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2 Chemistry -- Volumetric Analysis

How many mL of a 0.1 M HCl are required to react completely with a 1gm mixture of Na2CO3 and NaHCO3 containing the equimolar amount of the two.

How many mL of a 0.1 M HCl are required to react completely with a 1gm mixture of Na2CO3 and NaHCO3 containing the equimolar amount of the two.

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solution:let, the amount of Na2Co3 in the mixture = x gmNaHCo3 present in mixture = (1-x) gmMolar mass of Na2Co3 = 2*23+12+3*16 = 106 gm/molesimilarly, molar mass of NaHCo3 = 84 gm/moletherefore, moles of Na2Co3 in x gm = x/106 and moles of NaHCo3 in (1-x) gm = (1-x)/84According to the question,x/106 = (1-x)/84 so, x = 0.558gmThus the number of moles of Na2Co3 and NaHCo3 are 0.00526 moleAgain by reaction,


Na2CO3+2HCl→2NaCl+H2O+CO2

Na2CO3+2HCl→2NaCl+H2O+CO2
NaHCO3+HCl→NaCl+H2O+CO2∵ 1 mole of Na2CO3
 requires HCl = 2 mole
∴0.00526  mole of Na2C03 requires 0.01052 mole of HCl
similarly, 1 mole of NaHC03 requires 1 mole of HCl
so, 0.00526 mole of NaHCo3 requires 0.00526 mole of HCl
Total HCl required = 0.01578 mole
again, 0.1 mole of  0.1 M of HCl present in 1000 ml of HCl
therefore, 0.01578 mole of 0.1 M will be present = 157.8 ml . [1000/0.1 * 0.01578].





















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