Question

A 40.0 mL solution of 30.0 M of succinic acid (aq) is titrated with 0.300 M of KOH. Find the pH of the solution. (pKa=4.2).

A) initially

B) 1/2 way to the equivalence point

C) at the equivalence point

D) 1.00 mL past the equivalence post

2) sketch the titration curve that results from above titration. Label A) only a weak base present. B) Buffer. C) Only a weak acid present. D) Strong base in excess. E) pH = pKa

Answer 1

A) initial pH

acid dissociates in aqueous solution

H2A <==> HA- + H+

let x amount has dissociated

Ka = 6.31 x 10^-5 = [HA-][H+]/[H2A] = x^2/30

x = [H+] = 0.043 M

pH = -log[H+] = 1.36

B) 1/2 way to the equivalence point

moles of acid present = moles of salt formed

pH = pKa = 4.2

C) at the equivalence point

moles of acid = moles of base added

moles of acid = 30 M x 0.04 L =1.2 mols

Volume of base added = 1.2/0.3 = 4 L

[salt] = 1.2/4.04 = 0.30 M

salt hydrolyses as,

HA- + H2O <==> H2A + OH-

let x amount has hydrolyzed

Kb = Kw/Ka = 1 x 10^-14/6.31 x 10^-5 = x^2/0.3

x = [OH-] = 6.90 x 10^-6 M

pOH = -log[OH-] = 5.16

pH = 14 - pOH = 8.84

D) 1 ml past equivalence point

excess KOH = 0.3 M x 0.001 L/0.041 = 7.32 x 10^-3 M

pOH = 2.13

pH = 11.86

2) Given below is the plot of pH vs volume of KOH added in L.

Labels, C) only weak acid present

E) pH = pKa

B) Buffer

D) strong base in excess

A) no point for weak base only, this is a titration with strong base

Plot pH vs L of base added 15 10 4 4.001 volume of KOH (L)