Question

Titration Problem:

Lactic acid, HC₃H₅O₃ (see right for structure), builds up in muscles after vigorous exercise. It is a weak acid with a pK_a = 3.86.

Consider the following titration:

25.0 mL 0.500 M lactic acid is titrated with 1.00 M NaOH.

Answer the following questions:

1. What is the pH of the lactic acid solution before any NaOH is added?

2. What is the pH after 5.00 mL of 1.00 M NaOH is added?

3. What is the pH at the equivalence point?

4. What is the pH when 25.0 mL of 1.00 M NaOH has been added?

Answer 1

Let lactic acid be represented as HA (   A is C3H5O3)

The ionization of HA in water can be written as

HA+ H2O----àA-+ H3O+

Ka= [A-][H3O+]/[HA]

PKa= -log (Ka)= 3.86,

Ka= 10^(-3.86)= 0.000138

a. When no NaOH is added, initial concentration of HA=0.5M

Let x= drop in concentration of HA to reach equilibrium

At Equilibrium [A-]=[H3O+]=x and [HA]=0.5-x

Ka= Equilibrium constant = x²/(0.5-x)=0.000138

When solved using excel, x= 0.00825M

pH= -logx= -log (0.00825)= 2.083

b. When NaOH is added, the reaction is

NaOH + HA → NaA + H2O

The reaction suggests 1 mole of NaOH reacts with 1 mole of HA.

Theoretical molar ratio of HA : NaOH= 1:1

Moles of HA given = molarity* volume (L)= 0.5*25/1000 =0.0125

Moles of NaOH= 1*5/1000=0.005

This suggests excess is lactic acid and limiting reactant is NaOH, since their actual ratio is 0.00125:0.005

So all the NaOH reacts to give rise to 0.005 moles of sodium lactate (NaA)

Moles of latic acid left= 0.0125-0.005=0.0075

Volume after mixing HA with NaOH = 25+5= 30ml = 30/1000L= 0.03L

Concentrations after the reaction : HA= 0.0075/0.03 =0.25, NaA= 0.005/0.03= 0.167

Since pH= pKa+ log (acetate ion/ HA)

Acetate ion is obtained from NaA=0.167

Hence pH= 3.86+ log (0.167/0.25)= 3.68

c. At equivalence point, moles of HA= moles of NaOH= 0.0125

Volume of NaOH required= 0.0125/1=0.0125L= 12.5 ml

Moles of NaA formed= 0.0125

Volume after mixing of HA and NaOH= 25+12.5= 37.5ml= 37.5/1000L=0.0375L

Concentration of NaA after mixing =0.0125/0.0375 =0.33

NaA+ H2O-------->HA+ OH-

Kb= [HA][OH-]/[NaA]

Kb=10^-14/Ka = 10^-14/0.000138= 7.25*10^-11

Let x= drop in concentration of NaA to reach equilibrium

At Equilibrium [NaA]=0.33-x, [HA]=[OH-]=x

Kb= x²/(0.33-x)= 7.25*10^-11, when solved using excel, x=[OH-]= 4.89*10^-6

pOH= -log (4.89*10^-6)= 5.3, pH= 14-5.3= 8.7

d. Moles in 25ml of 1M= 1*25/1000=0.025

In this case excess reactant is NaOH and all the HA gets consumed.

Moles of NaOH remaining after the reaction =0.025-0.0125=0.0125

Volume after mixing = 25+25=50ml= 50/1000L=0.05L

Concentration of NaOH after mxing = 0.0125/0.05= 0.25

Since NaOH is strong base, it gets ionized completely and hence

NaOH------>Na+ OH-, [OH-]= 0.25

pOH= -log (0.25)= 0.6

pH= 14-0.6= 13.4