Answer to Question #312455 in Physical Chemistry for Rajani

Question

Solution #1 is a 0.025 M solution of hypochlorous acid, HOCl (Ka = 3.0 × 10–8). Calculate each of the

following quantities for Solution #1.

(a) [H+] in Solution #1

Answer 1

С(HOCl) = 0.025 M;

K_a(HOCl) = 3.0 * 10^-8;

HOCl = H⁺ + OCl^-

0.025 0 0

-x x x

0.025-x x x

K_a = [H⁺] * [OCl^-]/[HOCl];

K_a = x * x/(0.025 - x) = x²/(0.025 - x);

x² = K_a * 0.025 = 3.0 * 10^-8 * 0.025 = 7.5 * 10^-10;

x = 2.7 * 10^-5;

Therefore,

[H⁺] = [OCl^-] = 2.7 * 10^-5 M.

Answer: 2.7 * 10^-5 M.