Answer to Question #64828 in General Chemistry for Aylin Salas

Question

You have two 466.0 mL aqueous solutions. Solution A is a solution of silver nitrate, and solution B is a solution of potassium chromate. The masses of the solutes in each of the solutions 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.


M

(b) Calculate the concentration of the chromate ions in the final solution

Answer 1

Answer:
2AgNO3 + K2CrO4 = Ag2CrO4 + 2 KNO3
n = m/M M (Ag2CrO4) = 331.7 g/mol M (AgNO3)=170 g/mol M (K2CrO4)=194 g/mol
n (Ag2CrO4) = 331.8 / 331.7 = 1.00 mol

a) 1 mol of precipitate of Ag2CrO4 means two moles of AgNO3 were in 500 ml solution.
So its mass is 2·M (AgNO3) = 340 g which is also the mass of K2CrO4.
The number of moles of K2CrO4: 340 g/194 g/mol = 1.75 mol from potassium chromate formula you can see that there are two moles of potassium per mol of K2CrO4 so in 1.75 you would have 2 : 1=n (K) : 1,75 n (K) =3.5.
The concentration of K+ ions n (K)/0.5 dm3 = 7 mol/dm3
b) 1,75 moles of K2CrO4 reacting with 2 moles of AgNO3 the reaction product will be 1 mol of silver chromate and 0.75 moles of excess potassium chromate so the concentration of chromate ions in new solution is:
N (CrO42-)/V = 0.75/1 = 0.75 mol/dm3.