Project Of Christopher Boo

Chemistry Level 2

The solubility product of lead sulfate is 2.25 × 1 0 8 2.25 \times 10^{-8} at 2 5 C 25^\circ \text{C} . Calculate the solubility of the salt in grams/liter at 2 5 C . 25^\circ\text{C}.

Atomic Masses: O=16; S=32; Pb=207

0.0230 g/L 0.0230\text{ g/L} 0.0345 g/L 0.0345\text{ g/L} 0.0454 g/L 0.0454\text{ g/L} 0.8458 g/L 0.8458\text{ g/L}

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1 solution

Christopher Boo
Jan 27, 2014

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 = 2.25 × 1 0 8 \text{Solubility Product}=2.25\times10^{-8}

[ P b 2 + ] [ S O 4 2 ] = 2.25 × 1 0 8 [Pb^{2+}][SO_4^{2-}]=2.25\times10^{-8}

As the amount of P b 2 + Pb^{2+} is equal with S O 4 2 SO_4^{2-} ,

[ P b 2 + ] 2 = 2.25 × 1 0 8 [Pb^{2+}]^2=2.25\times10^{-8}

[ P b 2 + ] = 1.5 × 1 0 4 m o l / L [Pb^{2+}]=1.5\times10^{-4} mol/L

The molecular mass of P b S O 4 = 207 + 32 + 4 × 16 = 303 PbSO_4=207+32+4\times16=303

The solubility of the salt is,

1.5 × 1 0 4 m o l / L × 303 g / m o l = 0.04545 g / L 1.5\times10^-4 mol/L\times303g/mol=0.04545g/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!?

Good Job Christopher Boo !! Excuse-me , Sorry Christopher Boo !!

Gabriel Merces - 7 years, 4 months ago

I wonder. Which book do you follow? I guess most of the books on Physical Chemistry do introduce the Solubility Product.

Pranav Arora - 7 years, 4 months ago

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Oh, my mistake, I have checked again and the solubility product is in another chapter in my school textbook.

Christopher Boo - 7 years, 4 months ago

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