Question 7.60 The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Class XI Equilibrium Page 229
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HCNO + H2O <=> H3O+ + CNO-
given, pH = 2.34
we know, according to Arrhenius formula
pH = - log[H3O+]
2.34 = -log[H3O+]
log[H3O+] = -2.36 = 3(bar).66
[H3O+] = antilog (3bar.66) = 4.571 × 10^-3 M
hence,[CNO- ] = 4.571 × 10^-3 M
but we know, [H3O+] = Cα
where C is concentration and α is degree of Ionisation .
so, degree of Ionisation (α) = [H3O+]/C
= 4.571 × 10^-3/0.1 = 4.571 × 10^-2 M
now, Ionisation constant Ka = [H3O+][CNO-]/[HCNO]
Ka = 4.571 × 10^-2 × 4.571 × 10^-3/0.1
= 2.089 × 10^-4
given, pH = 2.34
we know, according to Arrhenius formula
pH = - log[H3O+]
2.34 = -log[H3O+]
log[H3O+] = -2.36 = 3(bar).66
[H3O+] = antilog (3bar.66) = 4.571 × 10^-3 M
hence,[CNO- ] = 4.571 × 10^-3 M
but we know, [H3O+] = Cα
where C is concentration and α is degree of Ionisation .
so, degree of Ionisation (α) = [H3O+]/C
= 4.571 × 10^-3/0.1 = 4.571 × 10^-2 M
now, Ionisation constant Ka = [H3O+][CNO-]/[HCNO]
Ka = 4.571 × 10^-2 × 4.571 × 10^-3/0.1
= 2.089 × 10^-4
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