Question

What is the pH of the solution which results from mixing 35 mL of 0.50 M NH3(aq) and 35 mL of 0.50 HCI(aq) at 25 degree C? (Kb for NH3 = 1.8*10^-5). Please show all work. The answer is pH=4.32

Answer 1

The molarities of NH₃ and HCl are the same, and same volume of each substance is added, so it is equivalence point. This means that the solution only contains the conjugate acid, NH₄⁺

At equilibrium:

NH₄⁺ <--> NH₃ + H⁺

Calculate Ka for NH₄⁺

At 25 degree C kw = 1.0 x 10^-14

Kb = Kw/Ka

Ka = Kw/Kb

Ka =1.0* x 10^-14/1.8*10^-5

      = 5.56 x10^-10

[NH₄⁺] = (Vol. of NH₄⁺) (Molarity of NH₄⁺) / Total Vol. of Solution

[NH₄⁺] = ( (35ml)(0.5M) ) / 70ml

          = 0.25M

Set up an equilibrium expression:

NH₄⁺ <--> NH₃ + H⁺

I 0.25        0         0

C -x        +x          +x

E 0.25-x x          x

Ka = [NH₃] + [H⁺]/[NH₄⁺]

5.6 x10^-10 = x*x/0.25-x

5.6 * 10^-10 = x^2/(0.25 –x)

5.6 * 10^-10 = - x^2/(x –0.25)

x = 1.18 x 10^-5

[H⁺] = 1.18 x 10^-5

pH = - log [H⁺ ]

    = - log (1.18 x 10^-5)

pH = 4.92

The answer is pH = 4.32 but I calculated 4.92