Calculate the solubility product of silver bromide if the solubility of the salt in saturated solution is `5.7xx10^(-7)` moles/litre.

Answer

To calculate the solubility product (Ksp) of silver bromide (AgBr) given its solubility in a saturated solution, follow these steps:

### Step 1: Understand the Dissociation of Silver Bromide
Silver bromide (AgBr) dissociates in water according to the following equilibrium equation:
\[ \text{AgBr (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Br}^- (aq) \]

### Step 2: Define the Solubility
Let the solubility of AgBr in moles per liter be denoted as \( S \). According to the problem, the solubility \( S \) is given as:
\[ S = 5.7 \times 10^{-7} \, \text{moles/liter} \]

### Step 3: Write the Expression for Ksp
The solubility product (Ksp) for the dissociation of AgBr can be expressed as:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \]

### Step 4: Substitute the Solubility into the Ksp Expression
Since each mole of AgBr that dissolves produces one mole of Ag\(^+\) and one mole of Br\(^-\), we can say:
\[ [\text{Ag}^+] = S \]
\[ [\text{Br}^-] = S \]

Thus, the Ksp expression becomes:
\[ K_{sp} = S \times S = S^2 \]

### Step 5: Calculate Ksp
Now, substitute the value of \( S \) into the Ksp expression:
\[ K_{sp} = (5.7 \times 10^{-7})^2 \]

### Step 6: Perform the Calculation
Calculating \( (5.7 \times 10^{-7})^2 \):
\[ K_{sp} = 5.7^2 \times (10^{-7})^2 = 32.49 \times 10^{-14} \]
\[ K_{sp} = 3.249 \times 10^{-13} \]

### Final Answer
The solubility product of silver bromide (AgBr) is:
\[ K_{sp} \approx 3.25 \times 10^{-13} \]

---