The solubility product of AgBr is `4.9xx10^(-9)` . The solubility of AgBr will be

Options

A. `7xx10^(-4)` mole /litre
B. `7xx10^(-5)` g / litre
C. `1.316 xx 10^(-2)` g/litre
D. `1 xx 10^(-3)` mole /litre

Answer

To find the solubility of AgBr given its solubility product (Ksp), we can follow these steps:

### Step 1: Write the dissociation equation
AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq)

### Step 2: Define solubility
Let the solubility of AgBr be \( S \) moles per liter. When AgBr dissolves, it produces \( S \) moles of Ag⁺ and \( S \) moles of Br⁻.

### Step 3: Write the expression for Ksp
The solubility product \( K_{sp} \) is given by the formula:
\[ K_{sp} = [Ag^+][Br^-] \]
Substituting the concentrations in terms of solubility \( S \):
\[ K_{sp} = S \cdot S = S^2 \]

### Step 4: Substitute the given Ksp value
We know that \( K_{sp} \) for AgBr is \( 4.9 \times 10^{-9} \):
\[ S^2 = 4.9 \times 10^{-9} \]

### Step 5: Solve for S
To find \( S \), take the square root of both sides:
\[ S = \sqrt{4.9 \times 10^{-9}} \]
Calculating this gives:
\[ S \approx 7.0 \times 10^{-5} \text{ moles per liter} \]

### Step 6: (Optional) Convert to grams per liter
To convert the solubility from moles per liter to grams per liter, use the molar mass of AgBr. The molar mass of AgBr is approximately \( 187.77 \) g/mol. Therefore:
\[ \text{Solubility in g/L} = S \times \text{Molar Mass} \]
\[ \text{Solubility in g/L} = 7.0 \times 10^{-5} \text{ moles/L} \times 187.77 \text{ g/mol} \approx 1.316 \times 10^{-2} \text{ g/L} \]

### Final Answer
The solubility of AgBr is approximately \( 7.0 \times 10^{-5} \) moles per liter or \( 1.316 \times 10^{-2} \) grams per liter.

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