JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The solubility product of \(\mathrm{Cr}(\mathrm{OH})_{3}\) at \(298\; \mathrm{K}\) is \(6.0 \times 10^{-31} .\) The concentration of hydroxide ions in a saturated solution of \(\mathrm{Cr}(\mathrm{OH})_{3}\) will be
- A \(\left(18 \times 10^{-31}\right)^{1 / 4}\)
- B \(\left(2.22 \times 10^{-31}\right)^{1 / 4}\)
- C \(\left(4.86 \times 10^{-29}\right)^{1 / 4}\)
- D \(\left(18 \times 10^{-31}\right)^{1 / 2}\)
Answer & Solution
Correct Answer
(A) \(\left(18 \times 10^{-31}\right)^{1 / 4}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Cr}(\mathrm{OH})_{3}(\mathrm{s}) \rightleftharpoons \mathrm{Cr}^{3+}(\mathrm{aq} .)+3 \mathrm{OH}^{-}(\mathrm{aq} .)\) \(\mathrm{k}_{\mathrm{sp}}=27(\mathrm{s})^{4}=6 \times 10^{-31}\) \(\Rightarrow[3(\mathrm{s})]^{4}=18 \times 10^{-31}\)…
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