Question

The pKa of phenylacetic acid is 4.32 at 25 oC. What is the pH of a 0.15 M aqueous solution of calcium phenylacetate, Ca(C7H7COO)2, at 25 oC?

Answer 1

[C7H7COO-] = 2*[Ca(C7H7COO)2]

= 2*0.15 M

= 0.30 M

use:

pKa = -log Ka

4.32 = -log Ka

Ka = 4.786*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)/4.786*10^-5

Kb = 2.089*10^-10

C7H7COO- dissociates as

C7H7COO- + H2O -----> C7H7COOH + OH-

0.3 0 0

0.3-x x x

Kb = [C7H7COOH][OH-]/[C7H7COO-]

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 ((2.089*10^-10)*0.3) = 7.917*10^-6

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

so, x = 7.917*10^-6 M

use:

pOH = -log [OH-]

= -log (7.917*10^-6)

= 5.1014

use:

PH = 14 - pOH

= 14 - 5.1014

= 8.8986

Answer: 8.90