1 dm3 solution containing `10^(-5)` moles each of `Cl^(-)` ions and `CrO_(4)^(2-)` ions is treated with `10^(-4)` moles of silver nitrate. Which one of the following observation is made?`[K_(sp)Ag_(2)CrO_(4)=4xx10^(-12)] , [K_(sp) AgCl=1xx10^(-10)]`

Options

A. Precipitation does not occur
B. Silver chromate gets precipitated first
C. Silver chloride gets precipitated
D. Both silver chromate and silver chloride start, precipitating simultaneously

Explanation

To solve the problem, we need to analyze the situation step by step:

### Step 1: Identify the ions present in the solution
We have a solution containing:
- `10^(-5)` moles of `Cl^(-)` ions
- `10^(-5)` moles of `CrO4^(2-)` ions

### Step 2: Add silver nitrate to the solution
We are treating this solution with `10^(-4)` moles of silver nitrate (`AgNO3`). This will provide `Ag^+` ions to the solution.

### Step 3: Calculate the concentrations of the ions
Since the volume of the solution is `1 dm^3`, the concentrations of the ions are:
- Concentration of `Cl^(-)` = `10^(-5) / 1 = 10^(-5) M`
- Concentration of `CrO4^(2-)` = `10^(-5) / 1 = 10^(-5) M`
- Concentration of `Ag^+` from silver nitrate = `10^(-4) / 1 = 10^(-4) M`

### Step 4: Calculate the ionic product (Q) for AgCl
The solubility product constant (`Ksp`) for `AgCl` is given as `1 x 10^(-10)`. The ionic product (Q) for the formation of `AgCl` can be calculated using the formula:
\[ Q = [Ag^+][Cl^-] \]
Substituting the values:
\[ Q = (10^{-4})(10^{-5}) = 10^{-9} \]

### Step 5: Compare Q with Ksp for AgCl
Now we compare the ionic product (Q) with the Ksp:
- \( Q = 10^{-9} \)
- \( Ksp = 10^{-10} \)

Since \( Q > Ksp \), a precipitate of `AgCl` will form.

### Step 6: Calculate the ionic product (Q) for Ag2CrO4
The Ksp for `Ag2CrO4` is given as `4 x 10^(-12)`. The ionic product (Q) for the formation of `Ag2CrO4` can be calculated using the formula:
\[ Q = [Ag^+]^2[CrO4^{2-}] \]
Substituting the values:
\[ Q = (10^{-4})^2(10^{-5}) = 10^{-8} \]

### Step 7: Compare Q with Ksp for Ag2CrO4
Now we compare the ionic product (Q) with the Ksp:
- \( Q = 10^{-8} \)
- \( Ksp = 4 x 10^{-12} \)

Since \( Q > Ksp \), a precipitate of `Ag2CrO4` will also form.

### Conclusion
Both `AgCl` and `Ag2CrO4` will precipitate when `10^(-4)` moles of silver nitrate is added to the solution containing `10^(-5)` moles of `Cl^(-)` and `CrO4^(2-)` ions.

### Final Observation
The observation made is that both `AgCl` and `Ag2CrO4` precipitates will form in the solution.

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