Back titration is used in volumetric analysis to find out excess of reagent added by titrating it with suitable reagent. It is also used to find out percentage purity of sample. For example in acid-base titration supposes we have added excess base in acid mixture. To find excess base we can titrate the solution with another acid of known strength.
Ex.2 20 g. of a sample of Ba(OH)2 is dissolved in 50 ml. of 0.1 N HCl solution. The excess of HCl was titrated with 0.1 N NaOH. The volume of NaOH used was 20 cc. Calculate the percentage of Ba(OH)2 in the sample.
Sol. Milli eq. of HCl initially =50×0.1=5
Milli eq. of NaOH consumed = Milli eq. of HCl in excess =20×0.1=2
∴ Milli eq. of HCl consumed = Milli eq. of Ba (OH)2=5-2=3
∴ eq. of Ba(OH)2 = 3 / 1000 = 3 × 10-3
Mass of Ba (OH)2 = 3 × 10-3 × (171/2) = 0.2565 g
% Ba (OH)2=(0.2565/20)×100=1.28% Ans.
Ex.3 4.0 g of monobasic, saturated carboxylic acid is dissolved in 100 mL water and its 10 mL portion required 8.0 mL 0.27 M NaOH to reach the equivalence point. In an another experiment, 5.0 g of the same acid is burnt completely and CO2 produced is absorbed completely in 500 mL of a 2.0 N NaOH solution. A 10 mL portion of the resulting solution is treated with excess of BaCl2 to precipitate all carbonate and finally titrated with 0.5 N H2SO4 solution. Determine the volume of the acid solution that would be required to make this solution neutral.
Sol. Meq. Of NaOH=8×0.27=Meq. Of acid for 10 ml of acid solution
Meq. Of acid for 100 ml of solution
= (8 × 0.27 × 1000)/10 = (4/M) × 103
M (acid) = 4/(8 × 0.27 × 10) × 1000 = 185.2
Formula of acid. = CnH2nO2
Þ M = 14n + 32 = 185.2
Þ n = 11
Now 5g acid will produce (5/185.2) × 11 = (55/185.2) mol CO2 after complete combustion.
2NaOH + CO2 → Na2CO3 + H2O
Total mass of NaOH available = 500 × 2 × 10-3 = 1.0 mole
Moles of NaOH left unreacted = 1-(2 × 55/185.2) = (76/185.2) in 500 mL
Þ Molarity of NaOH after precipitation of Na2CO3 = 0.812
Therefore, 0.812 × 10 = 0.5 × V
Þ V = 16.24 mL Ans.
Precipitation titration:
In ionic reaction we can know strength of unknown solution of salt by titrating it against a reagent with which it can form precipitate. For example NaCl strength can be known by titrating it against AgNO3 solution with which it forms white ppt. of AgCl.
So Meq. Of NaCl at equivalence point =Meq. Of AgNO3 used= Meq. Of AgCl formed
Ex.4 A complex of cobalt with ammonia is analyzed for determining its formula, by titrating it against a standardized acid as follows:
Co (NH3) x Cl3 (aq) + HCl → NH+4(aq) + Co3+ (aq) + Cl-(aq)
A 1.58 g complex required 23.63 mL 1.5 M HCl to reach the equivalence point. Determine formula. If the reaction mixture at equivalence point is titrated with excess of AgNO3 solution, what mass of AgCl will precipitate out?
Sol. The balanced chemical reaction is:
Co(NH3)x Cl3 + xHCl → xNH+4 + Co3+ + (x + 3) Cl-
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