Which Ion Precipitates Last as Na2SO4 Is Added?

Click For Summary
SUMMARY

The discussion centers on the precipitation order of ions in a 1L solution containing 0.1M of Pb2+, Ca2+, and Sr2+ when Na2SO4 is added. PbSO4 precipitates first due to its lowest solubility product constant (Ksp = 1.8 x 10-8), followed by SrSO4 (Ksp = 3.4 x 10-7). The concentration of sulfate ions ([SO42-]) required for SrSO4 to precipitate is calculated to be 3.4 x 10-6M. The discussion concludes that as Na2SO4 is added, the concentration of sulfate ions increases, allowing for the sequential precipitation of the cations based on their Ksp values.

PREREQUISITES
  • Understanding of solubility product constants (Ksp)
  • Knowledge of precipitation reactions in aqueous solutions
  • Ability to perform concentration calculations
  • Familiarity with ionic equations and equilibrium concepts
NEXT STEPS
  • Study the concept of solubility product constants (Ksp) in detail
  • Learn how to construct and interpret ICE tables for equilibrium reactions
  • Explore the effects of common ion effect on solubility
  • Investigate the precipitation sequences of other ionic compounds
USEFUL FOR

Chemistry students, educators, and professionals involved in analytical chemistry or environmental science, particularly those focusing on precipitation reactions and solubility equilibria.

Macroer
Messages
27
Reaction score
0

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.
 
Physics news on Phys.org
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
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 13 ·
Replies
13
Views
3K
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
6K
Replies
2
Views
7K
  • · Replies 2 ·
Replies
2
Views
3K