Question

What is the pKb of a weak base KA if the pH of a 0.077-M solution of KA is 9.31?

Answer 1

we have below equation to be used:

pH = -log [H+]

9.31 = -log [H+]

log [H+] = -9.31

[H+] = 10^(-9.31)

[H+] = 4.898*10^-10 M

we have below equation to be used:

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(4.898*10^-10)

[OH-] = 2.042*10^-5 M

Lets write the dissociation equation of A-

A- +H2O -----> KA + OH-

7.7*10^-2 0 0

7.7*10^-2-x x x

Kb = [KA][OH-]/[A-]

Kb = x*x/(c-x)

Kb = 2.042*10^-5*2.042*10^-5/(0.077-2.042*10^-5)

Kb = 5.415*10^-9

we have below equation to be used:

pKb = -log Kb

= -log (5.415*10^-9)

= 8.27

Answer: 8.27