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Determine the molar solubility of al(oh)3 in a solution containing 0. 05 m alcl3. Ksp for al(oh)3 is 1. 3 × 10-33.

Answer :

PSTNONSONJOKUGO

From the calculations carried out , we can see that the molar solubility of Al(OH)3 is 8.6 x 10^-12 M  

What is solubility?

The term solubility refers to the amount of substance that dissolves in a solution measured in moles per liter.

From the question we have initial  [Al+3] = 0.05 M

Setting up an ICE table:

             Al(OH)3(s) → Al3+  +  3OH-

I                                    0.05         0

C                                 +x           +3x

E                                0.05+x         3x

Know that;

Ksp = [Al3+] [OH-]^3

Ksp = 1.3 x 10^-33

So;

[Al3+] = (0.05+X) and [OH-]^3 = 3X

If x = molar solubility of Al(OH)3

Then:

1.3 x 10^-33 = (0.05 + X) (3X)^3 by solving for X

x = 8.6 X 10^-12 M

Learn more about Ksp: https://brainly.com/question/4736767

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