# chemistry

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What is the silver ion concentration in a solution prepared by mixing 385 mL of 0.380 M silver nitrate with 467 mL of 0.409 M sodium chromate? The Ksp of silver chromate is 1.2 × 10-12

I ended up making Ag+ = 2x. = 1.3 x10^-4 Which is incorrect so I'm not sure how to go about doing this.

• chemistry - ,

This is a limiting reagent and a common ion problem rolled into one.
2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NaNO3

385 mL x 0.380M = 146.3 mmols = 0.1463 mols AgNO3
467mL x 0.409M = 191 mmols = 0.191 mol Na2CrO4.
Which is the limiting reagent?
0.1463 mols AgNO3 x (1 mol Ag2CrO4/2 mol AgNO3) = 0.1463 x 1/2 = 0.0732 mol Ag2CrO4 formed.
Do the same for Na2CrO4. That will be
0.191 mol Na2CrO4 x (1 mol Ag2CrO4/1 mol Na2CrO4) = 0.191 x 1/1 = 0.191 mols Ag2CrO4 formed.
Both of these values can't be right; one must be wrong. In limiting reagent problems, the smaller value is ALWAYS the correct one and the reagent producing that value is the limiting reagent. Therefore, AgNO3 is limiting, meaning that all of the AgNO3 is used and Na2CrO4 is in excess.
Next determine the amount of Na2CrO4 in excess.
0.1463 mol AgNO3 x (1 mol Na2CrO4/2 mol AgNO3) = 0.0732 mols Na2CrO4 used.
initial = 0.191
used = 0.0732
Na2CrO4 remaining = 0.118.
Now we have a solubility problem with a common ion of CrO4^2-
........Ag2CrO4 ==> 2Ag^+ + CrO4^2-
........solid.......2x........x
Ksp = (Ag^+)^2(CrO4^2-)
(Ag^+)^2(CrO4^2-) = 1.2E-12
Substitute into Ksp expression the following:
(Ag^+) = 2x
(CrO4^2-) = x + (0.118 mol/0.385L+0.467L) = xM + 0.138M

Solve for x and obtain 2x. 2x is (Ag^+)
My best guess is that you were ok until you came to the common ion but didn't take that into account. I came up with 2.95E-6M = (Ag^+) but I could have punched the wrong buttons. Check my work above carefully. The chemistry is ok but sometimes I hit the wrong keys on the calculator.

• chemistry - ,

Thank you so much. I went back to my notes where I was trying to work it out. I did mess up with the common ion. Ag2+. It was set up pretty much just as you had it. Thanks again.