4.3 Back titration

BACK TITRATION

BACK TITRATION Back titration is a process in which the excess of a standard solution used to consume an analyte is determined by titration with a second standard solution.

Example: Determination of acetylsalicylic acid in aspirin.

Sometime direct titration of an analyte with a reagent is not FEASIBLE. This is due to the following reason:  The reaction kinetics is slow or the rate of reaction is slow.  No suitable indicator in the direct titration.  The colour change at the end point is slow or delay or not sharp.  The end point is far from the equivalence point.  Standard solution lacks stability.

The titration of insoluble organic acid with NaOH is not practical because the reaction is slow.

CH3COO-C6H4-COOH + 2NaOH

CH3COONa + HO-C3H4-COONa

To overcome it, add NaOH in excess and allow the reaction to reach completion and then titrate the excess NaOH with a standard solution of HCl. NaOH + HCl

NaCl + H2O

The system has gone from being ACID , past the equivalence point to the BASIC side (excess base), and then back to the equivalence point.

The final titration to the equivalence point is called a BACK TITRATION.

Calculation 150.0 mL of 0.2105 M nitric acid was added in excess to 1.3415 g calcium carbonate. The excess acid was back titrated with 0.1055 M sodium hydroxide. It required 75.5 mL of the base to reach the end point. a) Name a suitable indicator used in above titration b) Draw the titration curve for the above titration. c)Calculate the percentage (w/w) of calcium carbonate in the sample.

Answer a) Suitable indicator: phenolfthalein/methyl orange b)

c) First write a balance equation for the above reactions.

2HNO3 + CaCO3 HNO3 + NaOH

Ca(NO3)2 + CO2 + H2O NaNO3 + H2O

From Equations above: 2 mole HNO3 = 1 mole CaCO3 (initial) 1 mole HNO3 = 1 mole NaOH (back titration) Initial amount of acid: mole of acid = 0.2105 x 0.150 = 31.58 x 10-3 mol

1 mole HNO3 = 1 mole NaOH (back titration) So, remaining/excess acid during back titration: mole of excess acid = 0.1055 x 0.0755 = 7.965 x 10-3 mol.

Then, mole of acid reacted with CaCO3 = ( 31.58 x 10-3 – 7.965 x 10-3 ) = 23.61 x 10-3 mole HNO3

2 mole HNO3 = 1 mole CaCO3 (initial) 23.61 x 10-3 mole = ? mole of CaCO3 = ½ x mole acid = ½ x 23.61x 10-3 = 11.805 x 10-3 mol CaCO3

Gram CaCO3 = mole x molar mass = 11.805 x 10-3 x 100 = 1.1805 g. % CaCO3 = weight CaCO3 weight of sample = 1.1805 x 100% 1.3415 = 87.99 % (w/w)

x 100%