degree of hydrolysis of N BY 100 solution of KCN is (given ka = 1.4 multiplied by 10 power -9)
The question is missing the concentration unit for KCN. The question is solved using the 100 mM (0.1M).
Ka values are acid dissociation constant while KCN is not an acid. The question is solved assuming Ka is the hydrolysis dissociation constant for KCN.
The dissociation equation is KCN K+ + CN-
Assume α to the degree of dissociation due to hydrolysis
The concentration of the 3 species at various times is:
KCN K+ CN-
Initial 0.1 0 0
Change 0.1 α 0.1 α 0.1 α
Equilibrium 0.1-0.1 α 0.1 α 0.1 α
As dissociation constant is low, we can assume that 0.1 α <<< 0.1 and hence 0.1-0.1 α = 0.1
Kd =
Substituting values,
1.4 10-9 =
1.4 10-9 = 0.1α2
α2 = 1.4 10-8
α = 1.18 10-4
Ka values are acid dissociation constant while KCN is not an acid. The question is solved assuming Ka is the hydrolysis dissociation constant for KCN.
The dissociation equation is KCN K+ + CN-
Assume α to the degree of dissociation due to hydrolysis
The concentration of the 3 species at various times is:
KCN K+ CN-
Initial 0.1 0 0
Change 0.1 α 0.1 α 0.1 α
Equilibrium 0.1-0.1 α 0.1 α 0.1 α
As dissociation constant is low, we can assume that 0.1 α <<< 0.1 and hence 0.1-0.1 α = 0.1
Kd =
Substituting values,
1.4 10-9 =
1.4 10-9 = 0.1α2
α2 = 1.4 10-8
α = 1.18 10-4