# The potentiometric titration o

The potentiometric titration of 25.00 mL of malonic acidC3H4O4 (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 thecurve) and label each of the following points with the major acidor base species and the type of acid/base solution(e.g., strong acid, weak base, amphiprotic, buffer, pkavalues, endpoints,):

i.)the beginning (before any titranthas been added)

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

iii.)the first equivalence point

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

v.)the second equivalence point

vi.)any point beyond the secondequivalence point

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

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

Answer:

Potentiometric titration

A) Drawn below is a representative curve for the titration ofmalonic 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 toreach 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

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