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Calculate the concentration of fluroacetic acid when $\mathrm{[H^+]= 2\times 10^{-3}\ M}$ .

$\text{Given, Ka of acid}= 2.8\times 10^{-3}$
Option: 1
$\mathrm{2\times10^{-3} \ M}$

Option: 2
$\mathrm{2.8 \times10^{-3}\ M}$

Option: 3
$\mathrm{3.4 \times10^{-3}\ M}$

Option: 4
$\mathrm{4.8 \times10^{-5}\ M}$

Answers (1)

best_answer

Given:

$\mathrm{[H^+]= 2\times 10^{-3}\ M}$

$\text{Ka of acid}= 2.8\times 10^{-3}$

For Fluoroacetic acid:

$\mathrm{CH _{2} FCOOH \rightleftharpoons H ^{+}+ CH _{2} FCOO ^{-}}$

$\text{c}$ $0$ $0$

$\mathrm{c (1-\alpha)}$ $\mathrm{c \alpha}$ $\mathrm{c \alpha}$

From the above reaction it is clear that

$\mathrm{ [H^+] = c\alpha = 2\times10^{-3}}$

Now,

$\mathrm{K_{a}=\frac{c \alpha \cdot c \alpha}{c(1-\alpha)}=\frac{c \alpha^{2}}{(1-\alpha)}}$

$2.8\times 10^{-3}=\frac{2\times 10^{-3} \alpha}{(1-\alpha)}\ \ \ \longrightarrow (i)$

Here we can see $\alpha$ is not small and hence it cannot be neglected in comparison to 1.

Solving equaiton (i) we get,

$\alpha = 0.583$

Now, it is given that

$\mathrm{[H^+] = c \alpha = 2 \times 10^{-3}}$

$\Rightarrow c = \frac{2 \times 10^{-3}}{0.583}$

$\mathrm{\Rightarrow c = 3.43\times 10^{-3} \ M}$

Hence, the correct answer is Option (3)

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manish

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