Question

If the Kb of a weak base is 3.2 × 10^-6, what is the pH of a 0.50 M solution of this base?

Answer 1

Assume base as B

Show ICE and its reaction with water.

            B(aq) + H₂O (l)   ---- > BH⁺ (aq) + OH^- (aq)

I           0.50                                         0                      0

C         -x                                             +x                    +x

E          (0.50-x)                                   x                     x

Equilibrium constant expression.

Kb = [BH⁺ ] [OH^-] / [B]

Plug in the equilibrium concentration values and kb values in equilibrium expression .

3.2E-6 = x²/(0.50-x)

Since the value of Kb is very small we can neglect the x from the denominator.

Therefore.,

3.2E-6 = x²/0.50

Calculate for x

x = 0.0012649 M

we know from equilibrium concentration of OH-

x = [OH^-] = 0.0012649 M

Calculate pOH

pOH = - log [OH^-]

            = - log 0.0012649

            = 2.8979

Calculate pH

pH = 14 – pOH         

            = 14 – 2.8979

            = 11.1

pH of this solution would be 11.1