Question

Five ml of 8N nitric acid, 4.8 ml of 5N hydrochloric acid and a certain volume of 17M sulphuric acid are mixed together and made upto 2litre. Thirty ml. of this acid mixture exactly neutralise 42.9 ml of sodium carbonate solution containing one gram of Na2CO3 .10H2O in 100 ml. of water. Calculate the amount in gram of the sulphate ions in solution.

Answer 1

For sodium carbonate solution,

100 ml of solution contains 1 g of Na2CO3.10H2O.

So, 42.9 ml will contain Na2CO3.10H2O = 42.9/100 = 0.429 g

Molar mass of Na2CO3.10H2O = 286 g/mol

Moles of Na2CO3.10H2O = mass/molar mass = 0.429/286 = 0.0015 mol

To neutralise 1 mole of sodium carbonate, 2 moles of acid is required. So, moles of acid used to neutralise 0.0015 mol of sodium carbonate, 0.003 mol of acid is used.

The acid present is an acid mixture. The mixture contains 5 ml of 8N nitric Acid, 4.8 mL of 5 N hydrochloric acid and a certain volume of 17 M sulphuric acid.

Suppose volume of sulphuric acid = x L

Nitric acid => volume = 5 ml = 0.005 L, concentration = 8 N = 8 M

Moles of nitric acid = concentration*volume = 0.005*8 = 0.04 mol

hydrochloric acid=> volume = 4.8 ml = 0.0048 L, concentration = 5 N = 5 M

Moles of hydrochloric acid = concentration*volume = 0.0048*5 = 0.024 mol

Sulfuric acid=> volume = x L, concentration = 17 M

Moles of Sulfuric acid = concentration*volume = x*17 = 17x mol

Each mole of sulfuric acid contains 2 moles of H+, so moles of H+ present is 34x moles.

Total moles of H+ = 0.04+0.024+34x = (0.064+34x) mol

The whole mixture was made upto 2 L. Now 30 ml of this mixture was taken. 2 L of the mixture contains these number of moles of acid. So, 30 ml will contain = (0.064+34x)(30/2000) = 0.015(0.064+34x)

These are equal to twice the moles of sodium carbonate.

So, 0.015(0.064+34x) = 0.003

0.064+34x = 0.2

34x = 0.136

x = 0.004 L

Volume of sulfuric acid = 0.004 L = 4 ml

Volume = 0.004 L

Concentration = 17 M

moles = concentration*volume = 0.004*17 = 0.068 mol

Molar mass of sulfuric acid = 96 g/mol

mass of sulfuric acid = moles*molar mass = 0.068*96 = 6.528 g

Mass = 6.528 g

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