JEE MAIN - Chemistry (2019 - 12th April Morning Slot - No. 20)
What is the molar solubility of Al(OH)3 in 0.2 M NaOH solution ? Given that, solubility product of Al(OH)3 = 2.4 × 10–24
:
3 × 10–22
3 × 10–19
12 × 10–21
12 × 10–22
Explanation
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$${K_{sp}} = $$ [Al3+][OH-]3
$$ \Rightarrow $$ 2.4 × 10–24 = s(0.2)3
$$ \Rightarrow $$ s = $${{2.4 \times {{10}^{ - 24}}} \over {8 \times {{10}^{ - 3}}}}$$
$$ \Rightarrow $$ s = 3 × 10–22
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