Question

The Kb for A- is 4.9x10^-10. What is the pOH of a 0.0727 M aqueous solution at 25 degrees C.

Answer 1

Lets write the dissociation equation of A-

A- +H2O -----> HA + OH-

7.27*10^-2 0 0

7.27*10^-2-x x x

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

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 ((4.9*10^-10)*7.27*10^-2) = 5.969*10^-6

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

so, x = 5.969*10^-6 M

So, [OH-] = x = 5.969*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (5.969*10^-6)

= 5.22

Answer: 5.22