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3 years ago

What is the percent ionization of a 1.8 M HC2H3O2 solution (Ka = 1.8 10-5 ) at 25°C?

Chemistry

1 answer:

xz_007 [3.2K]3 years ago

8 0

Answer:

B) 0.32 %

Explanation:

Given that:

$K_{a}=1.8\times 10^{-5}$

Concentration = 1.8 M

Considering the ICE table for the dissociation of acid as:-

$\begin{matrix}&CH_3COOH&\rightleftharpoons &CH_3COOH&+&H^+\\ At\ time, t = 0 &1.8&&0&&0\\At\ time, t=t_{eq}&-x&&+x&&+x\\ ----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:-&1.8-x&&x&&x\end{matrix}$

The expression for dissociation constant of acid is:

$K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}$

$1.8\times 10^{-5}=\frac{x^2}{1.8-x}$

$1.8\left(1.8-x\right)=100000x^2$

Solving for x, we get:

x = 0.00568 M

Percentage ionization = $\frac{0.00568}{1.8}\times 100=0.32 \%$

Option B is correct.

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3 years ago

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<h3>Explanation</h3>

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