Question

Practical Q1(a) A is a solution containing 15.8 gdm⁻³ of Na2S2O3. B was obtained by dissolving 9.0g of an impure sample of I2 in aqueous KI and the solution made up to 1 dm³.
(i) Put A into the burette and titrate it against 20.0 cm³ or 25.0 cm³ portions of B, using starch solution as indicator. Repeat the titration to obtain concordant titre values.
Equation: I2+2S2O3²⁻→2I⁻+S4O6²⁻.
(b) From your results, calculate: (i) concentration of A in moldm⁻³; (ii) concentration of I2 in B in moldm⁻³; (iii) percentage by mass of I2 in the sample.
(c) Give a reason why the starch indicator was not added at the beginning of the titration. [23 marks]

Accepted Answer

Sample titration results: Rough=21.20, I=17.60, II=18.40, III=16.50 (final readings); average volume of A used (from concordant titres I–III) = (16.60+16.40+16.50)/3 = 16.50 cm³.
(b)(i) Molar mass Na2S2O3 = (23×2)+(32×2)+(16×3)=158 g/mol. Concentration in gdm⁻³ of Na2S2O3 = 15.8 gdm⁻³, so Concentration in moldm⁻³ = 15.8/158 = 0.1 moldm⁻³.
(ii) Using CaVa/Na = CbVb/Nb with Ca=0.1 moldm⁻³, Va=16.50 cm³, Na=2 (S2O3²⁻), Vb=25 cm³, Nb=1 (I2):
  Cb = (Ca×Va×Nb)/(Vb×Na) = (0.1×16.50×1)/(25×2) = 0.033 moldm⁻³.
(iii) Molar mass I2 = 254. Concentration of I2 in gdm⁻³ = 0.033×254 = 8.38 gdm⁻³ (mass of I2 in the sample). Given mass of impure I2 sample = 9.0 gdm⁻³.
  Percentage by mass of I2 = (8.38/9.0)×100 = 93.1%.
(c) The starch indicator is added only near the end-point because if added at the beginning, the iodine is strongly adsorbed onto the starch, making it difficult to desorb, which reduces the accuracy/sharpness of the final colour-change reading.

WAEC Chemistry 2019 Theory — Question 26

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