Given that for acetic acid () is at , determine the equilibrium constant () for its neutralization r — Ionic Equilibrium Chemistry Question
Question
Given that $K_a$ for acetic acid ($CH_3COOH$) is $2 \times 10^{-5}$ at $25^\circ C$, determine the equilibrium constant ($K_{eq}$) for its neutralization reaction with aqueous $NaOH$. If $K_{eq} = x \times 10^9$, find the value of $x$.
Answer: 2
💡 Solution & Explanation
The neutralization reaction is $CH_3COOH (aq) + OH^- (aq) \rightleftharpoons CH_3COO^- (aq) + H_2O (l)$. The equilibrium constant for this reaction is $K_{eq} = \frac{[CH_3COO^-]}{[CH_3COOH][OH^-]}$. Multiplying numerator and denominator by $[H^+]$ yields $K_{eq} = \frac{[CH_3COO^-][H^+]}{[CH_3COOH]} \times \frac{1}{[H^+][OH^-]} = \frac{K_a}{K_w}$. Given $K_w = 10^{-14}$ at $25^\circ C$, $K_{eq} = \frac{2 \times 10^{-5}}{10^{-14}} = 2 \times 10^9$. Thus, $x = 2$.
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