Ionic EquilibriumHard

Question

Which of the following result(s) is/ are correct for the equilibrium state in a solution originally having 0.1 M-CH3COOH and 0.1 M-HCl? The value of Ka for CH3COOH = 1.8 × 10−5.

Options

A.[H+] = 0.1 M
B.[CH3COO−] = 1.8 × 10−5 M
C.Degree of dissociation of acetic acid = 1.8 × 10−4
D.[H+] from water = 10−13 M

Solution

$\underset{(0.1 - x)\text{ M} \simeq \text{0.1 M}}{CH_{3}COOH} \rightleftharpoons \underset{x\text{ M}}{CH_{3}COO^{-}} + \underset{(0.1 + x)\text{ M = 0.1 M}}{H^{+}}$

Now, $1.8 \times 10^{- 5} = \frac{x \times 0.1}{0.1} \Rightarrow x = 1.8 \times 10^{- 5}$

and $\alpha = \frac{x}{0.1} = 1.8 \times 10^{- 4}$

Now, $\left\lbrack H^{+} \right\rbrack_{\text{from water}} = \left\lbrack OH^{-} \right\rbrack = \frac{Kw}{\left\lbrack H^{+} \right\rbrack_{\text{acid}}} = 10^{- 13}\text{ M}$

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