Which Ion Precipitates Last as Na2SO4 Is Added?

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Discussion Overview

The discussion centers on a homework problem involving the precipitation of ions in a solution as sodium sulfate (Na2SO4) is added. Participants explore the order of precipitation for lead (Pb2+), calcium (Ca2+), and strontium (Sr2+) ions based on their solubility product constants (Ksp) and the concentrations of sulfate ions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant asserts that Pb2+ will precipitate first due to its lower Ksp value compared to Sr2+ and Ca2+.
  • Another participant agrees with the order of precipitation, suggesting that the concentration of sulfate ions at which each cation precipitates can be calculated based on their Ksp values.
  • A participant calculates the sulfate concentration required for SrSO4 to precipitate and finds it to be 3.4x10-6M.
  • There is a question about whether an ICE table is necessary for further calculations regarding Pb2+ precipitation.
  • Another participant challenges the previous calculation, indicating that once SrSO4 precipitates, the concentration of sulfate ions will increase, allowing for the next cation to precipitate.

Areas of Agreement / Disagreement

Participants generally agree on the order of precipitation (Pb2+ first, followed by Sr2+), but there is some disagreement regarding the necessity of an ICE table and the implications of sulfate concentration changes after precipitation occurs.

Contextual Notes

Participants express uncertainty about the need for an ICE table and the implications of changing sulfate concentrations on subsequent precipitations. The discussion does not resolve these uncertainties.

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Homework Statement


1. A 1L solution contains 0.1M of each Pb+2,Ca+2, and Sr+2. Which ion precipitates last as Na2SO4 is slowly added with no change in volume?
2. What is the concentration of the ion that precipitates first when the second ion precipitates?

Given:
Ksp PbSO4=1.8x10^-8
Ksp CaSO4=7.1x10^-5
Ksp SrSO4=3.4x10^-7


Homework Equations





The Attempt at a Solution


1. Pb2+ precipitates first as it has the lowest Ksp value.
2. This is where i get stuck. I know SrSO4 is the second ion to precipitate. I get:
ksp=[Sr2+][SO42-]
4x10^-7=[Sr2+][SO42-]

Don't know where to go from here. Help is appreciated.
 
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Not sure what your problem is - so far you are right, Pb will precipitate first, Sr second. Not many possibilities left at this stage.

Think about it this way - calculate (knowing cation concentrations) at what concentration of SO42- they will precipitate. Obviously when you add sulfate concentration of SO42- can only go up starting from zero, so when you sort cations according to concentration of sulfate at which they start to precipitate, you have your answer. Not surprisingly, as concentrations of cations are identical, list you are looking for will not differ from just the sorted list of Ksp values.

--
methods
 
So...
ksp=[Sr2+][SO42-]
3.4x10^-7=[Sr2+][SO42-]
3.4x10^-7=[0.1M][SO42-]
[SO42-]=3.4x10^-6M

Now in Pb2+
1.8x10^-8=[Pb2+][SO42-]
1.8x10^-8=[Pb2+][3.4x10^-6M]
[Pb2+]=5.29x10^-3M

Is this how i do it, or do i have to set up an ICE table?
 
No, you got it wrong. Strontium sulfate starts to precipitate when sulfate concentration is as you have calculated, that's OK. At this moment nothing else can precipitate. But then, when you add more sodium sulfate, strontium is removed from the solution and concentration of free SO42- goes up - till it is high enough to precipitate next cation.

--
methods
 

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