Posted by CP on Monday, March 11, 2013 at 7:23pm.
This is a limiting reagent problem, a common ion problem, and a Ksp problem all rolled into one.
345 mL x 0.352M AgNO3 = 121.44 mmoles. We'll round this to 121 mmols AgNO3.
443 mL x 0.511M Na2CO3 = 226.373 mmols. We'll round this to 226 mmols Na2CO3.
You can round them differently and/or carry more places if you wish.
2AgNO3 + Na2CO3 ==> Ag2CO3 + 2NaNO3
First we find the limiting reagent
Convert 121 mmols AgNO3 to mmols Ag2CO3. That's 121 x (mols Ag2CO3/2 mols AgNO3) = 121 x (1/2) = 60.5 mmols Ag2CO3.
Convert 226 mmols Na2CO3 to Ag2CO3. That's 226 x (1 mol Ag2CO3/1 mol Na2CO3) = 226 x 1/1 = 226.
Obviously both answers can be right; the correct answer in limiting reagent problems is ALWAYS the smaller one and the reagent producing that value is the limiting reagent. Therefore, AgNO3 is the limiting reagent.
Next we determine how much of the Na2CO3 is used.
mmols Na2CO3 used = 121 mmols AgNO3 x (1 mol Na2CO3/2 mol AgNO3) = 121 x (1/2) = 60.5 mmols used. So how much is left unused? That's 226 - 60.5 = 165.5.
That 165.5 mmols Na2CO3 is the common ion. What's the concn now? We have 165.5 mmols and it's in a volume of 345 mL + 443 mL = 788 mL so the concn is
M = mmols/mL = 165.5/788 = about 0.2 but you do it more accurately than that.
.........Ag2CO3 ==> 2Ag^+ + CO3^2-
I.........solid......0......0.2
C.........solid.....+2x.....+x
E.........solid.....2x....0.2+x
Ksp = (Ag^+)^2(CO3^2-)
8.1E-12 = (2x)^2(0.2+x)
Solve for x, then 2x = (Ag^+).
These problems are too long. I wonder if this is for a general chemistry class or an advanced class.