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Question

$$K_{a}$$, for $$HCN$$ is $$5\times10^{-10}$$ at $$25^{0} C$$. For maintaining a constant $$pH$$ of $$9$$, the volume of $$5\ M\ KCN$$ solution required to be added to $$10\ mL$$ of $$2\ M\ HCN$$ 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 $$2\ mL$$
$$pH=pK_a+log\left(\cfrac{Salt}{Acid}\right)$$
$$K_a=5\times 10^{-10}$$

$$pK_a=-log(5\times 10^{-10})=10-log5$$

$$pH=9$$

Let, $$\cfrac{Salt}{Acid}=K$$

$$pH=pK_a+logK$$ 

$$9=10-log5+logK$$

$$log\left( \cfrac { 5 }{ K }  \right) =1$$

$$K\cong 0.5$$

$$\cfrac { Salt }{ Acid } =0.5$$

Volume of $$5M, KCN$$ is required to be added to $$10ml$$ of $$2M$$ of $$HCN$$ is  $$V$$

$$Salt=5\times V$$
$$Acid=10\times2$$

$$V=2ml$$

Volume of $$5M, KCN=2ml$$

Chemistry

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