In an aqueous reaction, barium chloride plus potassium chromate yields potassium

Question

In an aqueous reaction, barium chloride plus potassium chromate yields potassium chloride and barium chromate, 100. grams of barium chloride is mixed with sufficient water to make the volume exactly 1000 ml. It is titrated with a solution of 100 grams of potassium chromate dissolved in sufficient water to make exactly 2000. ml of solution. Calculate the molarity of the potassium chromate solution and if 200. grams of barium chloride were dissolved in sufficient water to make 2.000 liters, how many ml of the potassium chromate solution would be required to react completely with 35.00 ml of the barium chloride solution?

Explanation / Answer

BaCl2 Molar mass: 208.23 g/mol

100 g Bacl2= 100/208.23 = 0.480 mol

Molarity = 0.48 mol/L = 0.480 M

K2CrO4 Molar mass: 194.19 g·mol1

Moles of K2CrO4 = 100 /194.19 = 0.515 moles

MoLarity = no.of moles / Volume in litres = 0.515/2 = 0.258 M

200 g BaCl2 = 200/208.23 = 0.960 mol

Molarity = 0.960 mol/2L = 0.480 M

BaCl2 + K2CrO4 ----------->BaCrO4 + 2KCl

Since for complete reaction 1 mole BaCl2 needs 1 mole K2CrO4. We can use equation

V1 M1 = V2 M2

35 ml x 0.480 M = V2 x 0.258 M

V2 = (35 ml x 0.480 M ) / 0.258 M = 65.1 ml

Volume of K2 CrO4 = V2 = 65.1 ml

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