acid base titrations

Question

for the titration of 50.00 mL of 0.1317 M angelic acid with 0.1506 M KOH:

pKa = 4.29

a) Determine the equivalence volume and pH for this titration

b) calculate the pH at each of the following volumes: 0.00mL 10.00mL 40.00mL 50.00mL

Answer 1

C₅H₈O₂ + KOH ==> C₅H₇O₂K + H₂O

moles angelic acid = 50.00 ml x 1 L/1000 ml x 0.1317 mol/L = 0.006585 moles

moles KOH needed = 0.006585 moles KOH since the reaction requires 1:1 mol ratio of acid and base

a) volume of KOH needed: (x L)(0.1506 mol/L) = 0.006585 moles KOH

x = 0.04373 L = 43.73 mls KOH needed

To determine the pH at equivalence, we look at the hydrolysis of the conjugate base of angelic acid:

C5H7O2^- + H2O ==> C5H8O2 + OH^-

Since pKa of the acid = 4.29, the pKb of the conjugate base = 14 - 4.29 = 9.71 and Kb = 1.95x10^-10

The volume will be 50.00 ml + 43.73 ml = 93.73 ml = 0.09373 L and final [conj. base] = 0.006585 mol/0.09373 L = 0.0703 M

Kb = 1.95x10^-10 = [C5H8O2][OH^-]/[C5H7O2^-] = (x)(x)/0.0703 - x and assuming x is small we neglect it...

1.95x10^-10 = x²/0.0703

x² = 1.36997x10^-11

x= 3.7013x10^-6 M = [OH^-]

pOH = -log 3.7013x10^-6 = 5.432

pH = 14 - 5.432 = 8.5684

pH when 0.00 ml KOH added:

Ka = [H⁺][C5H7O2^-]/[C5H8O2] and Ka = 1x10^-4.29 = 5.13x10^-5 (calculated from pKa)

5.13x10^-5 = (x)(x)/0.1317 - x ... assume x is small and neglect it in the denominator

x² = 6.76x10^-6

x = H+ = 2.6x10^-3 (this value is only about 2% of 0.1317 so above assumption is valid)

pH = -log 2.6x10^-3 = 2.59

For the rest of the problem, you'd be best to set up an ICE table, letting angelic acid be HA....

HA + KOH ===> A^- + H2O

0.006585.....0.001506..................0..................Initial

-0.001506..-0.001506............+0.001506........Change

0.005079.......0.......................+0.001506.......Final

Final volume = 50 + 10 = 60 ml = 0.060 L

Convert moles in above table to concentration by dividing by 0.060 L

To find pH use pH = pKa + log [A-]/[HA] because you have formed a buffer

Repeat above procedure for addition of 40 and 50 ml. Note that when you add 50 ml of KOH, all of the angelic acid is neutralized and there will be OH- left over. You won't need pH = pKa + log A-/HA to find pH for this volume. Just find molarity of excess OH-, find pOH and then pH.