The increasing order of `Ag^(+)` ion concentration in
I. Saturated solution of `AgCl`
II. Saturated solution of `Agl`
III. `1MAg(NH_(3))_(2)^(+)" in "0.1 M NH_(3)`
IV. `1MAg(CN)_(2)^(-)" in "0.1 M KCN`
Given :
`K_(sp)" of "AgCl=1.0xx10^(-10)`
`K_(sp)" of "Agl=1.0xx10^(-16)`
`K_(d)" of "Ag(NH_(3))_(2)^(+)=1.0xx10^(-8)`
`K_(d)" of "Ag(CN)_(2)^(-)=1.0xx10^(-21)`

Options

A. `I lt II lt III lt IV`
B. `IV lt III lt II lt I`
C. `IV lt II lt III lt I`
D. `IV lt II lt I lt III`

Answer

To determine the increasing order of `Ag^+` ion concentration in the given solutions, we need to calculate the concentration of `Ag^+` in each case. Let's analyze each solution step by step.

### Step 1: Saturated solution of `AgCl`
- The dissociation of `AgCl` can be represented as:
\[ AgCl \rightleftharpoons Ag^+ + Cl^- \]
- Let the solubility of `AgCl` be \( S \). Therefore, at equilibrium:
\[ [Ag^+] = S \quad \text{and} \quad [Cl^-] = S \]
- The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [Ag^+][Cl^-] = S \cdot S = S^2 \]
- Given \( K_{sp} \) of `AgCl = 1.0 \times 10^{-10}`:
\[ S^2 = 1.0 \times 10^{-10} \implies S = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} \, M \]
- Thus, the concentration of `Ag^+` in the saturated solution of `AgCl` is:
\[ [Ag^+] = 1.0 \times 10^{-5} \, M \]

### Step 2: Saturated solution of `AgI`
- The dissociation of `AgI` can be represented as:
\[ AgI \rightleftharpoons Ag^+ + I^- \]
- Let the solubility of `AgI` be \( S \). Therefore, at equilibrium:
\[ [Ag^+] = S \quad \text{and} \quad [I^-] = S \]
- The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [Ag^+][I^-] = S^2 \]
- Given \( K_{sp} \) of `AgI = 1.0 \times 10^{-16}`:
\[ S^2 = 1.0 \times 10^{-16} \implies S = \sqrt{1.0 \times 10^{-16}} = 1.0 \times 10^{-8} \, M \]
- Thus, the concentration of `Ag^+` in the saturated solution of `AgI` is:
\[ [Ag^+] = 1.0 \times 10^{-8} \, M \]

### Step 3: `1 M Ag(NH3)2^+` in `0.1 M NH3`
- The equilibrium for the complexation can be represented as:
\[ Ag(NH_3)_2^+ \rightleftharpoons Ag^+ + 2NH_3 \]
- The dissociation constant \( K_d \) is given by:
\[ K_d = \frac{[Ag^+][NH_3]^2}{[Ag(NH_3)_2^+]} \]
- Given \( K_d = 1.0 \times 10^{-8} \) and \( [NH_3] = 0.1 \, M \):
\[ K_d = \frac{[Ag^+](0.1)^2}{1} \implies 1.0 \times 10^{-8} = [Ag^+] \cdot 0.01 \]
\[ [Ag^+] = \frac{1.0 \times 10^{-8}}{0.01} = 1.0 \times 10^{-6} \, M \]

### Step 4: `1 M Ag(CN)2^-` in `0.1 M KCN`
- The equilibrium for the complexation can be represented as:
\[ Ag(CN)_2^- \rightleftharpoons Ag^+ + 2CN^- \]
- The dissociation constant \( K_d \) is given by:
\[ K_d = \frac{[Ag^+][CN^-]^2}{[Ag(CN)_2^-]} \]
- Given \( K_d = 1.0 \times 10^{-21} \) and \( [CN^-] = 0.1 \, M \):
\[ K_d = \frac{[Ag^+](0.1)^2}{1} \implies 1.0 \times 10^{-21} = [Ag^+] \cdot 0.01 \]
\[ [Ag^+] = \frac{1.0 \times 10^{-21}}{0.01} = 1.0 \times 10^{-19} \, M \]

### Summary of Concentrations
1. Saturated solution of `AgCl`: \( [Ag^+] = 1.0 \times 10^{-5} \, M \)
2. Saturated solution of `AgI`: \( [Ag^+] = 1.0 \times 10^{-8} \, M \)
3. `1 M Ag(NH3)2^+` in `0.1 M NH3`: \( [Ag^+] = 1.0 \times 10^{-6} \, M \)
4. `1 M Ag(CN)2^-` in `0.1 M KCN`: \( [Ag^+] = 1.0 \times 10^{-19} \, M \)

### Increasing Order of `Ag^+` Concentration
Now, we can arrange these concentrations in increasing order:
- \( 1.0 \times 10^{-19} \, M \) (Ag(CN)2^-)
- \( 1.0 \times 10^{-8} \, M \) (AgI)
- \( 1.0 \times 10^{-6} \, M \) (Ag(NH3)2^+)
- \( 1.0 \times 10^{-5} \, M \) (AgCl)

Thus, the increasing order of `Ag^+` ion concentration is:
**IV < II < III < I**