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2 years ago

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.

SAT

1 answer:

Mariulka [41]2 years ago

3 0

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

<h3>What is solubility?</h3>

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: brainly.com/question/4736767

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