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Determine the molar solubility for Al(OH)3 in pure water. Ksp for Al(OH)3 = 1.3 x 10-33 2.6 x 10-9 M is the answer,...

Determine the molar solubility for Al(OH)3 in pure water. Ksp for Al(OH)3 = 1.3 x 10-33


2.6 x 10-9 M is the answer, but how? Please explain.


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Use Al(OH)3(s) which has a Ksp = 1.3 x 10-33 M4

1. Write the chemical reaction.

Al(OH)3(s) <---> Al3+(aq) + 3OH-(aq)

2. Write the Ksp expression.

Ksp = [Al3+] [OH-]3

3. Let "s" equal the molar solubility of Al(OH)3(s).

From the reaction stoichiometry:

[Al3+] = s and [OH-] = 3s

Substitute into the Ksp expression:

Ksp = (s)(3s)3 = 1.3 x 10-33 M4

                      (s)(27s3) = 1.3 x 10-33 M4

                          27s4  = 1.3 x 10-33 M4

                            s4 = 4.8 x 10-35 M4

                             s = 2.6 x 10-9 M

So   [Al3+] = 2.6 x 10-9 M

and

[OH-] = 3(2.6 x 10-9 M) = 7.8 x 10-9 M


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