NEET · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
Find out the solubility of \(Ni ( OH )_{2}\) in \(0.1 M NaOH.\) Given that the lonic product of \(Ni ( OH )_{2}\) is \(2 \times 10^{-15}\)
- A \(1 \times 10^{8} \,M\)
- B \(2 \times 10^{-13}\, M\)
- C \(2 \times 10^{-8}\, M\)
- D \(1 \times 10^{-13}\, M\)
Answer & Solution
Correct Answer
(B) \(2 \times 10^{-13}\, M\)
Step-by-step Solution
Detailed explanation
\(\alpha=1\) for \(NaOH\) \(NaO{H_{(aq)}} \to \mathop {N{a^ + }}\limits_{0.1\,M} \left( {aq} \right) + \mathop {O{H^ - }}\limits_{0.1\,M} \left( {aq} \right)\)
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