Question

What is the pH of a 0.05 M sodium butyrate solution? Kb = 6.9 x 10-10

Answer 1

let butyrate anion be written as A-

A- dissociates as:

A- +H2O -----> AH + OH-

5*10^-2 0 0

5*10^-2-x x x

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

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((6.9*10^-10)*5*10^-2) = 5.874*10^-6

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

so, x = 5.874*10^-6 M

so.[OH-] = x = 5.874*10^-6 M

use:

pOH = -log [OH-]

= -log (5.874*10^-6)

= 5.23

use:

PH = 14 - pOH

= 14 - 5.23

= 8.77

Answer: 8.77