Question

Determine the degree of ionization and pH of a 0.05 M of ammonia solution. The ionization constant of ammonia is 1.77 x 10^-5 . Also calculate the ionization constant of conjugate acid of ammonia.

Answer 1

Given, K_b = 1.77 × 10⁻⁵

C = 0.05 M

The ionization of NH₃ in water is represented by the equation:

NH₃ + H₂O ⇌ NH₄⁺ + OH^-

Initial	C	0	0
Conc.
Equilibrium	C(1 - α)	Cα	Cα
Conc.

Where α = Degree of ionization

C = concentration of solution

Ionization constant of ammonia

K_b = \(\frac{[NH_4^+][OH^-]}{[NH_3]}\)

= \(\frac{Ca.Ca}{C(1-a)}\)

= \(\frac{Ca^2}{(1-a)}\)

The value of α is small, so, the equation can be simplified by neglecting α in comparison to 1 in the denominator on right hand side of the equation.

\(\therefore\) k_b = Cα²

or α = \(\sqrt{\frac{K_b}{C}}\)

= \(\sqrt{\frac{1.77\times10^{-5}}{0.05}}\)

= 0.018

[OH^-] = Cα

0.05 × 0.018

= 9.4 × 10⁻⁴ M

\(\therefore\) [H⁺][OH^-] = K_w

or  [H⁺] = \(\frac{K_w}{[OH^-]}\) = \(\frac{10^{-14}}{9.4\times10^{-4}}\)

1.06 × 10⁻¹¹

Now, pH = −log₁₀ [H⁺]

= -log(1.06 x 10^-11)

= -log1.06 + 11 log10

= - 0.0253 + 11

= 10.9747

Now, using the relation for conjugate acid – base pair,

K_a x K_b = K_w

K_a = \(\frac{K_w}{K_b}\)

= \(\frac{10^{-4}}{1.77\times10^{-5}}\)

= 5.64 x 10^-10