Question

$K_b$ for $HCN$ is $5\times {10}^{-10}$ at ${25}^{o}C$ For maintaining a constant $pH$ of $9$, the volume of $5M$ $KCN$ solution required to be added to $10ml$ of $2M$ $HCN$ solution is : ($\log 2=0.3$)

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

Answer 1

The correct option is C $2ml$

By using the formula for pH:

$pH=p{k}_a+log\frac{\left[salt\right]}{\left[acid\right]}$

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

$p{K}_{a=}-log(5\times 10^{-10})=(-log5-log10)$

$p{K}_a=-log5+log10andpH=9$

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

now, $pH=p{K}_a+logK$

$9=-log5+10+logK$

$-1+log5=logK$

$logK=\stackrel{-}{1}.06990$

$K=antilog\left(\stackrel{-}{1}.06990\right)$

$K\approx 0.5$

Hence, $\frac{\left[salt\right]}{\left[acid\right]}$= 5

Now, volume of 5 molar ,$KCN$ required to be added to 10ml of 2 molar of HCN is V

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

$MV=0.5mv$

$V=0.5mv/M$

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

Hence, the volume of 5M, $KCN$ =2ml