Equilibrio of a sparingly soluble ionic compound

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SUMMARY

The discussion centers on the solubility equilibrium of lead(II) sulfate (PbSO4) and the calculation of its concentration in a saturated solution. The participants debate the correct concentration of lead ions (Pb^2+) in solution, with one user asserting a value of 4.33 x 10^-5 mol/L while another cites 8.4 x 10^-5 mol/L. Discrepancies arise from differing solubility product constants (Ksp) for PbSO4, with values ranging from 1.06 x 10^-8 to 2.13 x 10^-8 reported from various sources. Accurate calculations depend on the correct Ksp value used in the solubility equilibrium expression.

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daphnelee-mh
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Homework Statement
Calculate the concentration of the first cation until second cation just starts to precipitate as sulfate
Relevant Equations
Ksp
I got the first question correct but I don’t understand the answer given for second question which is c (Pb^2+) = 8.4x10^-5 molL ^-1. It divided the concentrate for sulphate ion of Ag2SO4 into 2 , but it is the solubility right ?
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This is very difficult to read :mad:

daphnelee-mh said:
It divided the concentrate for sulphate ion of Ag2SO4 into 2

As far as I can tell - it didn't.
 
1577086669679.png

My answer is 4.33 x 10^-5, but answer given is 8.4x10^-5, how can I get it?
 
Next time please type the answer (takes less time than writing and taking pictures).

As far as I can tell answer given is wrong, your approach looks OK.
 
Sorry, my computer have problems that time, I can just upload using phone. Therefore, is it the given answer is wrong?
 
Looks like, I got the same answer you did.
 
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Were you given the Ksp for PbSO4?
Wikipedia has it at 2.13X10^-8 (20C), the CRC has it at 1.06X10^-8 (18C) and you are using 1.3X10^-8!
 

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