The minimum mass of NaBr which should be added in 200 ml of `0.0004M-AgNO_(3)` solution just to start precipitation of `AgBr.K_(sp)` of `AgBr=4xx10^(-13)(Br=80)`

Options

A. `1.0xx10^(-9)g`
B. `2xx10^(-10)g`
C. `2.06xx10^(-8)g`
D. `1.03xx10^(-7)g`

Explanation

To solve the problem of finding the minimum mass of NaBr required to start the precipitation of AgBr in a 200 ml solution of 0.0004 M AgNO3, we will follow these steps:

### Step 1: Understand the Reaction
The reaction between AgNO3 and NaBr produces AgBr as a precipitate:
\[ \text{AgNO}_3 + \text{NaBr} \rightarrow \text{AgBr (s)} + \text{NaNO}_3 \]

### Step 2: Determine Ksp Expression
The solubility product constant (Ksp) for AgBr is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \]
Given \( K_{sp} = 4 \times 10^{-13} \).

### Step 3: Find the Concentration of Ag+
The concentration of Ag+ from the AgNO3 solution is:
\[ [\text{Ag}^+] = 0.0004 \, \text{M} = 4 \times 10^{-4} \, \text{M} \]

### Step 4: Calculate the Required Concentration of Br-
Using the Ksp expression, we can find the required concentration of Br- to start precipitation:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \]
Substituting the values:
\[ 4 \times 10^{-13} = (4 \times 10^{-4})[\text{Br}^-] \]
Solving for \([\text{Br}^-]\):
\[ [\text{Br}^-] = \frac{4 \times 10^{-13}}{4 \times 10^{-4}} = 10^{-9} \, \text{M} \]

### Step 5: Calculate Moles of Br- Required
To find the moles of Br- needed in 200 ml (0.2 L) of solution:
\[ \text{Moles of Br}^- = [\text{Br}^-] \times \text{Volume} \]
\[ \text{Moles of Br}^- = 10^{-9} \, \text{M} \times 0.2 \, \text{L} = 2 \times 10^{-10} \, \text{moles} \]

### Step 6: Calculate the Molar Mass of NaBr
The molar mass of NaBr is calculated as follows:
- Sodium (Na) = 23 g/mol
- Bromine (Br) = 80 g/mol
\[ \text{Molar mass of NaBr} = 23 + 80 = 103 \, \text{g/mol} \]

### Step 7: Calculate the Mass of NaBr Required
Now, we can calculate the mass of NaBr needed using the formula:
\[ \text{Mass} = \text{Moles} \times \text{Molar Mass} \]
\[ \text{Mass of NaBr} = 2 \times 10^{-10} \, \text{moles} \times 103 \, \text{g/mol} \]
\[ \text{Mass of NaBr} = 2.06 \times 10^{-8} \, \text{g} \]

### Final Answer
The minimum mass of NaBr required to start the precipitation of AgBr is:
\[ \text{Mass of NaBr} = 2.06 \times 10^{-8} \, \text{g} \]

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