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 will calculate the concentration of `Ag^+` in each case step by step.

### Step 1: Calculate `Ag^+` concentration in saturated solution of `AgCl`
- The dissociation of `AgCl` is given by:
\[ \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]
- The solubility product constant \( K_{sp} \) for `AgCl` is given as \( 1.0 \times 10^{-10} \).
- Since the concentration of `Ag^+` and `Cl^-` are equal in a saturated solution, we can write:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = x^2 \]
where \( x \) is the concentration of `Ag^+`.
- Thus, we have:
\[ x^2 = 1.0 \times 10^{-10} \]
\[ x = \sqrt{1.0 \times 10^{-10}} = 1.0 \times 10^{-5} \, \text{M} \]

### Step 2: Calculate `Ag^+` concentration in saturated solution of `AgI`
- The dissociation of `AgI` is given by:
\[ \text{AgI} \rightleftharpoons \text{Ag}^+ + \text{I}^- \]
- The solubility product constant \( K_{sp} \) for `AgI` is given as \( 1.0 \times 10^{-16} \).
- Similarly, we can write:
\[ K_{sp} = [\text{Ag}^+][\text{I}^-] = y^2 \]
where \( y \) is the concentration of `Ag^+`.
- Thus, we have:
\[ y^2 = 1.0 \times 10^{-16} \]
\[ y = \sqrt{1.0 \times 10^{-16}} = 1.0 \times 10^{-8} \, \text{M} \]

### Step 3: Calculate `Ag^+` concentration in `1 M Ag(NH3)2^+` in `0.1 M NH3`
- The dissociation of the complex is given by:
\[ \text{Ag(NH}_3\text{)}_2^+ \rightleftharpoons \text{Ag}^+ + 2 \text{NH}_3 \]
- The dissociation constant \( K_d \) is given as \( 1.0 \times 10^{-8} \).
- Let \( z \) be the concentration of `Ag^+`. We can set up the equation:
\[ K_d = \frac{[\text{Ag}^+][\text{NH}_3]^2}{[\text{Ag(NH}_3\text{)}_2^+]} \approx \frac{z(0.1)^2}{1 - z} \]
- Assuming \( z \) is very small compared to 1, we can simplify:
\[ K_d \approx \frac{z(0.1)^2}{1} = z(0.01) \]
- Thus, we have:
\[ 1.0 \times 10^{-8} = z(0.01) \]
\[ z = \frac{1.0 \times 10^{-8}}{0.01} = 1.0 \times 10^{-6} \, \text{M} \]

### Step 4: Calculate `Ag^+` concentration in `1 M Ag(CN)2^-` in `0.1 M KCN`
- The dissociation of the complex is given by:
\[ \text{Ag(CN)}_2^- \rightleftharpoons \text{Ag}^+ + 2 \text{CN}^- \]
- The dissociation constant \( K_d \) is given as \( 1.0 \times 10^{-21} \).
- Let \( w \) be the concentration of `Ag^+`. We can set up the equation:
\[ K_d = \frac{[\text{Ag}^+][\text{CN}^-]^2}{[\text{Ag(CN)}_2^-]} \approx \frac{w(0.1)^2}{1 - w} \]
- Assuming \( w \) is very small compared to 1, we can simplify:
\[ K_d \approx \frac{w(0.1)^2}{1} = w(0.01) \]
- Thus, we have:
\[ 1.0 \times 10^{-21} = w(0.01) \]
\[ w = \frac{1.0 \times 10^{-21}}{0.01} = 1.0 \times 10^{-19} \, \text{M} \]

### Summary of Concentrations
1. `AgCl`: \( 1.0 \times 10^{-5} \, \text{M} \)
2. `AgI`: \( 1.0 \times 10^{-8} \, \text{M} \)
3. `Ag(NH3)2^+`: \( 1.0 \times 10^{-6} \, \text{M} \)
4. `Ag(CN)2^-`: \( 1.0 \times 10^{-19} \, \text{M} \)

### Increasing Order of `Ag^+` Concentration
From the calculated concentrations:
- \( 1.0 \times 10^{-19} \) (Ag(CN)2^-) < \( 1.0 \times 10^{-8} \) (AgI) < \( 1.0 \times 10^{-6} \) (Ag(NH3)2^+) < \( 1.0 \times 10^{-5} \) (AgCl)

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