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Q.

pOH of 0.1 molar aqueous NaCN solution found to be 2, then calculate value of dissociation constant of HCN in its aqueous solution.

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a

10−11

b

1.11×10−3

c

9×10−12

d

10−2

answer is D.

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Detailed Solution

CN−+H2O⇌HCN+OH−

0.1−x                x          x

10−14Ka=x20.1−x

x=10−2

Ka=9×10−12

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