The precipitate of `Ag_(2)CrO_(4)(K_("sp") = 1.9 xx 10^(-12))` is obtained when equal volumes of the following are mixed.

Options

A. `10^(-5) M Ag^(+)` and `10^(-3) M CrO_(4)^(2-)`
B. `10^(-5) M Ag^(+)` and `10^(-2) M CrO_(4)^(2-)`
C. `10^(-4) M Ag^(+)` and `10^(-2) M CrO_(4)^(2-)`
D. `10^(-7) M Ag^(+)` and `10^(-3) M CrO_(4)^(2-)`

Answer

To solve the problem of determining when the precipitate of `Ag2CrO4` will form, we need to compare the ionic product (IP) of the ions in solution with the solubility product constant (Ksp) of `Ag2CrO4`. The precipitate will form when the ionic product exceeds the Ksp value.

### Step-by-Step Solution:

1. **Understanding the Dissociation of Ag2CrO4**:
The dissociation of `Ag2CrO4` in water can be represented as:
\[ Ag_2CrO_4 (s) \rightleftharpoons 2Ag^+ (aq) + CrO_4^{2-} (aq) \]
From this equation, we can see that for every mole of `Ag2CrO4` that dissolves, it produces 2 moles of `Ag^+` and 1 mole of `CrO4^{2-}`.

2. **Expression for Ionic Product**:
The ionic product (IP) can be expressed as:
\[ IP = [Ag^+]^2 \times [CrO_4^{2-}] \]

3. **Calculating Ionic Product for Each Option**:
We will calculate the ionic product for each of the given options and compare it with the Ksp value of `1.9 \times 10^{-12}`.

- **Option 1**:
- Concentration of `Ag^+ = 10^{-5} M`
- Concentration of `CrO4^{2-} = 10^{-3} M`
\[ IP = (10^{-5})^2 \times (10^{-3}) = 10^{-10} \times 10^{-3} = 10^{-13} \]
(This is less than Ksp, so no precipitation.)

- **Option 2**:
- Concentration of `Ag^+ = 10^{-5} M`
- Concentration of `CrO4^{2-} = 10^{-2} M`
\[ IP = (10^{-5})^2 \times (10^{-2}) = 10^{-10} \times 10^{-2} = 10^{-12} \]
(This is equal to Ksp, so no precipitation.)

- **Option 3**:
- Concentration of `Ag^+ = 10^{-4} M`
- Concentration of `CrO4^{2-} = 10^{-2} M`
\[ IP = (10^{-4})^2 \times (10^{-2}) = 10^{-8} \times 10^{-2} = 10^{-10} \]
(This is greater than Ksp, so precipitation occurs.)

- **Option 4**:
- Concentration of `Ag^+ = 10^{-7} M`
- Concentration of `CrO4^{2-} = 10^{-3} M`
\[ IP = (10^{-7})^2 \times (10^{-3}) = 10^{-14} \times 10^{-3} = 10^{-17} \]
(This is less than Ksp, so no precipitation.)

4. **Conclusion**:
The only option where the ionic product exceeds the Ksp value is **Option 3**. Therefore, the precipitate of `Ag2CrO4` will form when the concentrations are as given in Option 3.