TS EAMCET · Chemistry · Ionic Equilibrium
If the solubility product of \(\mathrm{Ni}(\mathrm{OH})_2\) is \(4.0 \times 10^{-15}\), the solubility (in \(\mathrm{mol} \mathrm{L}^{-1}\) ) is
- A \(5.0 \times 10^{-5}\)
- B \(4.0 \times 10^{-5}\)
- C \(2.0 \times 10^{-5}\)
- D \(1.0 \times 10^{-5}\)
Answer & Solution
Correct Answer
(D) \(1.0 \times 10^{-5}\)
Step-by-step Solution
Detailed explanation
Given, solubility product \(\left(K_{\mathrm{sp}}\right)\) of \(\mathrm{Ni}(\mathrm{OH})_2\) is \[ =4.0 \times 10^{-15} \] Solubility \((s)=\) ?…
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