Calculate the molar solubility of aluminum hydroxide (AlOH)3) in a 0.027M solution of aluminum nitrate (Al(NO3)3). The Ksp of AlOH)3 is 2.0x10-32.

s = 5.22x10⁻⁹ M

Explanation:

Let's write the equilibrium reaction for this compound:

Al(OH)₃ <> Al³⁺ + 3OH⁻      Ksp = 2x10⁻³²

To get the solubility, we need to write the equilibrium expression:

Ksp = [Al³⁺] [OH⁻]³

Al(OH)₃ does not contribute to the equilibrium expression, because is a solid compound. Now, according to this, we have the following:

                   Al(OH)₃ <> Al³⁺ + 3OH⁻

I.                                             0          0

E.                                            s          3s

Replacing into the Ksp expression:

2x10⁻³² = (s) (3s)³

2x10⁻³² = s * 27s³

2x10⁻³² = 27s⁴

s = ⁴√2x10⁻³² / 27

This would be the solubility of the compound.

Hope this helps

the answer is b & c i believe : )

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