Question

The potentiometric titration of 25.00 mL of malonic acid C₃H₄O₄ (pKa values 2.83 and 5.69) with 0.2217 M NaOH gave endpoints at 16.47 mL and 32.01 mL.

A)Sketch the titration curve (roughly - don’t worry about exact pH values, just the general shape of the curve) and label each of the following points with the major acid or base species and the type of acid/base solution (e.g., strong acid, weak base, amphiprotic, buffer, pka values, endpoints,):

i.)the beginning (before any titrant has been added)

ii.)the midpoint of the curve between the beginning and the first equivalence point

iii.)the first equivalence point

iv.)the midpoint of the curve between the first and second equivalence points

v.)the second equivalence point

vi.)any point beyond the second equivalence point

B)Do a statistically valid calculation of the initial concentration of the malonic acid. (Hint: what information is contained in the first and second endpoints?)

C)Calculate the pH when 8.24 mL, 16.47 mL, and 20 mL of titrant have been added (3 answers should be clearly identified and the pH calculated.) **should be using the amphrprotic equation on the EP1 and the HH equation on the other two.

Answer 1

Potentiometric titration

A) Drawn below is a representative curve for the titration of malonic acid with NaOH.

Labelled are,

(i) Initial when no NaOH was added

(ii) 1/2 way to the first equivalence point

(iii) first equivalence point

(iv) 1/2 way to the second equivalence point

(v) second equivalence point

(vi) beyond second equivalence point

B) Initial concentration of malonic acid = moles of NaOH used to reach first equivalence point/malonic acid solution

                                                              = 0.2217 M x 16.47 ml/25 ml = 0.146 M

C) pH calculation

when NaOH = 8.24 ml

moles malonic acid = 0.146 M x 25 ml = 3.65 mmol

moles NaOH added = 0.2217 M x 8.24 ml = 1.83 mmol

[malonic acid] remained = 1.82 mmol/33.24 ml = 0.055 M

[malonate formed] = 1.83 mmol/33.24 ml = 0.055 M

This is 1st half equivalence point

pH = pKa1 = 2.83

When NaOH = 16.47 ml

moles malonic acid = 0.146 M x 25 ml = 3.65 mmol

moles NaOH added = 0.2217 M x 16.47 ml = 3.65 mmol

Ist equivalence point

pH = 1/2(pKa1 + pKa2) = 1/2(2.83 + 5.69) = 4.26

when NaOH = 20 ml

moles malonic acid = 0.146 M x 25 ml = 3.65 mmol

moles NaOH added = 0.2217 M x 20 ml = 4.434 mmol

[malonic acid] remained = 2.866 mmol/45 ml = 0.064 M

[malonate formed] = 0.784 mmol/45 ml = 0.017 M

pH = pKa2 + log(malonate/malonic acid)

     = 5.69 + log(0.017/0.064) = 5.114