The weak acid, $H A$ , has a $K_{a}$ of $1.00 \times 10^{- 5}$ . If $0.1 mol$ of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to:

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Solution

$\begin{matrix} H A & ⇌ & H^{+} & + & A^{-} t = 0 : & C & 0 & 0 t = t_{e q} : & C - C α & C α & C α \end{matrix}$
Concentration $(C) = 0.1 M$
$K_{a} = \frac{[H^{+}] [A^{-}]}{[H A]} = \frac{C^{2} α^{2}}{C - C α} \Rightarrow \frac{C α^{2}}{1 - α} = C α^{2} (∵ (1 - α) \approx 1) α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{10^{- 5}}{0.1}} = 10^{- 2}$
Percentage of ionisation $= α \times 100 = 1 %$

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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:

HA ⇌ H++A−t=0:C00t=teq:C−CαCαCα Concentration (C) = 0.1 M Ka=[H+][A−][HA]=C2α2C−Cα⇒Cα21−α=Cα2 (∵ (1−α)≈1)α=√KaC=√10−50.1=10−2 Percentage of ionisation =α×100=1%

The weak acid, $H A$ , has a $K_{a}$ of $1.00 \times 10^{- 5}$ . If $0.1 mol$ of this acid is dissolved in one litre of water, the percentage of acid dissociated at equilibrium is closest to: