Question

74. What is the molar solubility of Al(OH)3, in 0.2M NaOHsolution? Given that, solubility product ofAl(OH)3 = 2.4 x 10^-24
a) 3 x 10-19b) 12 x 10-21c) 3 x 10-22d) 12 x 10-23​

Answer 1

The molar solubility of $Al(OH)_3$ is $3\times 10^{-22}M$

Explanation:

The equation for the reaction will be as follows:

$Al(OH)_3\rightleftharpoons Al^{3+}+3OH^-$

By Stoichiometry,

1 mole of  $Al(OH)_3$ gives 1 mole of $Al^{3+}/tex] and 3 mole of [tex]OH^-$

Thus if solubility of $Al(OH)_3$  is s moles/liter, solubility of $Al^{3+}/tex]i s s moles\liter and solubility of [tex]OH^-$ is 3s moles/liter

Therefore,  

$K_sp=[Al^{3+}][OH^{-}]^3$

$K_sp=[s][3s]^3$[/tex]

$NaOH\rightarrow Na^++OH^-$

1 mole of NaOH gives 1 mole of $Na^+$ and 1 mole of $OH^-$

0.2 moles of NaOH gives 0.2 mole of $Na^+$ and 0.2 mole of $OH^-$

$K_sp=[s][3s+0.2]^3$

$2.4\times 10^{-24=[s][3s+0.2]^3$

$s=3\times 10^{-22}M$

Hence, the molar solubility of $Al(OH)_3$ is $3\times 10^{-22}M$

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