Question

5. In the titration of 15.00 mL of a 0.1000 M acetic acid solution by a 0.1000 M NaOH solution, find the pH of the solution being titrated when

a) 0.00 mL of NaOH has been added

b) 5.00 mL of NaOH has been added

c) 7.50 mL of NaOH has been added

d) 14.90 mL of NaOH has been added

e) 15.00 mL of NaOH has been added

f) 30.00 mL of NaOH has been added

Answer 1

a)

moles of acetic acid in 15ml of 0.1M= 0.1*15/1000 =0.0015 moles

Ka of acetic acid =1.75*10^-5 and Pka= 4.76

1. When no Sodium hydroxide is added,

The ionization of CH3COOH is CH3COOH+H2O<---------->CH3COO- + H3O+

                                         CH3COOH                  CH3COO-                                  H3O+

       Initial                       0.1                                 0                                                 0

     Change                    -x                                     x                                                 x

Equilibrium                   0.1-x                             x                                                   x

Ka= [CH3COO-] [H3O+]/ [CH3COOH] = x²/ (0.1-x) = 1.75*10^-5

When solved using excel, x = 0.00132, pH= 2.9

The reaction between CH3COOH and NaOH is represented as

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

i.e 1 mole of acetic acid requires 1 moles of sodium hydroxide

moles of acetic acid in 15ml of 0.1M= 0.1*15/1000 =0.0015 moles

b)Moles of NaOH in 5ml of 0.1M NaOH= 0.1*5/1000 = 0.0005. Stoichiometric requirement of NaOH is 0.0015.

since NaOH is the limiting reactant, moles of CH3COOH remaining = 0.0015-0.0005=0.001

Moles of sodium acetate (CH3COONa) formed = .0005

pH= pKa + log [Sodium acetate/ acetic acid]

mole ratio = concentration ratio since moles are divided by same volume

pH= 4.72+ log (0.0005/0.001)=4.42

c)

1. Moles of NaOH in 7.5ml of 0.1M NaOH= 0.1*7.5/1000 =0.00075 .

Excess CH3COOH= 0.0015-0.00075= 0.00075

Moles of sodium acetate formed = 0.00075

pH= pKa+ log [sodium acetate/ acetic acid ] = 4.72+log1= 4.72

d)

Moles of NaOH in 14.9 ml of 0.1M NaOH= 0.1*14.9/1000 =0.00149

Excess CH3COOH= 0.0015-0.00149= 1*10^-5

Moles of sodium acetate formed = 0.00149

pH= pKa+ log [sodium acetate/ acetic acid ] = 4.72+log(0.00149/1*10^-5) = 6.89

e)

when 15ml of NaOH is added, neutralization point is reached. So moles of NaOH is same as moles of CH3COOH= 0.0015, moles of Sodium acetate formed = 0.0015

volume after mixing = 15+15=30ml, concentration of sodium acetate = 0.0015*1000/30 =0.05

CH3COONa + H2O ---à CH3COO- + OH-

Kb= [CH3COO-] [OH-]/ [CH3COONa]

Kb= 10^-14/Ka = 10^-14/ 1.75*10^-5 =5.71*10^-10

                        CH3COONa                  CH3COO-                                  OH-

       Initial                       0.5                                 0                                                 0

     Change                    -x                                     x                                                 x

Equilibrium                   0.05-x                             x                                                   x

When solved using excel, x= 5.35*10-6

pOH= 5.27, pH= 14-5.27= 8.73

f)

moles of NaOH in 30ml 0f 0.1 = 0.1*30/1000 =0.003 moles

moles of CH3COOH=0.0015

So excess moles of NaOH= 0.003-0.0015= 0.0015

Volume of mixing = 30+15= 45ml= 45/1000L= 0.045L

Concentration of NaOH= 0.0015/0.045 Moles/L= 0.033M

pOH= 1.477, pH= 14-1.477= 12.523