[TEA1]

A city water supply has a lead ion (Pb2+) level of 0.0025 g/L in a tank that is 1.60e7 L. The city wants to remove 98% of the lead ion as PbSO4(s). How many grams of sodium sulfate (Na2SO4 FW 142.05) must be added to the water tank to remove 98% of the lead ions? The Ksp of PbSO4 is 6.3e-7. Amount of Pb2+ remaining will be 2% of the original.

 I got the grams of Pb2+ ions by multiplying 0.0025g/L and 1.60e7 L. But not sure what to do with Ksp.  Thanks

[Borek]

Calculate required final Pb²⁺ concentration. Then you will need K_so to calculate necessary concentration of SO₄^2-.

[TEA1]

 I am still a little bit confused on how to do this.  I found the final Pb2+ conc. by making an ICE chart using the Ksp.  From there can I use the Pb M conc. and convert it to grams of Sodium sulfate, or am I missing a step?

[TEA1]

So apparently no ICE chart is needed, figured that one out.  I found conc of Pb (1.21e-5M) and used Ksp to find conc. of SO4 (0.0521M).  So now my question is, can I use the Molar of SO4 found and convert that to grams of sodium sulfate?  Does the percent have to be involved in the calculation? Because if not I get 1.18e8 g as the answer.  I am wondering if that is correct?

[Borek]

Perhaps you have missed SO₄^2- in precipitate?