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GREYUIT [131]
3 years ago
11

In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.

Chemistry
1 answer:
Mamont248 [21]3 years ago
5 0

Answer:

Approximately 1.30 \times 10^{-2}, assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let \rm HA denote this acid.

\rm HA \rightleftharpoons H^{+} + A^{-}.

Initial concentration of \rm HA without any dissociation:

[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}.

After 12.5\% of that was dissociated, the concentration of both \rm H^{+} and \rm A^{-} (conjugate base of this acid) would become:

12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}.

Concentration of \rm HA in the solution after dissociation:

(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}.

Let [{\rm HA}], [{\rm H}^{+}], and [{\rm A}^{-}] denote the concentration (in \rm mol \cdot L^{-1} or \rm M) of the corresponding species at equilibrium. Calculate the acid dissociation constant K_{\rm a} for \rm HA, under the assumption that this acid is monoprotic:

\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}.

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