JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The molar solubility of \(Cd(OH)_2\) is \(1.84\times10^{-5}\, M\) in water. The expected solubility of \(Cd(OH)_2\) in a buffer solution of \(pH = 12\) is
- A \(6.23\times10^{-11}\, M\)
- B \(1.84\times10^{-9}\, M\)
- C \(\frac{{2.49}}{{1.84}} \times {10^{ - 9}}\,M\)
- D \(2.49\times10^{-10}\, M\)
Answer & Solution
Correct Answer
(D) \(2.49\times10^{-10}\, M\)
Step-by-step Solution
Detailed explanation
\({K_{sp}}\) of \(Cd{(OH)_2} = 4{s^3} = 4 \times {(1.84 \times {10^{ - 5}})^3}\) If \(pH = 12\) \(pOH = 2\) \([O{H^ - }] = {10^{ - 2}}\,M\) \({K_{sp}} = [C{d^{2 + }}]{[O{H^ - }]^2}\) \(4 \times {(1.84 \times {10^{ - 5}})^3} = [C{d^{2 + }}]{[O{H^ - }]^2}\)…
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