Solutions of chromate are yellow in colour. At concentrations above 2.5 × 10-3 M, the sample...

Solutions of chromate are yellow in colour. At concentrations above 2.5 × 10-3 M, the sample solution is sufficiently coloured such that the observance of the red precipitate is masked. Calculate the relative error in the volume of titrant added that would be observed for the titration of an unknown chloride solution containing 2.5 × 10-3 M chromate with 0.1 M AgNO3. Assume that 25 mL of titrant is required to reach the endpoint, where the total solution volume will be 1 litre.

Expert Solution

We have,

Ksp of AgCl = 1.82 x 10^-10

Therefore,

Concentration of Ag at the equivalence point = (1.82 x 10^-10)^1/2 = 1.34 x 10^-5M

Also,

Ksp of Ag₂CrO₄ = 1.1 x 10^-12

Therefore,

[Ag⁺]² = Ksp/[CrO₄^2-] = (1.1 x 10^-12)/(2.5 x 10^-3) = 4.4 x 10^-10

[Ag⁺] = 2.1 x 10^-5 M

Error in concentration = Moles of excess Ag = 2.1 x 10^-5 – 1.34 x 10^-5 M = 7.6 x 10^-6 M

Error in the volume of titrant added = [(Moles of excess Ag)/Concentration of titrant] x 1000ml/L

= [(7.6 x 10^-6 M)/0.1 M] x 1000 mL/L = 0.08 mL

If 25 mL was used to reach the endpoint, then

Percent error = (0.08/25) x 100% = 0.32 %

queen_honey_blossom answered 2 years ago

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Solutions of chromate are yellow in colour. At concentrations above 2.5 × 10-3 M, the sample...

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