Question

$K_p$ for $HCN$ is $5\times {10}^{-12}$. 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. $6mL$
• C. $2mL$
• D. $10mL$

Answer 1

The correct option is C $2mL$

Given

Ka of HCN is 2*10-12

Maintain pH 9

Molarity of KCN is 5

Molarity of HCN is 2

Volume of HCN is 10 ml

Solution

pKa= -log(Ka)

pKa= -log(2*10-12)

pKa=9.3

pH=pKa+$\frac{\left[KCN\right]}{\left[HCN\right]}$ eq (1)

Let V mL of 5M KCN is added

milli moles of KCN=5*V

Volume of KCN=V+10

Conc. of KCN=$\frac{5\ast V}{V+10}$

Similarly, Volume of HCN=10ml

Molarity of HCN=2M

milli moles of HCN=2*10=20

Conc. of HCN =$\frac{20}{V+10}$

Putting values in eq (1)

$9=9.3+log\frac{\frac{5\ast V}{10+V}}{\frac{20}{10+V}}$

On solving

V=2ml

The correct option is C