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 for AgBr
AgBr(s) ⇌ Ag⁺(aq) + Br⁻(aq)

### Step 2: Define the solubility (s)
Let the solubility of AgBr be 's' mol/L. This means that at equilibrium:
- The concentration of Ag⁺ = s
- The concentration of Br⁻ = s

### Step 3: Write the expression for Ksp
The solubility product (Ksp) expression for AgBr is given by:
Ksp = [Ag⁺][Br⁻]

Substituting the concentrations from Step 2:
Ksp = (s)(s) = s²

### Step 4: Substitute the given Ksp value
We know that Ksp for AgBr is 4.9 × 10^(-9). Therefore:
s² = 4.9 × 10^(-9)

### Step 5: Solve for s
To find 's', take the square root of both sides:
s = √(4.9 × 10^(-9))

Calculating this gives:
s = √(4.9) × √(10^(-9)) = 2.2136 × 10^(-4.5)

Since √(10^(-9)) = 10^(-4.5):
s ≈ 7 × 10^(-5) mol/L

### Step 6: Convert solubility to grams per liter
To convert the solubility from moles per liter to grams per liter, we need the molar mass of AgBr.
- Molar mass of Ag = 108 g/mol
- Molar mass of Br = 80 g/mol
- Molar mass of AgBr = 108 + 80 = 188 g/mol

Now, using the formula:
Weight (g/L) = solubility (mol/L) × molar mass (g/mol)
Weight = (7 × 10^(-5) mol/L) × (188 g/mol)

Calculating this gives:
Weight ≈ 1.316 × 10^(-2) g/L

### Final Answer
The solubility of AgBr is approximately 1.316 × 10^(-2) g/L.

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