Question

Consider the titration of 171 mL of 0.961 M lactic acid, CH3CH9OH)COOH, Ka = 1.38E-4, with 1.203 M NaOH for the next three problems.

Problem #1

What is the pH of the solution when 83.33 mL of the 1.203 M NaOH are added to the 0.961 M lactic acid solution?

Problem #2

What is the pH of the solution at the equivalence point?

Problem #3

What is the pH of the solution when 178.9 mL of the 1.203 M NaOH has been added to the 0.961 M Lactic Acid solution?

Answer 1

171 mL of 0.961 M lactic acid = (171 x 0.96M)/1000 = 0.164 moles NaOH

Equation

NaOH + C3H6O3 → NaC3H5O3 + H2O

83.33 mL of the 1.203 M NaOH = (83.33 x 1.203M)/1000 = 0.100 moles of NaOH

So the amount of lactic acid remaining unreacted is 0.164 - 0.100 = 0.064 moles in 171 + 83.33 mL = 254.33 mL

= (0.064/254.33) * 1000 = 0.25 M

Concentration of sodium lactate 0.1 moles in 254.33 mL = (0.1/254.33)*1000 = 0.39 M

Lactic acid ionizes as

Lactic acid Lac- + H⁺

Ka = [Lac^-][H⁺]/[Lactic Acid]

1.38 x 10^-4 = 0.39 x [H⁺]/0.25

[H+] = (1.38 x 10^-4 x0.25)/0.39

[H+] = 8.84 x 10^-5

pH = -log [H+]

pH = -log (8.84 x 10^-5)

pH = 4.05

At equivalence point all lactic acid is converted to sodium lactate, this sodium lactate ionizes in water to lactic acid.

Amount of NaOH neede for equivalence point is

N1V1 = N2V2

171 x 0.961 = 1.203 x V2

V2 = (171 x 0.961)/1.203 = 136.6

Total volume of the material = 171 +136.6 = 307.6

moles of sodium lactate = 0.164, in 307.6 mL it is 0.533 M

C2H3O2Na(aq)

+ H2O(l) =

HC2H3O2(aq)

+

NaOH(aq)

Initial 0.533 M 0 0

change -x +x +x

equilibrium 0.533 -x x x

Kb = x²/(0.533 -x)

Kb = Kw/Ka

Kb = 1 x 10^-14/1.38 x 10^-4

Kb = 7.24 x 10^-11=x²/(0.533 - x) (ignoring x in denominator)

7.24 x 10^-11x (0.533) = x²

3.86 x 10^-11 = x²

x = 6.21 x 10^-6 = [OH^-]

[H⁺] = K_w/[OH^-(aq)] = 1.0x10^-14/6.21x10^-6 = 1.61 x 10^-9

pH = -log [H+] = -log (1.61 x 10^-9) = 8.79

178.9 mL of the 1.203 M NaOH = (178.9 x 1.203)/1000 = 0.215 moles

so NaOH in excess is 0.215 - 0.164 moles = 0.051 moles in 178.9 + 171mL = 349.9 = (0.051/349.9)*1000 = 0.146 M

pOH = - log 0.146 =0.83
pH = 14 - 0.83 = 13.16