A solution contains 0.0500 M Ca2 (aq) and 0.0350 M Ag (aq). What will be the concentration of Ca2 (aq) when Ag2SO4(s) begins to precipitate?
What percentage of the Ca2 (aq) can be separated from the Ag (aq) by selective precipitation?
-Solving the 1st one I got [Ca2+]= 5.03x10^-3 M
But, now how do I get the percentage by selective precipitation?
What Ksp values are you using? Did they come from your text/notes?
From my textbook,
Ksp(Ag2SO4)= 1.20x10^-5
Ksp(CaSO4)= 4.93x10^-5
I'm actually not sure how to solve it, now I think the [Ca+]= 0.0500M
Thanks. The values in my old text are different. I can't get your answer to the first part of 5.03E-5. How did you do that?
I think I did it wrong...
Ksp[Ag2SO4]= [SO42-][Ag+]^2
1.20x10^-5 = [SO42-][0.0350]^2
[SO42-}=9.80x10^-3M
Then I substituted into:
Ksp[CaSO4]= [SO42-][Ca2+]
4.93x10^-5 = [9.80x10^-3][Ca2+]
[Ca2+]= 5.03x10^-3M
You did it correctly. I forgot and used my Ksp values and not yours.
The way you do part b is this.
You have Ca^2+ when Ag2SO4 first starts to ppt (that's the 0.00503M). If we had a liter of that stuff, it would contain 0.00503 mols Ca^2+. So you have pptd at that point all but 0.00503; therefore, 0.05-0.00503 is the amount pptd at that point so the percent pptd is [(0.05-0.00503)/0.05]*100 = ? If I punched in the right numbers its approximately 90% (not a good separation technique if our goal was 100% selectivity). But then we should have realized from the close proximity of the Ksp values that we probably could not separate them completely. And we can't.
Thank you for your help!
What is the silver ion concentration in a solution prepared by mixing 375 mL of 0.363 M silver nitrate with 381 mL of 0.430 M sodium chromate? The Ksp of silver chromate is 1.2 × 10-12
To determine the percentage of Ca2+ that can be separated from Ag+ by selective precipitation, you need to calculate the amount of calcium sulfate (CaSO4) formed when the precipitation occurs.
Here are the steps to solve the problem:
1. Write the balanced chemical equation for the precipitation reaction:
Ag2SO4(s) + Ca2+(aq) → 2Ag+(aq) + CaSO4(s)
2. Determine the limiting reactant:
The reactant that will be completely consumed and limits the amount of product formed will be the limiting reactant.
In this case, Ca2+(aq) is the limiting reactant because there is less Ca2+(aq) present (0.0500 M) compared to Ag+(aq) (0.0350 M).
3. Calculate the moles of Ca2+ that can react:
moles of Ca2+ = concentration of Ca2+ × volume of the solution
In this case, the volume of the solution is not provided, so we cannot directly calculate the moles of Ca2+. Therefore, you need to assume a specific volume for the solution.
4. Calculate the moles of Ca2+ reacted with Ag2SO4:
According to the balanced chemical equation, 1 mole of Ca2+ reacts with 1 mole of Ag2SO4. Therefore, the moles of Ca2+ reacted will be equal to the moles of Ag2SO4 formed.
5. Calculate the moles of Ag2SO4 formed:
From the balanced chemical equation, the mole ratio between Ca2+ and Ag2SO4 is 1:1. Hence, the moles of Ag2SO4 formed will be the same as the moles of Ca2+ reacted.
6. Calculate the mass of Ca2+ that reacted with Ag2SO4:
To calculate the mass of Ca2+ that reacted, use the formula:
mass = moles × molar mass
The molar mass of Ca2+ is 40.08 g/mol.
7. Calculate the percentage of Ca2+ that can be separated:
The percentage of Ca2+ that can be separated is given by the formula:
percentage = (mass of Ca2+ reacted / mass of initial Ca2+) × 100%
Now, go back to step 3 and choose a specific volume for the solution to continue the calculations and find the percentage of Ca2+ that can be separated from Ag+.