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in Chemistry by (15 points)
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The solubility product of Al(OH)3 is 1×10-36.Calculate the solubility of Al(OH)3​.

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1 Answer

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Al(OH)3 \(\leftrightharpoons\) Al+3(aq)+ + 3OH-(aq)

using law of mass action

K = \(\frac{[Al^{+3}][OH^-]^3}{[Al(OH^-)_3]}\) 

⇒ K = equilibrium constant

\(\because\) [al(OH)_3] = constant

\(\therefore\) Rearranging above equation.

K x [Al(OH)3] = [Al+3] [OH-]3

Ksp = [Al+3] [OH-]3

where Ksp = [S] [3S]3

= s x 27S3

Ksp = 27s4

We have given, 

Ksp = 1 x 10-36

\(\therefore\) S4 = \(\frac1{27}\times10^{-36}\)

= 3.7 x 10-38

S4 = 370 x 10-40

S = 4.39 x 10-10 g/L

Hence, solubility of Al(OH)3 will be 4.39 x 10-10 g/L

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