JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
What is the molar solubility of \(Al(OH)_3\) in \(0.2\, M\, NaOH\) solution? Given that, solubility product of \(Al(OH)_3 = 2.4 \times 10^{-24}\)
- A \(3 \times {10^{ - 19}}\)
- B \(12 \times {10^{ - 21}}\)
- C \(12 \times {10^{ - 23}}\)
- D \(3 \times {10^{ - 22}}\)
Answer & Solution
Correct Answer
(D) \(3 \times {10^{ - 22}}\)
Step-by-step Solution
Detailed explanation
\(Al(O{H_3}) \rightleftharpoons \mathop {A{l^{3 + }}}\limits_s + \mathop {3O{H^ - }}\limits_{(3s + 0.2)} \) \(2.4 \times {10^{ - 24}} = s{(3s + 0.2)^3}\)
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