Methylamine (CH_3NH_2) is a weak base with K_b = 4.4 × 10^-4. Calculate the pH of a 0.10 M aqueous solution of methylamine at 25°C.

Methylamine ( $C H_{3} N H_{2}$ ) is a weak base with $K_{b} = 4.4 \times 1 0^{- 4}$ . Calculate the pH of a $0.10 M$ aqueous solution of methylamine at $2 5^{\circ} C$ .

easy 2 marks Class 12 · Chemistry · Amines

Answer

pH = 11.82

The correct answer is pH = 11.82 — Finally, $p H = 14 - pO H = 14 - 2.179 \approx 11.82$ .

Write the expression for the base dissociation: $K_{b} = \frac{[ C H _{3} N H _{3}^{+} ] [ O H ^{-} ]}{[ C H _{3} N H _{2} ]}$ .
Let the degree of dissociation be $x$ ; then $[O H^{-}] = x$ , $[C H_{3} N H_{3}^{+}] = x$ , and $[C H_{3} N H_{2}] \approx 0.10 - x \approx 0.10$ (since $x ≪ 0.10$ ).
Substitute into the $K_{b}$ expression: $4.4 \times 1 0^{- 4} = \frac{x ^{2}}{0.10}$ .
Solve for $x$ : $x = 4.4 \times 1 0^{- 4} \times 0.10 = 4.4 \times 1 0^{- 5} \approx 6.63 \times 1 0^{- 3} M$ .
The hydroxide ion concentration is $[O H^{-}] = 6.63 \times 1 0^{- 3} M$ , so $pO H = - lo g [O H^{-}] = - lo g (6.63 \times 1 0^{- 3}) \approx 2.179$ .
Finally, $p H = 14 - pO H = 14 - 2.179 \approx 11.82$ .

Common mistake: Using

[C H_{3} N H_{2}] = 0.10 - x

without the approximation

x ≪ 0.10

, leading to a complicated quadratic that is unnecessary.

Concept tested: Basicity of aliphatic amines (pH calculation)