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JEE Advanced 2011Paper 2
Exam
JEE Advanced
Year
2011
Paper
Paper 2
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Advanced 2011 · Paper 2 · Chemistry · Ionic Equilibrium

In 1 ~L saturated solution of AgCl[K_ sp AgCl=1.6 × 10^ -10 ], 0.1...

In 1 L1 \mathrm{~L} saturated solution of AgCl[KspAgCl=1.6×10−10],0.1\mathrm{AgCl}\left[K_{\mathrm{sp}} \mathrm{AgCl}=1.6 \times 10^{-10}\right], 0.1 mole of CuCl[Ksp(CuCl)=1.0×10−6]\mathrm{CuCl}\left[K_{\mathrm{sp}}(\mathrm{CuCl})=1.0 \times 10^{-6}\right] is added. The resultant concentration of Ag+\mathrm{Ag}^{+}in the solution is 1.6×10−x1.6 \times 10^{-x}. The value of xx is

Correct answer

0

Step-by-step solution

It is a case of simultaneous solubility of salts with a common ion. Here solubility product of CuCl\mathrm{CuCl} is much greater than that of AgCl\mathrm{AgCl}, it can be assumed that Cl−\mathrm{Cl}^{-}in solution comes mainly from CuCl\mathrm{CuCl}. \Rightarrow \quad[Cl−]=Ksp(CuCl)=\left[\mathrm{Cl}^{-}\right]=\sqrt{K_{s p}(\mathrm{CuCl})}=10^{-3} M\mathrm{M}Now for AgCl:Ksp=1.6×10−10=[Ag+][Cl−]=[Ag+]×10−3⇒[Ag+]=1.6×10−7\mathrm{AgCl}: K_{s p}=1.6 \times 10^{-10}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right]=\left[\mathrm{Ag}^{+}\right] \times 10^{-3} \Rightarrow\left[\mathrm{Ag}^{+}\right]=1.6 \times 10^{-7}
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