Question

A 70.0mL sample of .2M Iactic acid (HC3H5O3; Ka: 1.38*10^-4) is titrated with .2M of LiOH. Calculate the pH of the solution after the following volumes of LiOH are added to the solution. Assume that the small x approximation is valid in all cases for this question. A) Before any LiOH added (0.00mL LiOH added) B) 35mL LiOH added C) 70mL LiOH added D) 90mL LiOH added

Answer 1

A) Before any LiOH added (0.00mL LiOH added)

HA ---------------------------> H+ + A-

0.2                                       0        0 ----------> I

0.2-x                                    x          x -----------> E

Ka = [H+][A-]/[HA]

1.38 x 10^-4 = x^2 / 0.2 -x

x^2 + 1.38 x 10^-4 x - 2.76 x 10^-5 = 0

x = 5.18 x 10^-3

[H+] = x = 5.18 x 10^-3 M

pH = -log [H+]

pH = -log ( 5.18 x 10^-3 )

pH = 2.29

B) 35mL LiOH added

it half equivalence point . here pH = pKa

Ka = 1.38 x 10^-4

pKa = 3.86

pH = 3.86

C) 70mL LiOH added

it is equivalence point

here salt only formed

slat conncentration = 0.2 x 70 / (70 + 70) = 0.1 M

pH = 7 + 1/2 [pKa + log C]

pH = 7 + 1/2 [3.86 + log 0.1]

pH = 8.43

D) 90mL LiOH added

[OH-] = 90 x 0.2 - 70 x 0.2 / (90 +70 ) = 0.025

pOH = 1.60

pH + pOH = 14

pH = 12.40