1.) 50 mL of an unknown monoprotic weak acid was titrated with 0.1 M NaOH and th
ID: 1045105 • Letter: 1
Question
1.) 50 mL of an unknown monoprotic weak acid was titrated with 0.1 M NaOH and the results below were obtained.
a.What is the concentration of the weak acid?_________________________
b.Using the graph, what is the pKa of the weak acid?_________________________
c.Calculate the pH of the solution before and base is added._________________________
d.Calculate the pH of the solution when 9mL of base are added._________________________
e.Calculate the pH at the equivalence point._________________________
f.Calculate the pH when 14 mL of of base are added._________________________
14 12 10 6 4 2 0 10 15 20 ViNaOH), mLExplanation / Answer
From the titration plot shown above
a. concentration of weak acid = 0.1 M x 10 ml/50 ml = 0.02 M
b. pKa = 4.5
c. initial pH
HA + H2O <==> H3O+ + A-
le x amount dissociated
Ka = [A-][H3O+]/[HA]
pKa = -log[Ka] = 5
Ka = 1 x 10^-5
so,
1 x 10^-5 = x^2/0.02
x = [H3O+] = 4.47 x 10^-4 M
pH = -log[H3O+] = 3.35
d. after 9 ml NaOH added
moles HA = 0.02 M x 50 ml = 1 mmol
moles NaOH = 0.1 M x 9 ml = 0.9 mmol
A- formed = 0.9 mmol
HA remained = 0.1 mmol
pH = pKa + log(A-/HA)
= 5 + log(0.9/0.1)
= 5.95
e. pH at equivalence point
[A-] formed = 0.02 M x 50 ml/60 ml = 0.0167 M
A- + H2O <==> HA + OH-
let x amount hydrolyzed
Kb = [HA][OH-]/[A-]
1 x 10^-14/1 x 10^-5 = x^2/0.0167
x = [OH-] = 4.1 x 10^-6 M
[H3O+] = 1 x 10^-14/4.1 x 10^-6 = 2.45 x 10^-9 M
pH = -log[H3O+] = 8.61
f. after 14 ml NaOH added
excess [NaOH] = [OH-] = 0.1 M x 4 ml/64 ml = 0.00625 M
[H3O+] = 1 x 10^-14/0.00625 = 1.6 x 10^-12 M
pH = -log[H3O+] = 11.79
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