Question

a. Enough of a monoprotic acid is dissolved in water to produce a 0.0163 M solution. The pH of the resulting solution is 2.51. Calculate the Ka for the acid.

Ka= ________

b. The Ka of a monoprotic weak acid is 6.61 × 10-3. What is the percent ionization of a 0.108 M solution of this acid?

Percent ionization= ___________

Answer 1

a) moprotic acid dissociate as

HA + H₂O H₃O⁺ + A^-

K_a  = [H₃O⁺] [ A^-] / [HA]

[H₃O⁺] = 10^-pH = 0.00309 M

[H₃O⁺] = [A^-] = 0.00309 M

substitute thsse value

K_a = (0.00309) (0.00309) / 0.0163

K_a = 0.000586 = 5.86 10^-4

b)

Weak acid dissociates as AH ⇌ A- + H+

K_a = [A^-][H⁺] / [AH]

but  [A^-] = [H⁺] = x

K_a = [x][x] / [AH]

Substitute the value in equation

6.6110^-3 = [x]²/ 0.108

[x]² = 6.6110^-30.108 = 0.00071388

[x] = 0.0267185

Concentration of H⁺ = 0.0267 M

% dissociation = [H⁺] 100 / [HA]

% dissociation = 0.0267 100 / 0.108 = 24.72 %