Question

18. Calculate the ionization constants for each of the following solutes from the percentage ionization and the concentration of the solute.
(a) 0.050 M HClO, 8.4 x 10 -2 % ionized
(b) 0.010 M HNO2 , 19 % ionized

Answer 1

HCLO ---- > H++ ClO-

K = ionization constant= [H+] [ClO-]/[HClO]

molatiy of HClO initially =0.05, percent ionized = 8.4*10^-2% =(8.4/100)% =0.084% =0.05*0.084/100 =0.000042

[H+] =[ClO-] =0.000042

Percent of HClO remaining at equilibrium =(100-8.4/100)*0.05/100=0.0499M

K= 0.000042*0.000042/ 0.0499=3.5*10^-8

b) HNO2----> H+ + NO2-  

Ionization constant = K = [H+] [NO2-]/ [HNO2]

percentage HNO2 ionized = 0.01*0.19=0.0019

[H+] = [NO2-] =amount of HNO2 decomposed =0.0019

Concentration at equilibrium =0.01-0.0019=0.0081M

K = 0.0019*0.0019/0.0081=0.000446

Remaining HClO at equilibrium = 0.05