You have two 407.0 mL aqueous solutions...?

Question

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

(a) Calculate the concentration of the potassium ions in the original potassium chromate solution in M

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

Answer 1

silver chromate is Ag2CrO4 Mwt=331.8
2AgNO3 +K2CrO4 = Ag2CrO4 + 2 KNO3
You have 1 mole of Ag2CrO4 so 1 mole K2CrO4 194.2g
& 2 mole AgNO3=2 x169.9=339.8g - since same wt of both then excess of 145.6 g K2CrO4 left = 0.7497 moles/.407 l =1.842M
[CrO4] = 1mole per 0.914 L = 1.094M

Answer 2

I presume you have formed Ag2(CrO4), and the ppt. contains all of the original silver, but not necessarily all of the chromate. With this you first have to find the moles of silver chromate. This is 331.8/mol wt Ag2CrO4. Call this B. Then the g-atoms of silver in the initial AgNO3 solution is 2B. Then we can compute the MASS of AgNO3 in this solution as 2B x mole wt AgNO3.

This is, by the problem statement, the initial mass of K2CrO4. So we can find the initial moles of CrO4- from this (initial mass * molwt CrO4/mol wt K2CrO4). Call this C. C should be > than B, and B-C is the moles of CrO4 in final solution. The volume is 814 mL (assuming no change for reaction products), so you can compute the CrO4- concentration. Since you know C, you can solve part (a).