Question

A sample of 0.105 M CH3CO2H is titrated with 0.125 M KOH. What is the approximate pH of the resulting solution when the equivalence point is reached? Ka= 1.8 x 10^-5. Why does .105 have to be divided by 2?

Answer 1

at equivalence point all acid reatcs with base i.e all CH3CO2H reacts with OH- to form CH3CO2- and H2O

Hence moles of CH3CO2H = moles of KOH

Let CH3CO2H solution volume =1 L , Moles of CH3CO2H = M x V = 0.105 x 1 = 0.105

then KOH volume can be found out from formula Moles of KOH = M x V

             0.105 = 0.125 x V

V of KOH = 0.84 L

now solution volume = 1+0.84 = 1.84L

Moles of CH3CO2- formed = CH3CO2H moles reacted = 0.105

[CH3CO2-] = moles / volume = 0.105 /1.84 = 0.057

now we have reverse equilibrium     CH3CO2- (aq) + H2O (l) <--> Ch3CO2H (aq) + OH- (aq)

at equilibrium [CH3CO2-] = 0.057-X , [CH3CO2H] =[OH-] = X

Kb of reaction = Kw / ( Ka of acid) = 10^-14 / ( 1.8x10^-5) = 5.56 x 10^-10

now 5.56 x 10^-10 = X^2 / ( 0.057-X)                ( since Kb is small we get X small , hence 0.057-X is nearly 0.057)

X = [OH-] = 5.63 x 10^-6 , pOH =-log [OH-] = -log ( 5.63 x 10^-6) = 5.25

pH = 14-pOH = 14 - 5.25 = 8.75