Question

Determine the pH during the titration of 66.6 mL of 0.457 M acetic acid (Ka = 1.8×10-5) by 0.457 M KOH at the following points. (Assume the titration is done at 25 °C.)

a.) Before the addition of any KOH

b.)  After the addition of 16.0 mL of KOH

c.) At the half-equivalence point (the titration midpoint)

d.) At the equivalence point

e.) After the addition of 99.9 mL of KOH

Thank you!!

Answer 1

The dissociation of acetic acid can be written as

CH3COOH---------------> CH3COO- + H+

Ka= [CH3COO-] [H+]/[CH3COOH]

let x= drop in concentration of Acetic acid to reach equilibrium

Hence at Equilibrium [H+]= [CH3CO-] =x and [CH3COOH] =0.457-x

Ka= x²/(0.457-x) =1.8*10^-5

where x= [H+] when sovled for x using excel, we get x= 0.000906

pH= -log(0.000906)=3.04

b) moles of acetic acid in 66.6ml of 0.457M= 0.457*66.6/1000=0.0304

moles of KOH in 0.457 of 16ml =0.457*16/1000 =0.007312

The reaction between acetic acid and KOH can be represented as

CH3COOH+ KOH---> CH3COOK+ H2O

moles of KOH used will be same as moles of acetic acid. Moles of KOH is less and this is the limting reactant.

moels of Acetic acid remaining after reaction with KOH= 0.0304-0.007312=0.0231

Volume after mixing = 66.6+16= 82.6 ml =82.6/1000=0.0826L

concentration of acetic acid after mixing= 0.0231/0.0826=0.279 Concentration of CH3COOK= 0.007312/0.0826=0.0885

pH= pKa+log [Salt/acid] = 4.74+log (0.0885/0.279)=4.24

c) at half equivalence point half the moles of CH3COOH are consumed due to reaction with KOH

so moles of CH3COOH used= 66.6*0.457/2000=0.0152 moles

volume of KOH used= 0.0152/0.457=33.3 ml moles of CH3COOK formed =0.0152

total volume after miixng = 66.6+33.3 = 99.9ml =0.0999L

Concentration of acetic acid after mixing = 0.0152/0.0999=0.152 and KOH= 0.152

pH= pKa=4.74

c) at Equilvalenece point molesof KOH= 0.0304

moles of Sodium acetate formed = 0.0304

CH3COOH+ KOH--> CH3COO- + K+ H2O

CH3COO- + H2O---> CH3COOH+ OH-

volume after mixing =66.6+66.6= 133.2ml =0.1332L ,concentration = 0.0304/0.1332=0.228M

Kb= x²/(0.228-x)= Kw/Ka = 10^-14/ (1.8*10^-5) =5.55*10^-10, x =[OH-]

when solved using excel gives [OH-] = 0.0000112, pOH= 4.95, pH= 14-4.95= 9.05

e) when 99.9 ml of KOH is added =0.457*99.9/1000= 0.045654 moles

moles of acetic acid = 0.0304 exces is KOH by = 0.045654-0.0304=0.015254

Volume of mixing = 66.6+99.9= 166.5 ml =0.1665 L

Concentration of KOH= 0.015254/0.1665=0.092 ( KOH dissociates completely and   gives OH- ions)

pOH= 1.04, pH= 14-1.04=12.96