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Equal volume of the following Ca^(+2) an...

Equal volume of the following `Ca^(+2)` and `F^(-)` solutions are mixed. In which case will percipitation occur? (`K_(sp)` of `CaF_(2) = 7 xx 10^(-10)`).

A

`10^(-2) M Ca^(+2) + 10^(-5)MF^(-)`

B

`10^(-3) M Ca^(+2) + 10^(-5)MF^(-)`

C

`10^(-4) M Ca^(+2) + 10^(-2)MF^(-)`

D

`10^(-2) M Ca^(+2) + 10^(-6)MF^(-)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine whether precipitation occurs when equal volumes of `Ca^(+2)` and `F^(-)` solutions are mixed, we need to apply the solubility product constant (Ksp) for calcium fluoride (CaF2). The Ksp for CaF2 is given as \( 7 \times 10^{-10} \). ### Step-by-step Solution: 1. **Understanding the Reaction**: The dissolution of calcium fluoride can be represented as: \[ CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2F^{-} (aq) \] The solubility product expression is: \[ K_{sp} = [Ca^{2+}][F^{-}]^2 \] 2. **Condition for Precipitation**: Precipitation occurs when the product of the ion concentrations exceeds the Ksp value: \[ [Ca^{2+}][F^{-}]^2 > K_{sp} \] 3. **Analyzing Each Case**: We will analyze each option provided in the question to see if precipitation occurs. **Option 1**: \( [Ca^{2+}] = 10^{-2} \, M \) and \( [F^{-}] = 10^{-5} \, M \) \[ [Ca^{2+}][F^{-}]^2 = (10^{-2})(10^{-5})^2 = (10^{-2})(10^{-10}) = 10^{-12} \] Since \( 10^{-12} < 7 \times 10^{-10} \), precipitation does not occur. **Option 2**: \( [Ca^{2+}] = 10^{-3} \, M \) and \( [F^{-}] = 10^{-5} \, M \) \[ [Ca^{2+}][F^{-}]^2 = (10^{-3})(10^{-5})^2 = (10^{-3})(10^{-10}) = 10^{-13} \] Since \( 10^{-13} < 7 \times 10^{-10} \), precipitation does not occur. **Option 3**: \( [Ca^{2+}] = 10^{-4} \, M \) and \( [F^{-}] = 10^{-2} \, M \) \[ [Ca^{2+}][F^{-}]^2 = (10^{-4})(10^{-2})^2 = (10^{-4})(10^{-4}) = 10^{-8} \] Since \( 10^{-8} > 7 \times 10^{-10} \), precipitation occurs. **Option 4**: \( [Ca^{2+}] = 10^{-2} \, M \) and \( [F^{-}] = 10^{-6} \, M \) \[ [Ca^{2+}][F^{-}]^2 = (10^{-2})(10^{-6})^2 = (10^{-2})(10^{-12}) = 10^{-14} \] Since \( 10^{-14} < 7 \times 10^{-10} \), precipitation does not occur. 4. **Conclusion**: Precipitation occurs only in **Option 3** where \( [Ca^{2+}] = 10^{-4} \, M \) and \( [F^{-}] = 10^{-2} \, M \).

To determine whether precipitation occurs when equal volumes of `Ca^(+2)` and `F^(-)` solutions are mixed, we need to apply the solubility product constant (Ksp) for calcium fluoride (CaF2). The Ksp for CaF2 is given as \( 7 \times 10^{-10} \). ### Step-by-step Solution: 1. **Understanding the Reaction**: The dissolution of calcium fluoride can be represented as: \[ CaF_2 (s) \rightleftharpoons Ca^{2+} (aq) + 2F^{-} (aq) ...
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