At `25^(@)C` the dissociation constant of HCN is `4.9 xx 10^(-10) M.` Calculate the degree of dissociation of HCN if the concentration is 0.1 M.

Options

A. `7 xx 10^(-5)`
B. `5 xx10^(-5)`
C. `6 xx 10^(-5)`
D. `8 xx 10^(-5)`

Answer

A

Explanation

The reaction can be represented
`{:(HCN +,H_(2)O,hArr,H_(3)O+,CN^(-),),(C,0,,0,,),(C(1-alpha),,,alpha C,,alpha C):}`
Therefore , `K_(C) = (Calpha^(2))/(1-alpha) , 1-alpha = 1, alpha = sqrt((K_(a))/(c )), alpha = sqrt((4.9 xx 10^(-10))/(0.1)) = 7 xx 10^(-5)`