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The precipitate of Ag(2)CrO(4)(K("sp") =...

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

A

`10^(-4)M Ag^(+) + 10^(-4) M CrO_(4)^(2-)`

B

`10^(-2)M Ag^(+) + 10^(-3) M`

C

`10^(-5)M Ag^(+)+ 10^(-3) M`

D

`10^(-4)M Ag^(+) + 10^(-5) M`

Text Solution

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The correct Answer is:
To determine when the precipitate of `Ag2CrO4` will form upon mixing equal volumes of solutions containing `Ag^+` and `CrO4^2-`, we need to calculate the ionic product (IP) and compare it with the solubility product constant (Ksp) of `Ag2CrO4`. ### Step-by-Step Solution: 1. **Understand the Reaction**: The formation of the precipitate can be represented by the following equilibrium: \[ Ag^+ + CrO_4^{2-} \rightleftharpoons Ag_2CrO_4 (s) \] 2. **Write the Expression for Ksp**: The solubility product expression for `Ag2CrO4` is given by: \[ K_{sp} = [Ag^+]^2 [CrO_4^{2-}] \] Given that \( K_{sp} = 1.9 \times 10^{-12} \). 3. **Define Molar Solubility**: Let the molar solubility of `Ag2CrO4` be \( S \). When `Ag2CrO4` dissolves, it produces: - 2 moles of `Ag^+` for every mole of `Ag2CrO4` - 1 mole of `CrO4^{2-}` for every mole of `Ag2CrO4` Therefore, at equilibrium: \[ [Ag^+] = 2S \quad \text{and} \quad [CrO_4^{2-}] = S \] 4. **Mixing Equal Volumes**: When equal volumes of solutions are mixed, the concentrations of the ions will be halved. If the initial concentrations of `Ag^+` and `CrO4^{2-}` are \( C_{Ag} \) and \( C_{CrO4} \) respectively, after mixing: \[ [Ag^+] = \frac{C_{Ag}}{2} \quad \text{and} \quad [CrO_4^{2-}] = \frac{C_{CrO4}}{2} \] 5. **Calculate the Ionic Product (IP)**: The ionic product can be calculated using the concentrations after mixing: \[ IP = \left(\frac{C_{Ag}}{2}\right)^2 \left(\frac{C_{CrO4}}{2}\right) \] Simplifying gives: \[ IP = \frac{C_{Ag}^2}{4} \cdot \frac{C_{CrO4}}{2} = \frac{C_{Ag}^2 \cdot C_{CrO4}}{8} \] 6. **Compare IP with Ksp**: For precipitation to occur, the ionic product must exceed the solubility product: \[ IP > K_{sp} \] Substitute the calculated IP into this inequality to determine if precipitation occurs. 7. **Evaluate Different Cases**: You will need to evaluate the concentrations given in the options to find the correct one where \( IP > K_{sp} \).

To determine when the precipitate of `Ag2CrO4` will form upon mixing equal volumes of solutions containing `Ag^+` and `CrO4^2-`, we need to calculate the ionic product (IP) and compare it with the solubility product constant (Ksp) of `Ag2CrO4`. ### Step-by-Step Solution: 1. **Understand the Reaction**: The formation of the precipitate can be represented by the following equilibrium: \[ Ag^+ + CrO_4^{2-} \rightleftharpoons Ag_2CrO_4 (s) ...
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