Answer to Question #308167 in General Chemistry for Sofia

Question

Enough of a monoprotic weak acid is dissolved in water to produce a 0.0133 M solution. The pH of the resulting solution is 2.38. Calculate the K_a for the acid.

Answer 1

HX = H⁺ + X^-

C₀ 0.0133 0 0

$\Delta$C -x x x

[C] (0.0133-x) x x

K_a = [H⁺] * [X^-] / [HX] = x * x / (0.0133-x) = x²/(0.0133-x);

pH = 2.38, therefore [H⁺] = 10^-2.38 = 0.0042 M;

For reaction of dissociation: [H⁺] = [X^-] = 0.0042 M;

K_a = (0.0042)² / (0.0133-0.0042) = 1.76 *10^-5 / 0.0091 = 1.93 *10^-3 = 0.00193.

Answer: 1.93 *10^-3 or 0.00193.