Answer to Question #147520 in General Chemistry for gia

Question

The Ka for benzoic acid is 6.5 x 10^-5.

Calculate the pH of a 0.15 M benzoic acid solution.

Answer 1

The formula for K_a is:

K_a = [H⁺][B^-]/[HB]

where:

[H⁺] = concentration of H⁺ ions

[B^-] = concentration of conjugate base ions

[HB] = concentration of undissociated acid molecules

for a reaction HB → H⁺ + B^-

Benzoic acid dissociates one H⁺ ion for every C₆H₅COO^- ion, so [H⁺] = [C₆H₅COO^-].

Let x represent the concentration of H⁺ that dissociates from HB, then [HB] = C - x where C is the initial concentration.

Enter these values into the K_a equation:

K_a = x · x / (C -x)

K_a = x²/(C - x)

(C - x)K_a = x²

x² = CK_a - xK_a

x² + K_ax - CK_a = 0

Solve for x using the quadratic equation:

x = [-b ± (b² - 4ac)^½]/2a

x = [-K_a + (K_a² + 4CK_a)^½]/2

K_a = 6.5 x 10^-5

C = 0.15 M

x = {-6.5 x 10^-5 + [(6.5 x 10^-5)² + 4(0.15)(6.5 x 10^-5)]^½}/2

x = (-6.5 x 10^-5 + 1.6 x 10^-3)/2

x = (1.5 x 10^-3)/2

x = 7.7 x 10^-4

Find pH:

pH = -log[H⁺]

pH = -log(x)

pH = -log(7.7 x 10^-4)

pH = -(-3.11)

pH = 3.11