# Get Answers to all your Questions

#### 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 $\mathrm{4.8 \times10^{-5}\ M}$

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)