OL 8 87. You have two 500.0 mL aqueous solutions. Solution A is a solution of si

Question

OL 8 87. You have two 500.0 mL aqueous solutions. Solution A is a solution of silver nitrate, and solution B is a solution of potas- sium chromate. The masses of the solutes in each of the so- lutions are the same. When the solutions are added together, a blood-red precipitate forms. After the reaction has gone to completion, you dry the solid and find that it has a mass of 331.8 g. a. Calculate the concentration of the potassium ions in the original potassium chromate solution. b. Calculate the concentration of the chromate ions in the final solution.

Explanation / Answer

Balanced reaction when mix the two solution is

2 AgNO3 (aq) + K2CrO4 (aq) ---> Ag2CrO4 (s) + 2 KNO3 (aq)

2 mole 1mole 1 mole 2 mole

Given

Mass of solid ( Ag2CrO4 ) = 331.8 g

Molar mass of Ag2CrO4 = 331.73 g/mol

No. of moles of Ag2CrO4 = Mass / molar mass = 331.8 g / 331.73 g/mol = 1 moles of Ag2CrO4

according the reaction stoichometry

1 mole of K2CrO4 will produce 1 mole of Ag2CrO4  

so there was 1 mole of K2CrO4 in 500 ml of solution

diassociation reaction for K2CrO4

K2CrO4 (aq) ----> 2 K+ (aq) + CrO42- (aq)

1 mole 2 mole 1 mole

1 mole fo K2CrO4 will be diassociated into 2 moles of K+ in the solution A

Volume of solution A = 500 ml = 0.5 L

so concentration of K+ (potassium) ions in A = 2 moles / 0.5 L = 4 mol/L (or M) Answer (a)

1 mole of K2CrO4 willl be diassociated into 1 mole of CrO42-

chromate ion (CrO42- ) will not form any preciptate during reaction so it will remain same in the final solution also

Volume of final solution = 1 L

Concentration of Chromate ion (CrO42- ) in final solution = 1 moles / 1L = 1 mol/L or 1M Answer (b)

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