Question

Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5.

Part A

What is the pH of a 0.270 M ammonia solution?

Part B

What is the percent ionization of ammonia at this concentration?

Answer 1

Answer –

In this problem we are given, Kb value for NH₃ = 1.8*10^-5

[NH₃] = 0.270 M

We know ammonia is weak acid, so we need to put ICE chart

      NH₃ + H₂O -------> NH₄⁺ + OH^-

I    0.270                        0           0

C      -x                           +x         +x

E 0.270-x                      +x        +x

We know the Kb expression for this reaction

Kb = [NH₄⁺] [OH^-] / [NH₃]

1.8*10^-5 = x*x / (0.270-x)

1.8*10^-5*(0.270-x) =x²

We can neglect x in the 0.270-x, because 5 % rule

x² = 4.86*10^-6

x = 0.00221 M

so, x = [OH^-] = 0.00221 M

so, pOH = -log [OH^-]

              = -log 0.00221

              = 2.66

Now we know,

pH + pOH = 14

so, pH = 14- pOH

     = 14-2.66

     = 11.3

The pH of a 0.270 M ammonia solution is 11.3