The solubility product of lead sulfate is at . Calculate the solubility of the salt in grams/liter at
Atomic Masses: O=16; S=32; Pb=207
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Great! I haven't learn anything about Solubility Product before, it does not appear in my textbook either! However, this is what I hope to learn from brilliant users. So, I went searching some information about solubility product and solved this problem.
Solubility Product = 2 . 2 5 × 1 0 − 8
[ P b 2 + ] [ S O 4 2 − ] = 2 . 2 5 × 1 0 − 8
As the amount of P b 2 + is equal with S O 4 2 − ,
[ P b 2 + ] 2 = 2 . 2 5 × 1 0 − 8
[ P b 2 + ] = 1 . 5 × 1 0 − 4 m o l / L
The molecular mass of P b S O 4 = 2 0 7 + 3 2 + 4 × 1 6 = 3 0 3
The solubility of the salt is,
1 . 5 × 1 0 − 4 m o l / L × 3 0 3 g / m o l = 0 . 0 4 5 4 5 g / L
I hope Gabriel can write a note about solubility product for other users who are not familiar with this concept. By the way, why use my name as the title of your problem!?