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1.2g sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml solution. 20ml of this solution required 40ml of 0.1N HCl for complete neutralization. Calculate the weight of Na2CO3 in the mixture. If another 20ml of this solution is treated with excess of BaCl2, what will be the weight of the precipitate?

1.2g sample of Na2CO3 and K2CO3 was dissolved in water to form 100ml solution. 20ml of this solution required 40ml of 0.1N HCl for complete neutralization. Calculate the weight of Na2CO3 in the mixture. If another 20ml of this solution is treated with excess of BaCl2, what will be the weight of the precipitate?

Grade:12th pass

1 Answers

Manish Tiwari
86 Points
6 years ago
Given: Total mass of mixture = 1.2 gVolume of solution = 100 mLVolume of solution taken for Neutralisation, V2 = 20 mLVolume of HCl, V1 = 40 mLNormality of HCl, N1 = 0.1 NTo find: Mass of Na2CO3 in mixtureSolve: As we know N1V1 = N2V20.1× 40 = N2 ×20N2 = 0.2 NNormailty of solution = 0.2 N1 L of solution contains number of gram equivalents = 0.2100 mL or 0.1 L of solution contains number of gram equivalents = 0.2× 0.1= 0.02 gram equivalentsNumber of gram-equivalents is equal to the summation of gram-equivalents of Na2CO3 and K2CO3Number of gram-equivalents = mass / equivalent massEquivalent mass = Molar mass / valence (number of electrons gained or lost by one molecule or ion of the substance in the reaction).Molar mass of Na2CO3 = 106 gMolar mass of K2CO3 = 138 gEquivalent- mass of Na2CO3 = 106 / 2= 53 gEquivalent mass of K2CO3 = 138 / 2= 69 gConsider the mass of Na2CO3 = x gMass of K2CO3 = (1.2 - x)gWe can write that 0.02 = (x / 53) + (1.2-x) / 690.02 = 69 x+ 63.6−53x365773.14 = 16x + 63.6x = 0.59 = 0.6 g (approx.)Mass of Na2CO3 = 0.6 gIn the second part of query Na2CO3 nad K2CO3 react with Barium chloride to produce precipitate of barium carbonate.100 mL of solution contains 0.02 g equivalents20 mL of solution contains (0.02 / 100) × 20= 0.004 g equivalentMass of Barium carbonate formed will be equivalent-mass X number of gram-equivalentsEquivalent mass of Barium carbonate = 137.3 / 2= 68.65 gMass of barium carbonate = 0.004 ×68.650.2746 g Hope it helps!

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