AP EAMCET · Chemistry · Chemical Equilibrium
A \(1.0 \mathrm{~L}\) of aqueous solution contains \(1 \times 10^{-8} \mathrm{M} \mathrm{NaBr}, 1 \times 10^{-8} \mathrm{M} \mathrm{NaCl}\) and \(1 \times 10^{-8} \mathrm{M}\) NaI. To this solution, \(1 \times 10^{-10} \mathrm{M}\) aqueous \(\mathrm{AgNO}_3\) solution is added drop wise. The order of precipitation of \(\operatorname{Ag} X(X=\mathrm{Cl}, \mathrm{Br}, \mathrm{I})\) is
\(\begin{aligned} & \left(K_{\mathrm{sp}}(\mathrm{AgCl})=1.8=10^{-10} ; K_{\mathrm{sp}}(\mathrm{AgBr})=5 \times 10^{-13} ;\right. \\ & \left.K_{\mathrm{sp}}(\mathrm{AgI})=8.3 \times 10^{-17}\right)\end{aligned}\)
- A \(\mathrm{AgBr}, \mathrm{AgCl}, \mathrm{Agl}\)
- B \(\mathrm{AgCl}, \mathrm{AgBr}, \mathrm{Agl}\).
- C \(\mathrm{Agl}, \mathrm{AgBr}, \mathrm{AgCl}\)
- D \(\mathrm{AgBr}, \mathrm{Agl}, \mathrm{AgCl}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{Agl}, \mathrm{AgBr}, \mathrm{AgCl}\)
Step-by-step Solution
Detailed explanation
\(K_{\mathrm{sp}}\) of any salt is equal to the multiplication of concentration of its ions. For \(\mathrm{AgCl} \longrightarrow \mathrm{Ag}^{+}+\mathrm{Cl}^{-}\) \(K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]\)…
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