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.

Answer 1

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) + CrO₄^2- (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 CrO₄^2-

chromate ion (CrO₄^2- ) 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 (CrO₄^2- ) in final solution = 1 moles / 1L = 1 mol/L or 1M Answer (b)