Question

Ka for HCN is 5 × 10–10 at 25⁰C. For maintaining a constant pH of 9, the volume of 5 M KCN solution required to be added to 10 ml of 2M HCN solution is (log 2 = 0.3)

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

Answer 1

Answer: (C)

2 ml

Answer 2

Ka = 5 × 10–10 pKa = 10 log 5 = 9.3

pH = pKb + log $\left\lbrack \frac{CN^{-}}{HCN} \right\rbrack$

9 = 9.3 + log $\left\lbrack \frac{5 \times V_{ml}}{10 \times 2} \right\rbrack$

$\Rightarrow$ – 0.3 = log $\left\lbrack \frac{V_{ml}}{4} \right\rbrack$

0.3 = log $\left\lbrack \frac{4}{V_{ml}} \right\rbrack$ 

$\Rightarrow$ $\frac{4}{V_{ml}}$ = 2

$\Rightarrow$ Vml = 2 ml