Question

A 0.198 M weak acid solution has a pH of 4.12. Find Ka for the acid.
Express your answer using two significant figures.

Find the [OH−] of a 0.44 M  aniline (C6H5NH2) solution. (The value of Kb for aniline (C6H5NH2) is 3.9×10−10.)
pH=9.12

Answer 1

#

[H⁺] = 10^-pH = 10^-4.12 = 7.586 10^-5

Weak acid dissociates as AH ⇌ A- + H+

K_a = [A^-][H⁺] / [AH]

but  [A^-] = [H⁺] = 7.586 10^-5

substitute value

K_a = (7.586 10^-5)(7.586 10^-5) / 0.198

Ka = 2.906 10^-8

#

Aniline dissociate as

C₆H₅NH₂ + H₂O C₆H₅NH₃⁺ + OH^-

K_b = [C₆H₅NH₃⁺] [OH^-] / [C₆H₅NH₂]

aniline is monobesic base therefore

[C₆H₅NH₃⁺] = [OH^-] = x

K_b = [x][x] /  [C₆H₅NH₂]

K_b =[x]² /  [C₆H₅NH₂]

[x]² = K_b  [C₆H₅NH₂]

[x]² = 3.9 10^-10 0.44 = 1.716 10^-10

[x] = 1.31 10^-5

[C₆H₅NH₃⁺] = [OH^-] = x = 1.31 10^-5

[OH^-] = 1.31 10^-5