Question

What is the pH of a solution prepared by dissolving 0.5 mol of sodium acetate in enough water to make 2.0 L solution. pKa of acetic acid is 4.74/

Answer 1

pH of a solution prepared by dissolving 0.5 mol of sodium acetate in enough water to make 2.0 L solution

Molarity of sodium acetate(C_weakbase) = (0.5mol)/2.0 L = 0.25 M

pK_a of acetic acid is 4.74 = -log K_a

K_a of acetic acid = antilog(-4.74)= 1.81 *10^-5

Sodium acetate is weak base, a conjugate base of acetic acid

therefore, K_a^{acetic acid} * K_b^acetate = K_w

K_b^acetate = ( K_w/K_a^{acetic acid} ) = (10^-14/1.81 *10^-5) = 5.52*10^-10

C_weakbase>>>K_b therefore equation used must be

[OH^-] = (K_b*C_weakbase)^1/2 = ( 5.52*10^-10 * 0.25M)^1/2 =1.74*10^-5 M

pOH = -log(OH^- ) = -log(1.74*10^-5 M) =4.75

pH = 14 - pOH = 14 - 4.75 = 9.25

The pH of a solution prepared by dissolving 0.5 mol of sodium acetate in enough water to make 2.0 L solution​ = 9.25