Question

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

Answer 1

Answer:

Option b)

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Answer 2

The molar solubility of  $Al(OH)_3$ in 0.2 M NaOH  is $3\times 10^{-22}mol/L$

Explanation:

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

As 1 mole of $Al(OH)_3$ dissociates to give 1 mole of $Al^{3+}$ and 3 moles of $OH^-$  ions.

If solubility of $Al(OH)_3$ = x mol/L

solubility of aluminium ions = x mol/L

solubility of hydroxide ions = 3x mol/L

${K_{sp}}$ of $Al(OH)_3=(x)^1\times (3x)^3$

But as 0.2 M of NaOH is also present

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

Given solubility of $NaOH$ = 0.2 mol/L

solubility of sodium ions = 0.2 mol/L

solubility of hydroxide ions = 0.2 mol/L

$K_{sp}=(x)^1\times (3x+0.2)^3$

Solubility product of $Al(OH)_3=2.4\times 10^{-24}$

$2.4\times 10^{-24}=(x)^1\times (3x+0.2)^3$

$x=3\times 10^{-22}mol/L$

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