1 `dm^(3)` solution `10^(-5)` moles each of `Cl^(-)` ions and `CrO_(4)^(2-)` ions is treated with 10 moles of silver nitrate. Which one of the following observations is made?
`[K_(sp) Ag_(2) CrO_(4) = 4 xx 10^(-12)] & [K_(sp) AgCl = 1 xx 10^(-10)]`

Options

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

Answer

To solve the problem, we need to analyze the situation step by step, focusing on the solubility products (Ksp) of the salts involved and the ionic products (IP) formed when silver nitrate is added to the solution containing chloride and chromate ions.

### Step 1: Identify the ions and their concentrations
We have:
- Chloride ions (Cl⁻) = \(10^{-5}\) moles in \(1 \, dm^3\) solution
- Chromate ions (CrO₄²⁻) = \(10^{-5}\) moles in \(1 \, dm^3\) solution
- Silver nitrate (AgNO₃) is added in a quantity of 10 moles.

### Step 2: Calculate the ionic product (IP) for AgCl
The dissociation of AgCl can be represented as:
\[ \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- \]

The ionic product (IP) for AgCl is given by:
\[ \text{IP}_{AgCl} = [\text{Ag}^+][\text{Cl}^-] \]

When 10 moles of AgNO₃ are added, the concentration of Ag⁺ ions becomes 10 moles in \(1 \, dm^3\):
\[ [\text{Ag}^+] = 10 \, \text{mol/dm}^3 \]
\[ [\text{Cl}^-] = 10^{-5} \, \text{mol/dm}^3 \]

Thus, the ionic product for AgCl is:
\[ \text{IP}_{AgCl} = 10 \times 10^{-5} = 10^{-4} \]

### Step 3: Compare IP with Ksp for AgCl
The solubility product (Ksp) for AgCl is given as:
\[ K_{sp} = 1 \times 10^{-10} \]

Since:
\[ \text{IP}_{AgCl} = 10^{-4} \]
\[ K_{sp} = 1 \times 10^{-10} \]

We find that:
\[ \text{IP}_{AgCl} > K_{sp} \]

This indicates that precipitation of AgCl will occur.

### Step 4: Calculate the ionic product (IP) for Ag₂CrO₄
The dissociation of Ag₂CrO₄ can be represented as:
\[ \text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-} \]

The ionic product (IP) for Ag₂CrO₄ is given by:
\[ \text{IP}_{Ag_2CrO_4} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] \]

Using the same concentration for Ag⁺ (10 mol/dm³) and CrO₄²⁻ (10⁻⁵ mol/dm³):
\[ \text{IP}_{Ag_2CrO_4} = (10)^2 \times (10^{-5}) = 100 \times 10^{-5} = 10^{-3} \]

### Step 5: Compare IP with Ksp for Ag₂CrO₄
The solubility product (Ksp) for Ag₂CrO₄ is given as:
\[ K_{sp} = 4 \times 10^{-12} \]

Since:
\[ \text{IP}_{Ag_2CrO_4} = 10^{-3} \]
\[ K_{sp} = 4 \times 10^{-12} \]

We find that:
\[ \text{IP}_{Ag_2CrO_4} > K_{sp} \]

This indicates that precipitation of Ag₂CrO₄ will also occur.

### Step 6: Conclusion
Both AgCl and Ag₂CrO₄ will precipitate because their ionic products exceed their respective solubility products. Therefore, the correct observation is that both silver chromate and silver chloride start precipitating simultaneously.

### Final Answer:
Both silver chromate and silver chloride start precipitating simultaneously.