Mass of a precipitate

[realMelody]

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 if50 mL of 0.200 mol/L CH₃COONa is mixed with200 mL 0.100 mol/L AgNO₃?

This is what I have done so far:

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 gAnd then I subtracted the latter from the former and 1.836 -1.669 g ≈ 0.167 g but the correct answer is 0.484g

Can someone help me please?

[ChenBeier]

Calculation 4 and 5 are not logic and wrong.

You found 0,01 mol CH₃COOAg will be formed from 250 ml.
S = 0,044 mol/l what means 0,011 mol for 250 ml.
So if 0,011 mol is still soluble how can 0,01 mol then precipitate?

Calculate
Ksp = (cAg+ -x)*(cCH₃COO- -x)
c in mol/l convert.
Solve for x and divide by 4 back to 250 ml.
This times molare mass gives the right result.