[demonat0r]

You have two 500.0 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 hte solid and find that it has a mass of 331.8 grams.
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.

So first i wrote out the balanced equation:
2AgNO3 + K2CrO4 ~~> Ag2CrO4 + 2KNO3

so it tells you that the precipitate which is silver chromate has a mass of 331.8 grams. I calculated the molar mass of silver chromate using my periodic table and found that it's molar mass is also 331.8 grams. i know the equation: moles = (grams) / (molar mass).

so i did (331.8 ) / (331.8 ) = 1 mole of silver chromate. and based on the stoichiometry, if 1 mole of silver chromate is produced, then 1 mole of potassium chromate is used up. So i calculated to molar mass of potassium chromate and found it to be 194.2 grams. and i can rearrage the equation to solve for grams: grams = (molar mass) (moles)

so (1 mole) (194.2 g / mol) = 194.2 grams. now the equation for molarity is as follows: M = (moles of solute / Liters of solution)

so i only have 1 mole of potassium chromate and i divide it by the volume of the original solution of it: (1) / (.5 L) = 2 M

and since there are 2 potassium ions per chromate ion i multiply 2 M by 2 to get 4 M concentration of K+ ions.

now i thought i did it correctly but when i checked the back of the book for the answer, it said 7.000 M K+

i'm really confused. i went over the problem several times and still cannot figure out why it's 7.000 M instead of just 4 M. please help me on this problem.

[Borek]

You are told there were identical masses of silver nitrate and potassium chromate dissolved. This is in fact a limiting reagent question. Check, if 194.2 g of AgNO₃ is enough to precipitate all chromate from the solution.