A 2.0 g sample of a mixture containing sodium carb

Question

A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO2 ceases. The volume of CO2 at 750 mm Hg pressure and at 298 K is measured to be 123.9 ml. A 1.5g of the same sample requires 150 ml. of (M/10) HCI for complete neutralization. Calculate the % composition of the components of the mixture.

Navjyot Kalra · Accepted Answer

1.5 g of sample require = 150 ml. of M/10 HCI &there4; 2 g of sample require = 150/1.5 ml of M/10 HCI = 200 ml. of M/10 HCI On heating, the sample, only NaNCO3 undergoes decomposition as given below. 2NaHCO3 &rarr; Na2CO3 + H2O + CO2 &uarr; 2 moles 1 mole 1 mole 2 equ Neutralization of the sample with HCI takes place as given below. 2NaHCO3 + HCI &rarr; NaCI + H2O + CO2 1 eq 1 eq Na2CO3 + 2HCI &rarr; 2NaCI + H2O + CO2 1 mole 1 mole 2 eq 2 eq Hence, 2 g sample &equiv; 200 ml. of M/10 HCI = 200 ml. of N/10 HCI = 20 meq = 0.020 eq Number of moles of CO2formed, i.e. N = PV/RT = 750/760 * 123.9/0.082 = 0.005 Moles of NaHCO3 in the sample (2 g) = 2 * 0.005 = 0.01 Equivalent of NaHCO3 = 0.01 Wt. of NaHCO3 = 0.01 * 84 = 0.84 g % of NaHCO3 = 0.84 *100/2 = 42% Equivalent of Na2CO3 = 0.02 &ndash; 0.01 = 0.01 Wt. of Na2CO3 = 0.01 * 53 = 0.53 g &there4; % of Na2CO3 = 0.53 *100/2 = 26.5% &there4; % of Na2SO3 in the mixture = 100 &ndash; (42 + 26.5) = 31.5%

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