Question

Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 mL sample of 0.114 M lactic acid (HC₃H₅O₃,

pK_a = 3.86)

is titrated with 0.114 M NaOH solution. (Assume that all solutions are at 25°C.)

(a)

Calculate the pH after the addition of the following amounts of NaOH.

Vol. NaOH added	pH
0 mL
4.0 mL
8.0 mL
12.5 mL
20.0 mL
24.0 mL
24.5 mL
24.9 mL
25.0 mL
25.1 mL
26.0 mL
28.0 mL
30.0 mL

Answer 1

The given Lactic acid is a weak acid with pK_a = 3.86. Now when it reacts with a strong base like NaOH, then it forms the salt NaC₃H₅O₃ as-

HC₃H₅O₃ + NaOH ---------> NaC₃H₅O₃ + H2O

i.e from the reaction of 1 mole of HC₃H₅O₃ and 1 mole of NaOH, we get 1 mole NaC₃H₅O₃

Now the solution in both a weak acid and its salt are present is called a Buffer. And the pH of a buffer is calculated by Henderson Hasselblach equation-

pH = pKa + log [salt]/ [acid]

1- Now when we add no NaOH or 0 ml NaOH, then we have only HC₃H₅O₃ in the solution. Now since it is a weak acid, it cant dissociate completely in water. The dissociation reaction is-

HC₃H₅O₃ + H2O -----------> C₃H₅O₃^- + H3O+

Now the given pKa = 3.86

So Ka = 10^-pKa =  10^-3.86 = 0.000138

Now we know Ka is the dissociation constant of Weak acid and is calculated as

Ka = [C₃H₅O₃^-] * [H3O+] / [HC₃H₅O₃]

To calculate these concentrations, lets form the ICE table-

Reaction	[HC3H5O3]	[C3H5O3-]	[H3O+]
Initial	0.114 M	0	0
Change	-x	+x	+x
Equilibrium	0.114 M - x	x	x

Here

Ka = [C₃H₅O₃^-] * [H3O+] / [HC₃H₅O₃]

0.000138 = [x] * [x] / [0.114 M - x]

0.000138 * [0.114 M - x] = x²

0.000015732 - 0.000138 x = x²

x² +  0.000138 x - 0.000015732 = 0

Solving this-

x = 0.00389

So at equilibrium, [H3O+] = x = 0.00389 M

So pH = -log[H3O+] = -log [0.00389] = 2.41

So before addition of NaOH, the pH of the acid = 2.41

2- After addition of 4 ml NaOH

Initial acid volume taken (V1) = 25.0 mL

Initial acid concentration taken (M1) = 0.114 M

Initial NaOH volume taken (V1) = 4 mL

Initial NaOH concentration taken (M1) = 0.114 M

After addition of the two, the final volume becomes (V2) = 25.0 mL + 4 mL = 29 mL

So the

new initial concentration of acid = M2 = M1V1/V2 = 0.114 M * 25.0 mL / 29 mL = 0.098 M

new initial concentration of NaOH = M2 = M1V1/V2 = 0.114 M * 4.0 mL / 29 mL = 0.0157 M

To get the final volume, lets form the ICE table

Reaction	[HC3H5O3]	[NaOH]	NaC3H5O3]
Initial	0.098 M	0.0157 M	0
Change	-0.0157 M	-0.0157 M	+0.0157 M
Equilibrium	0.0823	0	0.0157 M

Now here the pH is

pH = pKa + log [salt]/ [acid]

pH = pKa + log [NaC₃H₅O₃]/ [HC₃H₅O₃]

= 3.86 + log [0.0157]/ [0.0823]

= 3.86 + log [0.19]

= 3.86 + (-0.719)

= 3.141

The rest calculations are done in similar way as calculation - 2