Determine the molar solubility of Al(OH)3 in a solution containing 0.0500M AlCl3, Ksp (Al(OH)3)=1.3*10^-33

Question

Question

Answer 1

Answer #1

AlCl3 here is Strong electrolyte

It will dissociate completely to give [Al3+] = 0.05 M

At equilibrium:

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

5*10^-2 +s 3s

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

1.3*10^-33=(5*10^-2 + s)*(3s)^3

Since Ksp is small, s can be ignored as compared to 5*10^-2

Above expression thus becomes:

1.3*10^-33=(5*10^-2)*(3s)^3

1.3*10^-33= 5*10^-2 * 27(s)^3

s = 9.875*10^-12 M

Answer: 9.9*10^-12 M

Answer 2

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