JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The molar solubility of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in \(0.1 \,\mathrm{M} \,\mathrm{NaOH}\) solution is \(\mathrm{x} \times 10^{-18} \,\mathrm{M}\). The value of \(\mathrm{x}\) is ...... . (Nearest integer) (Given : The solubility product of \(\mathrm{Zn}(\mathrm{OH})_{2}\) is \(\left.2 \times 10^{-20}\right)\)
- A \(1\)
- B \(3\)
- C \(2\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Zn}(\mathrm{OH})_{2}(\mathrm{~s}) \rightleftharpoons \mathrm{Zn}^{+2}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})\) \(\quad\quad\quad\quad\quad\quad\quad S\quad\quad\quad\quad (0.1+2S) \simeq\,0.1\) \(\mathrm{K}_{\mathrm{sp}}=\mathrm{S}(0.1)^{2}\)…
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