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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)



\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}}


 \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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