Question

20 ml of 0.2 M $HCN$ mix with 10 ml of 0.2 M $NaOH$, then calculate $pH$ of resulting mixture, $pKa$ value of $HCN$ is 5 :-

• A. $6$
• B. $7.5$
• C. $5$
• D. $11$

Answer 1

The correct option is C $5$

No. of mole $=$ concentration $\times$ volume

moles of $NaOH=\left(0.2H\right)\left(10ml\right)=2$ mollimoles

moles of $HCN=\left(0.2H\right)\left(20ml\right)=4$ millimoles

$HCN+NaOH⟶NaCN+H_{2}O$

limiting reagents is $NaOH$

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

$1$ mole $NaOH$ react with $HCN$ to give $1$ mole $NaCN$

$\Rightarrow$ $2$ millimoles of $NaOH$ gives $2$ millimoles $NaCN$

$\left[NaCN\right]=\frac{2\times {10}^{-3}}{30\times {10}^{-3}}=\frac{1}{15}$

$\Rightarrow$ acid also have the same concentration as the remaining moles of acid is too $0.002$

$\Rightarrow$ $pH=5+log\frac{\left[1/5\right]}{\left[1/5\right]}=5+0=5$

$\therefore$ $pH$ of resulting mixture $=5$