Question

A buffer solution is prepared by making a solution 0.1 M in NH3 and 0.2M in NH4Cl.

a) Write the equation for the ionization of NH3 in water.

b) Write the expression for the ionization constant, Kb, for NH3. The value of Kb at 25 celsius is 1.8E-5.

c) Calculate the [OH-], the [H+], and the pH for this buffer solution.

d) Calculate the [OH-], the [H+], and the pH for 0.1 M NH3.

Answer 1

Ans. 1. #a. Ionization of NH₃ in water:

NH₃(aq) + H₂O(l) ------------------> NH₄⁺(aq) + OH^-(aq)

#b. Base dissociation constant, Kb = [NH₄⁺] [OH^-] / [NH₃]

#c. For the buffer: Create an ICE table as shown in figure with following initial concentrations-

            Initial [NH3] = 0.10 M

            Initial [NH4⁺] = 0.20 M. [NH4Cl dissociates completely to yield NH₄⁺ in water].

Now,

            Base dissociation constant, Kb = [NH₄⁺] [OH^-] / [NH₃]

            Or, 1.8 x 10^-5 = (0.20 + X) (X) / (0.10 - X)

            Or, (1.8 x 10^-5) (0.10 - X) = 0.20X + X²

            Or. 0.0000018 - 0.000018X = 0.20X + X²

            Or, 0.20X + X² + 0.000018X - 0.0000018 = 0

            Or, X² + 0.199982X - 0.0000018 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 0.00000900                  ; X2 = - 0.19999

Since concentration can’t be negative, reject X2.

Therefore, X = 0.00000900 M = 9.00 x 10^-6 M

Therefore, [OH^-] = X = 9.00 x 10^-6 M

Putting the value of [OH^-] in the equation – “[H₃O⁺] [OH^-] = 10^-14”

            Or, [H₃O⁺] = 10^-14 / (9.00 x 10^-6) = 1.11 x 10^-9 M

And,

            pH of buffer = -log [H₃O⁺] = -log (1.11 x 10^-9) = 8.95

#d. For NH3 assuming the solution consists of NH3 alone.    

Create an ICE table as shown in figure with initial [NH₃] = 0.10 M but initial [NH₄⁺] = 0 M

Base dissociation constant, Kb = [NH₄⁺] [OH^-] / [NH₃]

            Or, 1.8 x 10^-5 = (X) (X) / (0.10 - X)

            Or, (1.8 x 10^-5) (0.10 - X) = X²

            Or. 0.0000018 - 0.000018X = X²

            Or, X² + 0.000018 X - 0.0000018 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 0.001333                      ; X2 = - 0.00135

Since concentration can’t be negative, reject X2.

Therefore, X = 0.001333 M = 1.33 x 10^-3 M

Therefore, [OH^-] = X = 1.33 x 10^-3 M

Putting the value of [OH^-] in the equation – “[H₃O⁺] [OH^-] = 10^-14”

            Or, [H₃O⁺] = 10^-14 / (1.33 x 10^-3) = 7.52 x 10^-12 M

And,

            pH = -log [H₃O⁺] = -log (7.52 x 10^-12) = 11.12