Question

What is solubility of a l ( o h ) 3 , ( k s p = 1 × 10 − 33 ) in a buffer solution ph = 4?

Answer 1

x(3x+10^-10) = 1×10^-33

x(10^-10) = 1×10^-33 ( here 3x negligible as compared to 10^-10 )

Than

x = 10^-3

Answer 2

Given:

Substance = $Al(OH)_3$

$k_{sp}$ = $1\times10^{-33}$

pH = 4

To find:

Solubility = ?

Calculation:

Dissociation of $Al(OH)_3$, yields

$Al(OH)_3$ ⇄ $Al^{3+} + 3OH^-$

Consider $Al^{3+}$ as S, whereas $3OH^-$ as 3S.

'S' is solubility

∴$k_{sp}=[Al^{3+}][OH^-]^3$

$k_{sp}$ = $[S]\times [3S]^3$

$k_{sp}$ = $27S^4$

$27S^4$ =  $1\times10^{-33}$

$S^4=\frac{1\times10^{-33}}{27}$

S = $\sqrt[4]{\frac{1\times10^{-33}}{27} }$

S = $3.08\times 10^{-8} mol L^{-1}$

Conclusion:

The solubility of solution is calculated as $3.08\times 10^{-8} mol L^{-1}$.

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