How many grams of `KBr` can be added to `1L` of `0.12 M` solution of `AgNO_(3)` just to start the precipitation of `AgBr. (Mw "of" KBr = 120, K_(sp) of AgBr = 10^(-13))`

Options

A. `10^(-10)g`
B. `10^(-9)g`
C. `0.5 xx 10^(-10)g`
D. `0.5 xx 10^(-9)g`

Explanation

To find out how many grams of KBr can be added to a 1L solution of 0.12 M AgNO3 just to start the precipitation of AgBr, we will follow these steps:

### Step 1: Understand the precipitation condition
The precipitation of AgBr will start when the ionic product (IP) of Ag+ and Br- ions equals the solubility product (Ksp) of AgBr.

### Step 2: Write the dissociation equations
- AgNO3 dissociates into Ag+ and NO3-:
\[ \text{AgNO}_3 \rightarrow \text{Ag}^+ + \text{NO}_3^- \]
- KBr dissociates into K+ and Br-:
\[ \text{KBr} \rightarrow \text{K}^+ + \text{Br}^- \]

### Step 3: Determine the concentration of Ag+
From the given information, the concentration of Ag+ from the 0.12 M AgNO3 solution is:
\[ [\text{Ag}^+] = 0.12 \, \text{M} \]

### Step 4: Set up the Ksp expression
The Ksp expression for AgBr is:
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \]
Given that \( K_{sp} = 10^{-13} \), we can substitute the known concentration of Ag+:
\[ 10^{-13} = (0.12)[\text{Br}^-] \]

### Step 5: Solve for [Br-]
Rearranging the equation to find the concentration of Br-:
\[ [\text{Br}^-] = \frac{10^{-13}}{0.12} = 8.33 \times 10^{-13} \, \text{M} \]

### Step 6: Relate [Br-] to the mass of KBr
Since KBr dissociates into K+ and Br-, the concentration of Br- is equal to the concentration of KBr. We can express the molarity of KBr in terms of mass:
\[ [\text{Br}^-] = \frac{W}{120 \times 1} \quad \text{(where W is the mass of KBr in grams)} \]
Thus,
\[ 8.33 \times 10^{-13} = \frac{W}{120} \]

### Step 7: Solve for W
Now, we can solve for W:
\[ W = 8.33 \times 10^{-13} \times 120 = 10^{-10} \, \text{grams} \]

### Final Answer
The mass of KBr that can be added to the solution just to start the precipitation of AgBr is:
\[ W = 10^{-10} \, \text{grams} \]

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