[VISUAL] Three saturated solutions (X, Y, and Z) are prepared at 25°C. Based on the information in t — Equilibrium Chemistry Question

STEPS:

1. Understand the chemical process of dissolution for each salt:
All three salts listed in the table—silver bromide ($\text{AgBr}$), silver chloride ($\text{AgCl}$), and silver iodide ($\text{AgI}$)—are $1:1$ ionic compounds that dissociate in water according to the following equilibrium equations:
* Solution X: $\text{AgBr}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Br}^-(aq)$
* Solution Y: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$
* Solution Z: $\text{AgI}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{I}^-(aq)$

2. Write the solubility product constant ($K_{sp}$) expressions:
For any $1:1$ salt ($\text{AgA}$) in a saturated solution with no common ions present:
$$K_{sp} = [\text{Ag}^+][\text{A}^-]$$
Since the salt dissociates in a $1:1$ ratio, the concentration of the silver ion is equal to the concentration of the anion ($[\text{Ag}^+] = [\text{A}^-]$). Substituting this relationship into the $K_{sp}$ expression yields:
$$K_{sp} = [\text{Ag}^+]^2 \implies [\text{Ag}^+] = \sqrt{K_{sp}}$$

3. Determine the mathematical relationship between $K_{sp}$ and $[\text{Ag}^+]$:
Because all three salts share the exact same stoichiometry ($1:1$), their silver ion concentrations are directly proportional to the square root of their respective $K_{sp}$ values. This means:
* The smaller the $K_{sp}$ value, the lower the $[\text{Ag}^+]$ in a saturated solution.
* The larger the $K_{sp}$ value, the higher the $[\text{Ag}^+]$ in a saturated solution.

4. Compare the $K_{sp}$ values from the table:
* Solution Z ($\text{AgI}$): $K_{sp} = 8.3 \times 10^{-17}$ (the smallest $K_{sp}$ value)
* Solution X ($\text{AgBr}$): $K_{sp} = 5.0 \times 10^{-13}$ (the intermediate $K_{sp}$ value)
* Solution Y ($\text{AgCl}$): $K_{sp} = 1.8 \times 10^{-10}$ (the largest $K_{sp}$ value)

5. Rank the solutions in order of increasing $[\text{Ag}^+]$:
Comparing the exponents of the $K_{sp}$ values ($-17 < -13 < -10$) allows us to rank the concentration of $[\text{Ag}^+]$ in the saturated solutions from lowest to highest:
$$\mathbf{Z < X < Y}$$
This ranking identifies Option D as the correct answer.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: This ranking suggests that Solution X has the lowest silver ion concentration and Solution Y has the highest. While it correctly places Y as the highest, it swaps the positions of X and Z.
• Option B is incorrect: This option ranks the solutions in order of *decreasing* solubility (highest $[\text{Ag}^+]$ to lowest $[\text{Ag}^+]$), which is the exact opposite of what the question asks for.
• Option C is incorrect: This option incorrectly places Solution Z (the least soluble salt with the smallest $K_{sp}$) as having the lowest concentration but incorrectly places Solution X as having a higher concentration than Solution Y.