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Question

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


A
4 mL
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B
7.95 mL
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C
2 mL
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D
9.3 mL
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Solution

The correct option is B $$2$$ mL
$$HCN$$ and $$KCN$$ will form Buffer solution,

$$pH =\displaystyle  pK_a + log\frac {[salt]}{[acid]}$$

$$pH=-\log { (5)\times { 10 }^{ -10 } } +\log { \left[ { \cfrac { 5\times V }{ V+10 }  }/{ \cfrac { 10\times 2 }{ V+10 }  } \right]  } $$

or $$9=-\log {( 5)\times { 10 }^{ -10 } } +\log\left( { \cfrac { V }{ 4 }  }\right) $$

$$\therefore$$ $$V=2$$ mL

Chemistry

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