Question

What is the pH of a 0.20 M solution of Ba(ClO2)2? For HClO2, Ka=1.1x10E-2​

Answer 1

we have below equation to be used:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/1.1*10^-2

Kb = 9.091*10^-13

[ClO2-] = 2*[Ba(ClO2)2] = 2*0.20 M = 0.40 M

ClO2- dissociates as

ClO2- + H2O -----> HClO2 + OH-

0.4 0 0

0.4-x x x

Kb = [HClO2][OH-]/[ClO2-]

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((9.091*10^-13)*0.4) = 6.03*10^-7

since c is much greater than x, our assumption is correct

so, x = 6.03*10^-7 M

we have below equation to be used:

pOH = -log [OH-]

= -log (6.03*10^-7)

= 6.22

we have below equation to be used:

PH = 14 - pOH

= 14 - 6.22

= 7.78

Answer: 7.78