Question

What is the pH of a 0.02M solution of potassium benzoate solution ( K^{+ -}OOCC₆H₅). The pka of benzoic acid is 4.19. (Write answer to the hundredths place)

please help me

Answer 1

use:
pKa = -log Ka
4.19 = -log Ka
Ka = 6.457*10^-5

use:
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)/6.457*10^-5
Kb = 1.549*10^-10
C6H5COO- dissociates as

C6H5COO-        + H2O   ----->     C6H5COOH +   OH-
0.02                        0         0
0.02-x                      x         x


Kb = [C6H5COOH][OH-]/[C6H5COO-]
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 ((1.549*10^-10)*2*10^-2) = 1.76*10^-6

since c is much greater than x, our assumption is correct
so, x = 1.76*10^-6 M



use:
pOH = -log [OH-]
= -log (1.76*10^-6)
= 5.7545


use:
PH = 14 - pOH
= 14 - 5.7545
= 8.2455
Answer: 8.25