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?
[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
The pKa of phenylacetic acid is 4.32 at 25 oC. What is the pH of a...
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