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$K_{a}$ for $H C N$ is $5 \times 10^{- 10}$ at $25^{0} C$ For maintaining a constant $p H$ of 9, the volume of 5M $K C N$ solution required to be added to 10mL of 2M $H C N$ solution is: $(\log 5 = 0.7)$

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2 mL

7.95 mL

4 mL

9.3 mL

answer is C.

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Detailed Solution

9 = 9.3 + \log [\frac{v \times 5}{20}] \Rightarrow 10^{- 0.3} = \frac{v}{4} \Rightarrow V = 4 \times 0.5 = 2

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