–6 M. What is the value of Ka for this acid?

Answer 1

3.77 x 10^-11

Step-by-step explanation:

So the equilibrium would be (0.725 - 0.00000523) ≈ 0.7249

Ka = [H+] x [CH3COO^-] / [CH3COOH]

Ka = [5.23 x 10^-11] x [5.23 x 10^-11] / 0.7249

Ka = 3.77 x 10^-11

Hope this helped : )