AP EAMCET · Chemistry · Ionic Equilibrium
If the ionic product of \(\mathrm{Ni}(\mathrm{OH})_2\) is \(1.9 \times 10^{-15}\), the molar solubility of \(\mathrm{Ni}(\mathrm{OH})_2\) in \(1.0 \mathrm{M} \mathrm{NaOH}\) is
- A \(1.9 \times 10^{-18} \mathrm{M}\)
- B \(1.9 \times 10^{-13} \mathrm{M}\)
- C \(1.9 \times 10^{-15} \mathrm{M}\)
- D \(1.9 \times 10^{-14} \mathrm{M}\)
Answer & Solution
Correct Answer
(C) \(1.9 \times 10^{-15} \mathrm{M}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{NaOH} \rightleftharpoons \underset{\mathrm{CM}}{\mathrm{Na}^{+}}+\underset{\mathrm{CM}}{\mathrm{OH}^{-}} \\ & \mathrm{Ni}(\mathrm{OH})_2 \rightleftharpoons \mathrm{Ni}_x^{2+}+\underset{x}{2 \mathrm{OH}^{-}} \\ & \therefore \quad…
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