Question

$K_a$, for $HCN$ is $5\times {10}^{-10}$ at ${25}^{0}C$. For maintaining a constant $pH$ of $9$, the volume of $5MKCN$ solution required to be added to $10mL$ of $2MHCN$ solution is:

• A. $4mL$
• B. $7.95mL$
• C. $2mL$
• D. $9.3mL$

Answer 1

Answer: (C)

$2mL$

$pH=p{K}_a+log\left(\frac{Salt}{Acid}\right)$

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

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

$pH=9$

Let, $\frac{Salt}{Acid}=K$

$pH=p{K}_a+logK$

$9=10-log5+logK$

$log\left(\frac{5}{K}\right)=1$

$K\cong 0.5$

$\frac{Salt}{Acid}=0.5$

Volume of $5M,KCN$ is required to be added to $10ml$ of $2M$ of $HCN$ is $V$

$Salt=5\times V$

$Acid=10\times 2$

$V=2ml$

Volume of $5M,KCN=2ml$