Question

14. Calculate the ionization constant for each of the following acids or bases from the ionization constant of its conjugate base or conjugate acid:

(a) HTe⁻ (as a base)

(b) (CH₃)₃NH⁺

(c) HAsO₄³⁻ (as a base)

(d) C₆H₅NH₃⁺

(e) HSO₃⁻(as a base)

Answer 1

14. Calculation of ionization constants for.

(a) HTe- (as a base)

Ka for H2Te = 2.512 x 10^-3

Kb for HTe- = 1 x 10^-14/2.512 x 10^-3 = 4.00 x 10^-12

(b) (CH3)3NH+

Kb for (CH3)3N = 1.02 x 10^-3

Ka of (CH3)3NH+ = 1 x 10^-14/1.02 x 10^-3 = 9.80 x 10^-12

(c) HAsO4^2- (as a base)

Ka for HAsO4^2- = 5.6 x 10^-8

Kb for HAsO4^2- = 1 x 10^-14/5.6 x 10^-8 = 1.786 x 10^-7

(d) C6H5NH3+ (as a base)

Kb for C6H5NH2 = 4.3 x 10^-10

Ka for C6H5NH3+ = 1 x 10^-14/4.3 x 10^-10 = 2.32 x 10^-5

(e) HSO3- (as a base)

Ka for H2SO3 = 1.2 x 10^-2

Kb for HSO3- = 1 x 10^-14/1.2 x 10^-2 = 8.33 x 10^-13