Calculate the pH of a $0.040 M$ aqueous solution of triethylamine at 25°C, given $K_{b} = 5.6 \times 1 0^{- 4}$ .

medium 2 marks Class 12 · Chemistry · Amines

Answer

pH ≈ 11.68

The correct answer is pH ≈ 11.68 — Use $p H = 14 - pO H = 11.675$ , round to $11.68$ .

Step-by-step solution

Write the base dissociation equilibrium: $C_{6} H_{15} N + H_{2} O ⇌ C_{6} H_{15} N H^{+} + O H^{-}$
Assume $x l l 0.040$ ; then $K_{b} = \frac{x ^{2}}{0.040}$
Solve for $x$ : $x = K_{b} \times 0.040 = 5.6 \times 1 0^{- 4} \times 0.040 = 4.73 \times 1 0^{- 3} M$
$[O H^{-}] = 4.73 \times 1 0^{- 3} M$
Compute $pO H = - l o g [O H^{-}] = 2.325$
Use $p H = 14 - pO H = 11.675$ , round to $11.68$ .

Common mistake: Using

[O H^{-}]

instead of

[O H^{-}]^{2}

in the expression for

K_{b}

Concept tested: pH of weak base solution