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Q. The increasing order of $Ag ^{+}$ion concentration in
I. Saturated solution of $AgCl$
II. Saturated solution of $AgI$
III. $1 \,M \,Ag \left(( NH )_{3}\right)_{2}^{+}$in $0.1 \,M\, NH _{3}$
IV. $1 \,M \,Ag ( CN )_{2}^{-}$in $0.1 \,M \,KCN$
Given:
$K _{ sp } $ of $ AgCl =1.0 \times 10^{-10}$
$ K _{ sp } $ of $ AgI =1.0 \times 10^{-16} $
$ K_{ d } $ of $ Ag \left(( NH )_{3}\right)_{2}^{+}=1.0 \times 10^{-8}$
$ K _{ d } $ of $ Ag ( CN )_{2}^{-}=1.0 \times 10^{-21}$

NTA AbhyasNTA Abhyas 2022

Solution:

I. $Ag ^{+}=\sqrt{ (K) _{ sp } (AgCl) }=\sqrt{1 \times 10^{-10}}=10^{-5} M$
II. $Ag ^{+}=\sqrt{ (K) _{ sp } (AgI) }=\sqrt{1 \times 10^{-16}}=10^{-8} M$
III. $[Ag(NH _{3})_{2}]^{+} \rightleftharpoons Ag ^{+}+2 NH _{3}$
$K _{ d }=1 \times 10^{-8}=\frac{ [Ag ]^{+} [NH _{3}]^{2}}{ [AgNH _{3}]_{2 aq }^+}=\frac{ [Ag] ^{+} 0.1^{2}}{1.0} $
$\Rightarrow [Ag] ^{+}=1 \times 10^{-6} M$
$\underset{0.1\,M}{KCN} \rightarrow K _{ aq }^{+}+ \underset{(0.1\,M)}{CN _{ aq }^{-1}}$
$K _{ d }=1 \times 10^{-21}=\frac{ [Ag ]^{+} [CN ]^{-2}}{ [Ag(CN) _{2}]^{-1}}=\frac{ Ag ^{+} 0.1^{2}}{1.0}$
$\left[ Ag ^{+}\right]=1 \times 10^{-19}$