The solubility of `AgBrO_(3)` (formula mass=236) is `0.0072` g in 1000 mL. What is the `K_(sp)`?

Options

A. `2.2xx10^(-8)`
B. `3.0xx10^(-10)`
C. `3.0xx10^(-5)`
D. `9.3xx10^(-10)`

Answer

To find the solubility product constant (Ksp) for AgBrO₃, we will follow these steps:

### Step 1: Calculate the molarity of AgBrO₃
Given that the solubility of AgBrO₃ is 0.0072 g in 1000 mL (1 L), we first convert the mass into moles using the formula mass.

**Calculation:**
1. Convert grams to moles:
\[ \text{Moles of AgBrO₃} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{0.0072 \text{ g}}{236 \text{ g/mol}} \]
\[ = 0.00003051 \text{ mol} \]

2. Since this is in 1 L, the molarity (M) is:
\[ \text{Molarity (M)} = \frac{\text{moles}}{\text{volume (L)}} = 0.00003051 \text{ mol/L} \]

### Step 2: Write the dissociation equation
AgBrO₃ dissociates in water as follows:
\[ \text{AgBrO₃ (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{BrO}_3^- (aq) \]

### Step 3: Determine the expression for Ksp
The solubility product constant (Ksp) is given by the product of the concentrations of the ions at equilibrium:
\[ K_{sp} = [\text{Ag}^+][\text{BrO}_3^-] \]
Since the dissociation produces one mole of Ag⁺ and one mole of BrO₃⁻ for each mole of AgBrO₃ that dissolves, we can denote the solubility as \( S \):
\[ K_{sp} = S \cdot S = S^2 \]

### Step 4: Substitute the value of S
From our earlier calculation, we found that:
\[ S = 0.00003051 \text{ mol/L} \]
Now, we can calculate \( K_{sp} \):
\[ K_{sp} = (0.00003051)^2 \]
\[ = 9.305 \times 10^{-10} \text{ mol}^2/\text{L}^2 \]

### Final Answer
Thus, the solubility product constant \( K_{sp} \) for AgBrO₃ is:
\[ K_{sp} \approx 9.3 \times 10^{-10} \]

---