Question

$K_a$ for HNC is $5\times {10}^{-10}$ at 25C for maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 ml of 2M HCN solution is _________.$\left(\log 2=0.3\right)$

• A. 4 ml
• B. 8 ml
• C. 2 ml
• D. 10 ml

Answer 1

The correct option is C 2 ml

Using the formula for pH

$pH=pKa+\log \frac{\left[salt\right]}{\left[acid\right]}$

here

$Ka=5\times {10}^{-10}$

so, $pKa=-\log (5\times {10}^{-10})=-(\log 5-\log 10)$

$pKa=-\log 5+10$ and $pH=9$

Let $K=\frac{\left[salt\right]}{\left[acid\right]}$

Now,

$pH=pKa+\log K$

$9=-\log 5+10+\log K$

$-1+\log 5=\log K$

$\log K=-1.6990$

$K=antilog(-1.6990)$

$K=0.5$

hence,

$\frac{\left[salt\right]}{\left[acid\right]}=0.5$

Means concentration of salt is 0.5 of concentration of acid

Now, volume of $5Mkcn$ required to be added to $10ml$ of $2M$ of $HCn$ is $V$

$\left[salt\right]=0.5\left[acid\right]$

$MV=0.5mv$

$V=\frac{0.5mv}{M}$

$V=\frac{0.5\times 2M\times 10ml}{5}$

$V=2ml$

Hence volume of 5M kcn is 2ml