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Topic: Mass of a precipitate  (Read 1302 times)

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Offline realMelody

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Mass of a precipitate
« on: September 17, 2023, 10:16:33 AM »
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?

Offline billnotgatez

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Re: Mass of a precipitate
« Reply #1 on: September 17, 2023, 10:33:02 AM »
to get help from someone you might show what you have done so far like
writing a balanced equation. there are buttons in the editor to help with this and latex is supported

edit
I have moved this entry to a forum board that may be more appropriate.
« Last Edit: September 17, 2023, 10:55:32 AM by billnotgatez »

Offline realMelody

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Re: Mass of a precipitate
« Reply #2 on: September 17, 2023, 11:22:37 AM »
I did try that but can't seem to find the right answer.
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.
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
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
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
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 :(

Offline mjc123

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Re: Mass of a precipitate
« Reply #3 on: September 17, 2023, 02:31:48 PM »
S = √(Ksp) only applies if [Ag+] = [AcO-]. Is that the case here?

Offline Hunter2

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Re: Mass of a precipitate
« Reply #4 on: September 17, 2023, 03:28:38 PM »
Last Calculations 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?
We have 0,02 mol Ag+ and 0,01 mol Acetate ions to combine.
« Last Edit: September 17, 2023, 03:42:44 PM by Hunter2 »

Offline Hunter2

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Re: Mass of a precipitate
« Reply #5 on: September 18, 2023, 03:59:31 AM »
Write the solubility product as a different with x.
Start concentration of Ag+ minus x and the same  with Acetate -x.
Concentration in mol/l . Mass calculated to 250 ml.
« Last Edit: September 18, 2023, 04:57:00 AM by Hunter2 »

Offline ProfOxidizer

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Re: Mass of a precipitate
« Reply #6 on: September 18, 2023, 08:54:09 AM »
Are you sure that answer is 0.48 g? Maybe 1.48 g?

Offline Hunter2

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Re: Mass of a precipitate
« Reply #7 on: September 18, 2023, 12:05:59 PM »
No its 0,485 g

Offline Aldebaran

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Re: Mass of a precipitate
« Reply #8 on: September 23, 2023, 07:02:44 AM »
@ ProfOxidizer

I agree with Hunter 2. 0.485g looks good to me.

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