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Afew drops of acetic acid are added to a large excess of water at $\pu{25 ^\circ C}$. In the limit of infinite dilution, what percentage of acid remains unionised? The $K_\mathrm{a}$ of acetic acid is $1.85\times10^{-5}$.

The answer is supposed to be 0.54%, but I'm not sure how to reach this value. I know that given the pH we can use the definition of $K_\mathrm{a}$ to calculate the ratio of $\ce{CH3COOH}$ to $\ce{CH3COO-}$, but how do I find the pH?

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As the dilution becomes infinite for the amount of acetic acid, then the acetic acid's contribution to the solution's acidity falls, and the pH of the water is controlled by the acid dissociation of water. So the pH falls to 7.00. Thus,

$$\begin{align} \frac{(10^{-7})[\ce{CH3COO-}]}{[\ce{CH3COOH}]} &= 1.85 \times 10^{-5} \\ [\ce{CH3COO-}] &= 185 [\ce{CH3COOH}] \end{align}$$

The fraction of unionised acid, $x$, is therefore given by

$$\begin{align} x &= \frac{[\ce{CH3COOH}]}{[\ce{CH3COOH}] + [\ce{CH3COO-}]} \\ &= \frac{1}{1+185} \\ &= 0.54\%. \end{align}$$

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