Question

(a) The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3 M, what is its pH?

(b) Calculate the pH of the solution in which 0.2 M NH₄Cl and 0.1 M NH₃ are present. The pK_b of ammonia solution is 4.75.

Answer 1

(a) [H⁺] = 3.8 x 10^-3 M

pH = -log₁₀ [H⁺]

= −log [13.8 × 10⁻³]

= -[log 3.8 x log 10^-3]

= -[(0.58)+(-3 log 10)]

= -[(0.58) +(-30)]

= -[-2.42]

= 2.42

(b) pK_b of ammonia = 4.75

For the reaction,

NH₃ + H₂O ⇌ NH₄⁺ + OH^-

The ionization constant of NH₃

k_b = antilog (−pk_b)

\(\therefore\) k_b = 10^-4.75

= 1.77 x 10^-5 M

NH₃ + H₂O ⇌ NH₄⁺ + OH^-

Initial	0.10	0.20	0
Conc. (M)
Change to	-x	+x	+x
reach
Equilibrium (M)
At equilibrium (M)	0.10 – x	0.20 + x	+x

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

1.77 × 10⁻⁵ = \(\frac{(0.20+X)(X)}{(0.1-X)}\)

As K_b is small, we can neglect in comparison to 0.1 M and 0.2 M. So,

1.77 x 10⁻⁵ = \(\frac{0.20X}{0.1}\)

or X = \(\frac{1.77\times10^{-5}}{2}\)

= 0.88 x 10^-5

^{\(\therefore\)} [OH^-] = x = 0.88 x 10^-5

pOH = -log₁₀[OH^-]

= -log0.88 x 10^-5

= -log 0.88 + 5 log 10

= -(-0.055)+5

= 0.055

pH + pOH = 14

pH + 5.05 = 14

pH = 14 - 5.05 = 8.95