Question

What is the pH of a 0.15 M solution of the weak base aniline (C6H5NH2; Kb = 3.8 x 10-10)

A) 5.12 B) 8.88 C) 13.18 D) 9.42 E) 10.24

Answer 1

a) C6H5NH2(l) + H2O(l) <---> C6H5NH3+(aq) + OH-(aq)

This equation shows that aniline is a base, because it attaches H+ ions of water and increase OH- concentration in the solution.

Kb = [C6H5NH3+][OH-] / [C6H5NH2] = 3.8×10^-10

Ionization equation;
C6H5NH2(l) + H2O(l) <---> C6H5NH3+(aq) + OH-(aq)
0.15 - x M ................................. + x M .............. + x M

Kb = (x) ( x) / (0.15 - x) = 3.8×10^-10

Assumption: x is very small compared to 0.15, then x can be neglected.

(x) ( x) / (0.15) = 3.8×10^-10
x^2 = 5.7×10^-11
x = 7.55×10^-6 = [OH-]

b) pOH = - log [OH-] = - log (7.55×10^-6) = 5.12

pH + pOH = 14.00
pH = 14.00 - pOH = 14.00 - 5.12 = 8.88

Answer is B