To find the `K_(sp)` of `AgBrO_(3)`, a student prepared one litre of a saturated solution by adding only sufficient `AgBrO_(3)` in water at `27^(@)C`. He found that a copper wire left in the solution overnight become covered with silver and `Cu^(2+)` ion were also formed in the solution. The wire was dried and found to weigh 15.25 mg more that its original weight. `Cu(s)+2Ag^(+)(aq)hArrCu^(2+)(aq)+2Ag(s)`,
`DeltaG^(@)=-110.4kJ`
[Given: `R=8J//K-"mol",Cu=63.5,Ag=108`]
The concentration of `Ag^(+)` in the original saturated solution was:

Options

A. `1.0xx10^(-2)`
B. `1.0xx10^(-4)`
C. `2.0xx10^(-4)`
D. `5.0xx10^(-5)`

Answer

To find the concentration of \( \text{Ag}^+ \) in the original saturated solution of \( \text{AgBrO}_3 \), we can follow these steps:

### Step 1: Calculate the moles of silver deposited
The reaction between copper and silver ions is given by:
\[ \text{Cu(s)} + 2\text{Ag}^+(aq) \rightleftharpoons \text{Cu}^{2+}(aq) + 2\text{Ag(s)} \]

From the problem, we know that the copper wire gained 15.25 mg of silver.

To find the moles of silver deposited, we can use the formula:
\[ \text{moles of Ag} = \frac{\text{mass of Ag}}{\text{molar mass of Ag}} \]

Given:
- Mass of silver deposited = 15.25 mg = \( 15.25 \times 10^{-3} \) g
- Molar mass of silver (\( \text{Ag} \)) = 108 g/mol

Calculating the moles of silver:
\[ \text{moles of Ag} = \frac{15.25 \times 10^{-3} \text{ g}}{108 \text{ g/mol}} = 1.4157 \times 10^{-4} \text{ mol} \]

### Step 2: Relate moles of silver to moles of copper
From the stoichiometry of the reaction, 1 mole of copper reacts with 2 moles of silver. Therefore, the moles of copper that reacted can be calculated as:
\[ \text{moles of Cu} = \frac{\text{moles of Ag}}{2} = \frac{1.4157 \times 10^{-4}}{2} = 7.0785 \times 10^{-5} \text{ mol} \]

### Step 3: Calculate the concentration of \( \text{Ag}^+ \)
Since the solution is saturated and we have 1 liter of the solution, the concentration of \( \text{Ag}^+ \) ions in the saturated solution is equal to the moles of \( \text{Ag}^+ \) ions in 1 liter of solution.

Thus, the concentration of \( \text{Ag}^+ \) is:
\[ [\text{Ag}^+] = \text{moles of Ag} = 1.4157 \times 10^{-4} \text{ mol/L} \]

### Step 4: Calculate \( K_{sp} \)
The solubility product constant \( K_{sp} \) for \( \text{AgBrO}_3 \) can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{BrO}_3^-] \]
Since the dissociation of \( \text{AgBrO}_3 \) in solution gives equal concentrations of \( \text{Ag}^+ \) and \( \text{BrO}_3^- \), we have:
\[ K_{sp} = [\text{Ag}^+]^2 = (1.4157 \times 10^{-4})^2 = 2.0073 \times 10^{-8} \]

### Final Result
Thus, the concentration of \( \text{Ag}^+ \) in the original saturated solution is:
\[ [\text{Ag}^+] = 1.4157 \times 10^{-4} \text{ mol/L} \]