Solubility product of silver bromide is `5.0xx10^(-13)`. The quantity of potassium bromide (molar mass taken as `120g mol^(-1)`) to be added to `1 L` of `0.05 M` solution of silver nitrate to start the precipitation of `AgBr` is

Options

A. `1.2xx10^(-1)g`
B. `1.2xx10^(-9)g`
C. `6.2xx10^(-5)g`
D. `5.0xx10^(-8)g`

Answer

To solve the problem, we need to determine the amount of potassium bromide (KBr) that must be added to a 1 L solution of 0.05 M silver nitrate (AgNO3) to initiate the precipitation of silver bromide (AgBr). The solubility product (Ksp) of AgBr is given as \(5.0 \times 10^{-13}\).

### Step-by-step Solution:

1. **Understand the Reaction**:
The precipitation reaction can be represented as:
\[ \text{Ag}^+ + \text{Br}^- \rightarrow \text{AgBr (s)} \]
Here, Ag+ ions come from the dissociation of AgNO3, and Br- ions will come from the KBr we are adding.

2. **Calculate the Concentration of Ag+**:
Since we have a 0.05 M solution of AgNO3, the concentration of Ag+ ions is also 0.05 M.

3. **Set Up the Expression for Ksp**:
The solubility product expression for AgBr is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \]
We know \(K_{sp} = 5.0 \times 10^{-13}\) and \([\text{Ag}^+] = 0.05\) M. We need to find \([\text{Br}^-]\) at the point of precipitation.

4. **Calculate the Required Concentration of Br-**:
Rearranging the Ksp expression to find \([\text{Br}^-]\):
\[ [\text{Br}^-] = \frac{K_{sp}}{[\text{Ag}^+]} = \frac{5.0 \times 10^{-13}}{0.05} \]
\[ [\text{Br}^-] = 1.0 \times 10^{-11} \text{ M} \]

5. **Convert Concentration to Moles**:
Since we are working with a 1 L solution, the number of moles of Br- required is:
\[ \text{Moles of Br}^- = [\text{Br}^-] \times \text{Volume} = 1.0 \times 10^{-11} \text{ moles} \]

6. **Calculate the Moles of KBr Needed**:
The moles of KBr needed will be the same as the moles of Br- since KBr dissociates completely in solution:
\[ \text{Moles of KBr} = 1.0 \times 10^{-11} \text{ moles} \]

7. **Calculate the Mass of KBr**:
Using the molar mass of KBr (120 g/mol), we can find the mass:
\[ \text{Mass of KBr} = \text{Moles of KBr} \times \text{Molar Mass} = 1.0 \times 10^{-11} \text{ moles} \times 120 \text{ g/mol} \]
\[ \text{Mass of KBr} = 1.2 \times 10^{-9} \text{ g} \]

### Final Answer:
The quantity of potassium bromide (KBr) to be added is \(1.2 \times 10^{-9}\) grams.

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