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
B
7.95 mL
C
2 mL
D
9.3 mL

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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