Question

Find the percentage of ionized substituted benzoic acid (substituted benzoate) which will be in the buffered aqueous solution with pH=5.8 if the pKa of the substituted benzoic acid is 4.7. Round your answer to the nearest integer, e.g. 1 %

Answer 1

pH = 5.8, H⁺ = 10^-pH = 10^-5.8 = 1.58 x 10^-6 = 0.00000158 M

pKa = 4.7, Ka = 10^-pKa = 10^-4.7 = 1.99 x 10^-5

C₆H₅COOH(aq) -------> C₆H₅COO^-(aq) + H⁺(aq)

   Initial    a 0    0

change -0.00000158    + 0.00000158    +0.00000158

   Equilibrium    a-0.00000158 0.00000158 0.00000158

Ka = 1.99 x 10^-5 = [ H⁺] [C₆H₅COO^-]/[C₆H₅COOH] = 0.00000158 x 0.00000158/(a-0.00000158)

0.0000199 (a-0.00000158) = 0.00000158²

0.0000199a - 0.0000000000314 = 0.000000000002496

0.0000199a = 2.496 x 10^-12 +3.14 x 10^-11 = 10^-11 (0.2496 + 3.14) = 10^-11 x 3.3896

a = 10^-11 x 3.3896 /1.99 x 10^-5 = 1.7 x 10^-6 M

Percent ionisation = [HA]ionised/[HA]initial) x 100 = 1.58 x 10^-6/1.7 x 10^-6 ) x 100 = 0.93 x 100 = 93 %