Mass of a precipitate: solubility and Ksp in a chemical reaction

[realMelody]

TL;DR Summary

Mass of a precipitate

I have a question related to solubility and Ksp in a chemical reaction. I don't want to seek answers to a specific homework problem, but I'm genuinely stuck and seeking a deeper understanding of the concept. Can someone please explain the process for determining the mass of a precipitate in a given scenario when considering the solubility product constant (Ksp)? I'm not sure how to proceed, and I'm looking for guidance on the general approach.

This specifically is the question I have been having some problems with:
How many grams of precipitate would we have if
50 mL of 0.200 mol/L CH3COONa is mixed with
200 mL 0.100 mol/L AgNO3?

 

[Borek]

Can you calculate (from a simple stoichiometry) how much precipitate would be produced if the reaction went to completion?

Can you calculate (from Ksp) how much of the substance will not precipitate and will stay on the solution?

If so, just subtract latter from the former.

 

[realMelody]

I did try that but can't seem to find the right answer.

1. How much precipitate would be produced if the reation went to completion:
   • Moles of CH3COONa = Concentration x Volume = 0.200 mol/L x 0.050 L = 0.010 mol
   • Since the reaction is 1:1 between CH3COONa and AgCH3COO, the moles of AgCH3COO produced will also be 0.010 mol.
2. Calculate the mass of AgCH3COO produced:
   • Molar mass of AgCH3COO = 107.87 g/mol (Ag) + 12.01 g/mol (C) + 3 x 1.01 g/mol (H) + 2 x 16.00 g/mol (O) = 166.912 g/mol
   • Mass of AgCH3COO produced = Moles x Molar mass = 0.010 mol x 166.912 g/mol ≈ 1.669 g
3. Calculate the molar solubility (amount that stays dissolved) using the corrected Ksp:
   • Ksp for AgCH3COO is given as 1.94 x 10^-3 mol^2/L^2.
   • To find the molar solubility (S), we'll take the square root of Ksp.
   • S = √(Ksp) = √(1.94 x 10^-3 mol^2/L^2) ≈ 0.044 mol/L
4. Calculate the moles of AgCH3COO that will stay dissolved (molar solubility):
   • Moles of AgCH3COO not precipitated = Molar solubility x Volume = 0.044 mol/L x 0.250 L = 0.011 mol
5. Calculate the mass of AgCH3COO that will not precipitate:
   • Mass of AgCH3COO not precipitated = Moles x Molar mass = 0.011 mol x 166.912 g/mol ≈ 1.836 g
     And then I subtracted the latter from the former as you said and 1.836 -1.669 g ≈ 0.167 g
     But the correct answer is 0.484g :(

 

[mjc123]

See my comment in chemicalforums.

 

[Borek]

As @mjc123 signaled at CF - you are on the right track, you are just incorrectly assuming whatever is left in the solution is given by the molar solubility of the salt. But that's not the case, solution contains excess of one of the ions involved.

 

[realMelody]

O I see. But how should I go about solving the problem then?

 

[Borek]

One of the reagents is in excess - you can at first assume reaction went to completion and concentration of one of the ions is that of the left excess and use it to calculate concentration of the other ion.

But that's only a first approximation. For cases where Ksp is low enough it will work OK, but not necessarily here. You will probably need to use ICE table.

You can think of the problem in a bit different way, perhaps that will make it easier. It doesn't matter how the solution was prepared - as long as it contains same amount of all ions. Imagine you have some amount of solid salt (calculated from the given data assuming reaction went to completion) and your drop it into solution made of whatever was left at the end of the reaction (so all excess ions). Now it is a matter of finding out how much solid will dissolve - that's where the ICE table comes handy, as it will help keeping a track of by how much concentration of the ion left in the solution goes up during the dissolution.