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
B
7.95 mL
C
2 mL
D
9.3 mL

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 $$9=-\log {( 5)\times { 10 }^{ -10 } } +\log\left( { \cfrac { V }{ 4 } }\right)$$$$\therefore$$ $$V=2$$ mLChemistry

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