Calculate the pH of a $0.015 M$ aqueous solution of diethylamine $((C H_{3} C H_{2})_{2} NH)$ at $2 5^{\circ} C$ given $K_{b} = 5.6 \times 1 0^{- 4}$ .

medium 2 marks Class 12 · Chemistry · Amines

Answer

pH = 11.20

The correct answer is pH = 11.20 — Finally, $p H = 14 - pO H = 14 - 2.84 = 11.20$ .

Step-by-step solution

Write the base dissociation equilibrium: $((C H_{3} C H_{2})_{2} NH) + H_{2} O ⇌ ((C H_{3} C H_{2})_{2} N H_{2}^{+}) + O H^{-}$ .
Set up the expression $K_{b} = \frac{[ ( C H _{3} C H _{2} ) _{2} N H _{2}^{+} ] [ O H ^{-} ]}{[ ( C H _{3} C H _{2} ) _{2} NH ]}$ .
Assume $[(C H_{3} C H_{2})_{2} N H_{2}^{+}] = [O H^{-}] = x$ and $[(C H_{3} C H_{2})_{2} NH] \approx 0.015 - x \approx 0.015$ (since $K_{b}$ is small).
Solve $5.6 \times 1 0^{- 4} = \frac{x ^{2}}{0.015}$ , giving $x = 5.6 \times 1 0^{- 4} \times 0.015 = 1.45 \times 1 0^{- 3} M$ .
Calculate $pO H = - lo g [O H^{-}] = - lo g (1.45 \times 1 0^{- 3}) = 2.84$ .
Finally, $p H = 14 - pO H = 14 - 2.84 = 11.20$ .

Common mistake: Using the approximation

x ≪ 0.015

incorrectly and obtaining a much larger

[O H^{-}]

value

Concept tested: Basicity of aliphatic amines