question_answer

Arrange the following solutions in decreasing order of \[[A{{g}^{\oplus }}]\] ion:

(I)  \[1\,M\,{{[Ag{{(CN)}_{2}}]}^{\Theta }}\]

(II) Saturated \[AgCl\]

(III) \[1\,M{{[Ag{{(N{{H}_{3}})}_{2}}]}^{\oplus }}\]in \[0.1M\,\,N{{H}_{3}}\]

(IV) Saturated \[0.1M\,\,N{{H}_{3}}\]

(\[{{K}_{sp}}\] of \[AgCl={{10}^{-10}},\,\,{{K}_{sp}}\] of \[AgI=8.3\times {{10}^{-17}}\] \[{{K}_{f}}\] (formation constant) \[{{[Ag{{(CN)}_{2}}]}^{\Theta }}={{10}^{21}},\] \[{{K}_{f}}{{[Ag{{(N{{H}_{3}})}_{2}}]}^{\oplus }}={{10}^{8}}\]

A) \[I>II>III>IV\]

B)        \[II>III>I>IV\]   

C)     \[IV>III>II>I\]  

D)        \[I>IV>III>II\]

Correct Answer: B

Solution :

[b] \[II>III>IV>I\] From the values of \[{{K}_{sp}}\] and \[{{K}_{f'}}\] it is evident that \[AgCl\]is more soluble than \[AgI\]and \[{{[Ag{{(CN)}_{2}}]}^{\Theta }}\]is more soluble than \[{{[Ag{{(N{{H}_{3}})}_{2}}]}^{\oplus }}\] Calculation of \[[A{{g}^{\oplus }}]\] in each case reveaels \[II>III>I>IV.\] (I) \[Ag{{(CN)}_{2}}^{\Theta }\xrightarrow{{}}A{{g}^{\oplus }}+2C\overset{\Theta }{\mathop{N}}\,\] \[K=\frac{1}{{{K}_{f}}}=\frac{[A{{g}^{\oplus }}]{{[C{{N}^{\Theta }}]}^{2}}}{[Ag(CN)_{2}^{\Theta }]}={{10}^{-21}}\] Let  \[[A{{g}^{\oplus }}]=x,\,\,[C{{N}^{\Theta }}]=2x,\] \[[Ag{{(CN)}_{2}}^{\Theta }]=1.0-x\simeq 1.0\] \[\therefore \,\,\,\,\,\,\,\,\,K=\frac{(x){{(2x)}^{2}}}{1.0}={{10}^{-21}}=4{{x}^{3}}\] \[\Rightarrow \,\,\,\,\,\,\,\,x=6.3\times {{10}^{-8}}M\] (II)  \[AgClA{{g}^{\oplus }}+C{{l}^{\Theta }};\] \[{{K}_{sp}}=[A{{g}^{\oplus }}][C{{l}^{\Theta }}]={{10}^{-10}}\] \[\therefore \,\,\,\,\,{{x}^{2}}={{10}^{-10}},\,\,\,\Rightarrow \,x={{10}^{-5}}M\] (III) \[K=\frac{1}{{{K}_{f}}}=\frac{[A{{g}^{\oplus }}]{{[N{{H}_{3}}]}^{2}}}{[Ag(N{{H}_{3}})_{2}^{\oplus }]}={{10}^{-8}}\] Let \[[A{{g}^{\oplus }}]=x,\,\,[N{{H}_{3}}]=(2x+0.1)\approx 0.1\] \[[Ag{{(N{{H}_{3}})}_{2}}^{\oplus }]=1.0-x\simeq 1.0\] \[K=\frac{x\times 0.1\times 0.1}{0.1}={{10}^{-8}}\Rightarrow x={{10}^{-6}}M\] (IV) \[AgIA{{g}^{\oplus }}+{{I}^{\Theta }},\] \[{{x}^{2}}=8.3\times {{10}^{-17}}\Rightarrow x\Rightarrow 2.8\times {{10}^{-9}}.\] (II) \[(AgCl={{10}^{-5}})>\]     (III) \[[Ag{{(N{{H}_{3}})}_{2}}^{\oplus })={{10}^{-6}}]>\] (I)\[[Ag{{(CN)}_{2}}^{\Theta })=6.3\times {{10}^{-9}}]>\] (IV) \[(AgI=2.8\times {{10}^{-9}})\] \[\therefore \,\,\,\,\,\,\,\,II>III>I>IV\]