Analysis of Titration (20 points!!!)

Mass of the sample in 500 mL of the solution = 1.127 g
Mass of the sample in 50 mL of the solution = 1.127 x (50/500) = 0.1127 g

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Consider the second titration:
The excess BaCl2 solution followed by filtration removes all the CO32- ions as BaCO3 solid. Therefore, the titration only involves the reaction between HCl and KOH.

HCl + KOH → KCl + H2O
Mole ratio HCl : KOH = 1 : 1

No. of moles of HCl used = MV = 0.05304 x (28.56/1000) = 0.001515 mol
No. of moles of KOH used = 0.001515 mol
Molar mass of KOH = 39 + 16 + 1 = 56 g mol-1
Mass of KOH in 50 mL of solution = mol x (molar mass) = 0.001515 x 56 = 0.08484 g

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Consider the first titration (back titration):
Part of the HCl in 40 mL of the solution is used to react with KOH and K2­CO3 in 50 mL of the solution, and then the excess is titrated with NaOH solution.

HCl + KOH → KOH + H2O ...... (*)
2HCl + K2CO3 → 2KCl + CO2 + H2O ...... (**)
Mole ratio HCl : K2CO3 = 2 : 1
HCl + NaOH → NaCl + H2O ...... (***)
Mole ratio HCl : NaOH = 1 : 1

No. of moles of NaOH used in (***) = MV = 0.04983 x (4.74/1000) = 0.0002362 mol
No. of moles of HCl used in (***) = 0.000236 mol
As calculated above, no. of moles of HCl used in (*) = 0.001515 mol
Total no. of moles of HCl used = MV = 0.05304 x (40/1000) = 0.002122 mol
No. of moles of HCl used in (**) = 0.002122 - (0.001515 + 0.0002362) = 0.0003708 mol
No. of moles of K2CO3 used in (**) = 0.0003708 x (1/2) = 0.0001854 mol
Molar mass of K2CO3 = 39x2 + 12 + 16x3 = 138 g mol-1
Mass of K2CO3 in 50 mL of solution = mass x (molar mass) = 0.0001854 x 138 = 0.02559 g

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In 50 mL of the solution:
Total mass of the sample = 0.1127 g
Mass of KOH = 0.08484 g
Mass of K2CO3 = 0.02559 g
% by mass of KOH = (0.08484/0.1127) x 100% = 75.28%
% by mass of K2CO3 = (0.02559/0.1127) x 100% = 22.71%
% by mass of H2O = 100% - (75.28% + 22.71%) = 2.01%