PLEASE HELP ASAP

Question

The solubility-product constant for Ag2CrO4 is 1.2 × 10-12. What is the silver ion concentration in a solution prepared by mixing 50.00 mL of 0.032 M AgNO3 with 50.00 mL of

0.010 M Na2CrO4

0.250 M Na2CrO4

PLEASE WRITE THE SOLUTION WITH STEPS

Answer 1

I had to do some research to do this problem. The first thing you have to do is determine the number of moles of AgNO₃ you have and the number of moles of Na₂CrO₄ you have after mixing the two solutions.

The number of moles of AgNO₃ is 0.032 mol/L / 1000 mL/L x 50.0 mL = 0.0016 moles.

The number of moles of Na₂CrO₄ is 0.10 mol/L / 1000 mL/L x 50.0 mL = 0.005 moles.

 The balanced equation for the reaction is  2AgNO₃ + Na₂CrO₄   →        Ag₂CrO₄ + 2NaNO₃

It takes 2 moles of AgNO₃ for every mole of Na₂CrO₄ ,so according to the balanced equation, you would need 0.005 x 2 moles = 0.01 moles of AgNO₃. But, there are only 0.0016 moles available, so AgNO₃ is a limiting reactant. Only 0.0016 /2 = 0.0008 moles of Na₂CrO₄  can react to form Ag₂CrO₄. There will be excess Na₂CrO₄ which would be 0.005 – 0.0008 = 0.0042 moles left over.

Next calculate the molarity of the excess Na₂CrO₄. This would be 0.0042 mol /100 mL x 1000 mL/L = 0.042 M.

The Ag₂CrO₄ concentration is the unknown.

Let X be the concentration of dissolved Ag¹⁺. Ksp = [Ag¹⁺]² [CrO₄^2-]. The CrO₄²- concentration is ½ the Ag¹⁺ + 0.042 M due to the excess chromate.

Ksp = [X]² *[0.5X + 0.042]. As Ag₂CrO₄ is very insoluble, the X in the second bracket can be dropped which leads to Ksp = [X]² * 0.042. Ksp is 1.2 x 10^-12 so, this equation 1.2 x 10^-12 = [X]² * 0.042 can be solved for X to give X = √(1.2 x 10^-12 /0.042) = 5.4 x 10^-6 M.

This is a very lengthy explanation, so you can work the second part yourself by doing the same steps but substituting 0.25M for the Na₂CrO₄ molar concentration.