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Question: The weak acid, HA, has a K<sub>a</sub> of 1.00 × 10<sup>–5</sup>. If 0.1 mol of this acid is dissolv...

The weak acid, HA, has a Ka of 1.00 × 10–5. If 0.1 mol of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to –

A

1%

B

99.9%

C

0.1%

D

99%

Answer

1%

Explanation

Solution

HA ⇌\rightleftharpoons H+ + A–-

t = 0 C 0 0

teq. C – Ca a a

Ka = [H+][A−][HA]=C2α2C−Cα\frac{\lbrack H^{+}\rbrack\lbrack A^{-}\rbrack}{\lbrack HA\rbrack} = \frac{C^{2}\alpha^{2}}{C - C\alpha}

= Cα21−α≈Cα2\frac{C\alpha^{2}}{1 - \alpha} \approx C\alpha^{2}

= KaC=10−50.1\sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{10^{- 5}}{0.1}} = 10–2