Question

Consider the titration of 10.00 mL of 0.10 M acetic acid (CH3COOH) with 0.10 M NaOH

a. What salt is formed during this reaction?

b. do you expect the salt solution at the equivalence point to be acidic, neutral, or basic?

Calculate the pH of this solution at the equicalence point.

Answer 1

CH3COOH+NaOH-----> CH3COONa +H2O

salt formed is sodium acetate.s that weak acid strong base titrations will produce a equivalence point give rise to pH above 7. Acidic acid is weak acid and sodium hydroxide is strong base. Hence at equivalence point the solution is basic.

Moles of acid = 0.1*10/1000=0.001

At equivalecne point same moles of NaOH must have been added, volume of NaOH=10ml

HC2H3O2 + OH- <=====> C2H3O2- + H2O

C2H3O2- formed = 0.001 moles

the volume after mixing =10+10=20ml

                              C2H3O2 + H2O <========> HC2H3O2 + OH-

Initial                     0.001*1000/=0.05                         0             0

x                           -x                                                x              x

Equilbrium          0.05-x                                           x               x

Kb= x²/(0.05-x)= 5.9*10-10

Considering x to be smaler than 0.05

x² =5.9*10^-10* 0.05 = 2.95*10^-11

x= 5.43*10-6 which is much less than 0.05

pOH= -log(5.43*10-6) =5.3

pH= 14-pOH= 14-5.3 =8.7