[kdbmvp]

Hello,

I am a bit confused re. solubility calculations.

Calculate the solubility of Pb(OH)₂ at pH 10. Setting up the expression for Ksp:

Ksp = [Pb²⁺][OH^-]² = 8* 10^-17 (Ksp value from SI chemical data)

pOH = 14 - 10 = 4, i.e. [OH^-] = 10^-4.

8*10^-17 = [Pb²⁺][OH^-]². Solving for [Pb²⁺] we get that [Pb²⁺] = 8*10^-9.

My question is: why couldn't we use the relationship between the concentrations of Pb²⁺ and OH^- like we do with ICE tables, to get the result? I mean, couldn't we just divide the concentration of OH^- by 2 and get the concentration of Pb²⁺..?

[chenbeier]

This would be possible, if lead hydroxide would be dissolved like sodium hydroxide. But it doesn't. You need the soloubility product, because the soloubility is poor.

[kdbmvp]

Thanks. Is it possible to deduce that from the Ksp value?

[chenbeier]

How you want to do it?

[kdbmvp]

I just wondered if you could from the Ksp value see that lead hydroxide has poor solubility, and that it wouldn't be possible to use the relationship between the concentrations.

[chenbeier]

Of course you can.  Check the formula between soloubility, concentration and soloubility product

[kdbmvp]

You mean the equation for Ksp?

[chenbeier]

Yes and No

Read here: https://chemistry.stackexchange.com/questions/75592/for-the-solubility-of-ab2-ksp-ab2-x2x2-why-is-2x2-not-considered-d