MHT CET · Chemistry · Ionic Equilibrium
The solubility of \(\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4\) is \(2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}\) at \(298 \mathrm{~K}\). What is it's solubility product?
- A \(1.6 \times 10^{-6}\)
- B \(3.2 \times 10^{-11}\)
- C \(1.6 \times 10^{-11}\)
- D \(3.2 \times 10^{-6}\)
Answer & Solution
Correct Answer
(B) \(3.2 \times 10^{-11}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_{4(\mathrm{~s})} \rightleftharpoons \underset{2 \mathrm{~S}}{2 \mathrm{Ag}_{(\mathrm{aq})}^{+}}+\mathrm{C}_2 \mathrm{O}_{4(\mathrm{aq})}^{2-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{C}_2 \mathrm{O}_4^{2-}\right] \\ & =(2 \mathrm{~S})^2 \mathrm{~S} \\ & =4 \mathrm{~S}^3=4 \times\left(2 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=3.2 \times 10^{-11}\end{aligned}\)
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