Question

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

• A. 4 mL
• B. 2.5 mL
• C. 2 mL
• D. 6.4 mL

Answer 1

The correct option is C 2 mL

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

$pH=p{K}_a+log\frac{\left[KCN\right]}{\left[HCN\right]}$ .............. (i)

$p{K}_a=-log{K}_a=-log(5\times {10}^{-10})=9.30$

Let the volume of 5 M KCN solution required to be V mL.

KCN is added to 10 mL of 2 M HCN solution.

$\therefore \left[KCN\right]=\frac{5M\times VmL}{VmL+10mL}$ and

$\left[HCN\right]=\frac{10mL\times 2M}{VmL+10mL}$

Now from eqn.(i) ,

$9=9.30+log\left[\frac{\frac{5\times V}{V+10}}{\frac{10\times 2}{V+10}}\right]$

$9=9.30+log\frac{V}{4}$

$log\frac{V}{4}=-0.30$

$\frac{V}{4}=0.5$

$V=2mL$

Hence, the correct option is $\text{C}$