Question

Calculate the pH at 25°C of a 0.31M solution of potassium acetate KCH3CO2 . Note that acetic acid HCH3CO2 is a weak acid with a pKa of 4.76 . Round your answer to 1 decimal place.

Answer 1

CH3CO2^-(aq) + H2O(l) <--------> CH3COOH + OH^-(aq)

K_b = [CH3COOH][OH^-]/[CH3CO^-]

K_{​​​​​​b} = K_w/K_b

K_w = ionic product of water , 1.00 ×10^-14

pK_a = -logK_a

- logK_a = 4.76

K_a = 1.74×10^-5

K_b = K_w/ K_a

K_b = 1.00 ×10^-14/1.74 ×10^-5 = 5.75 × 10^-10

initial concentration

  [CH3CO2^-] = 0.31

[CH3COOH] = 0

[OH^-] = 0

change in concentration

[CH3CO^2-] = - x

[CH3COOH] = + x

[OH^-] = +x

Equillibrium concentration

[CH3CO2^2-] = 0.31 - x

[CH3COOH] = x

[OH^- ] = x

so,

x²/(0.31 - x) = 5.75×10^-10

we can assume 0.31 - x = 0.31 because x is small value

x²/ 0.31 = 5.75×10^-10

x² = 1.78× 10^-10

x = 1.33×10^-5

[OH^- ] = 1.33 ×10^-5 M

pOH = -log(1.33×10^-5)

pOH = 4.88