Question

A solution contains $0.09MHCI$, $0.09M$ $CHC{l}_{2}COOH$, and 0$.1M$ $C{H}_{3}COOH$. The pH of this solution is $1$. K for $CHC{l}_{2}COOH$ is $1.25\times {10}^{-x}$. The value of $x$ is :
[Given : $K_a$ for $C{H}_{3}COOH={10}^{-5}$)

Answer 1

pH will be decided by $\left[H^{\oplus }\right]$ furnished by HCl and $CHC{l}_{2}COOH$ not by $C{H}_{3}COOH$ as it is weak acid.

$CHC{l}_{2}COOH\rightleftharpoons CHC{l}_{2}CO{O}^{⊝}+H^{\oplus }$
Initial cone 0.09 0 0.09 (from HCl)
Final cone (0.09-x) x (0.09+x)

$\therefore \left[H^{\oplus }\right]=0.09+x$

but, $pH=1,\therefore \left[H^{\oplus }\right]={10}^{-1}=0.1M$

$0.09+x=0.1,\therefore x=0.01$

$K_a$ for $CHC{l}_{2}COOH$ can be given as

$K_a=\frac{\left[H^{\oplus }\right]\left[CHC{l}_{2}CO{O}^{-}\right]}{\left[CHC{l}_{2}COOH\right]}=\frac{0.1\times 0.01}{(0.09-0.01)}=1.25\times {10}^{-2}$

So, value of $x$ is 2.