Question

$1M$ solution of $C{H}_{3}COOH$ is diluted to x-time so that pH of the solution is doubled. calculate x ? Given: $K_a=1.8\times {10}^{-5}$

• A. $4.55\times {10}^4$
• B. $5.55\times {10}^6$
• C. $5.55\times {10}^4$
• D. None of the above

Answer 1

The correct option is C $5.55\times {10}^4$

Given $\left[C{H}_{3}COOH\right]=1M,K_a=1.8\times {10}^{-5}$
We know that the relation between $pH$ and $p{K}_a$ is given by
$pH=\frac{1}{2}(p{K}_a-logC)$
$pH=\frac{1}{2}(p{K}_a-log(\left[C{H}_{3}COOH\right])$
$pH=\frac{1}{2}(p{K}_a-log1)$
$pH=\frac{1}{2}p{K}_a$
Also given,
$p{H}^{{}^{\prime }}=2\times pH=2\times \frac{1}{2}p{K}_a=p{K}_a$
$p{H}^{{}^{\prime }}=p{K}_a$
$p{H}^{{}^{\prime }}=p{K}_a=\frac{1}{2}(p{K}_a-log{C}^{{}^{\prime }})$
$p{K}_a=-log{C}^{{}^{\prime }}=-log{K}_a$
$-log{C}^{{}^{\prime }}=-log{K}_a$
$C^{{}^{\prime }}=K_a=1.8\times {10}^{-5}$
Dilution, $x=\frac{1}{C^{{}^{\prime }}}=5.55\times {10}^4$