Question

The increasing order of $A{g}^{+}$ion concentration in
I. Saturated solution of $AgCl$
II. Saturated solution of $AgI$
III. $1MAg{\left((NH{)}_3\right)}_2^{+}$in $0.1MN{H}_3$
IV. $1MAg(CN{)}_2^{-}$in $0.1MKCN$
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 $Ag(CN{)}_2^{-}=1.0\times 10^{-21}$

• A. I < II < III < IV
• B. IV < III < II < I
• C. IV < II < III < I
• D. IV < II < I < III

Answer 1

Answer: (C)

I. $A{g}^{+}=\sqrt{(K{)}_{sp}(AgCl)}=\sqrt{1\times 10^{-10}}=10^{-5}M$
II. $A{g}^{+}=\sqrt{(K{)}_{sp}(AgI)}=\sqrt{1\times 10^{-16}}=10^{-8}M$
III. $[Ag(N{H}_3{)}_2{]}^{+}\rightleftharpoons A{g}^{+}+2N{H}_3$
$K_d=1\times 10^{-8}=\frac{[Ag{]}^{+}[N{H}_3{]}^2}{[AgN{H}_3{]}_{2aq}^{+}}=\frac{[Ag{]}^{+}0.1^2}{1.0}$
$\Rightarrow [Ag{]}^{+}=1\times 10^{-6}M$
$\underset{0.1M}{\mathrm{KCN}}\to K_{aq}^{+}+\underset{(0.1M)}{C{N}_{aq}^{-1}}$
$K_d=1\times 10^{-21}=\frac{[Ag{]}^{+}[CN{]}^{-2}}{[Ag(CN{)}_2{]}^{-1}}=\frac{A{g}^{+}0.1^2}{1.0}$
$\left[A{g}^{+}\right]=1\times 10^{-19}$