[tex]$30.0 \text{ mL}$[/tex] of [tex]$0.400 \text{ M Zn}^{2+}$[/tex] is mixed with [tex]$100.0 \text{ mL}$[/tex] of [tex]$0.200 \text{ M F}^{-}$[/tex] [tex]$\left( K_{sp} \text{ ZnF}_2 = 0.030 \right)$[/tex].

In the mixture, you would observe:
A. a precipitate forming.
B. no reaction taking place.



Answer :

To determine whether a precipitate forms or a reaction takes place when mixing 30.0 mL of 0.400 M [tex]\( \text{Zn}^{2+} \)[/tex] with 100.0 mL of 0.200 M [tex]\( \text{F}^{-} \)[/tex], we need to follow several steps. Let’s break it down:

### Step 1: Calculate the Total Volume of the Mixture
The total volume of the mixture will be the sum of the volumes of the two solutions:
[tex]\[ \text{Total Volume} = 30.0 \, \text{mL} + 100.0 \, \text{mL} = 130.0 \, \text{mL} \][/tex]

### Step 2: Determine the Final Concentration of Each Ion
To find the final concentrations of [tex]\( \text{Zn}^{2+} \)[/tex] and [tex]\( \text{F}^{-} \)[/tex] after mixing, we use the formula for dilution:
[tex]\[ \text{Concentration Final} = \frac{\text{Concentration Initial} \times \text{Volume Initial}}{\text{Total Volume}} \][/tex]

#### For [tex]\( \text{Zn}^{2+} \)[/tex]:
[tex]\[ \text{Zn}^{2+} \text{ Final Concentration} = \frac{0.400 \, M \times 30.0 \, \text{mL}}{130.0 \, \text{mL}} = 0.09230769230769231 \, M \][/tex]

#### For [tex]\( \text{F}^{-} \)[/tex]:
[tex]\[ \text{F}^{-} \text{ Final Concentration} = \frac{0.200 \, M \times 100.0 \, \text{mL}}{130.0 \, \text{mL}} = 0.15384615384615385 \, M \][/tex]

### Step 3: Calculate the Ion Product (Q) for [tex]\( \text{ZnF}_2 \)[/tex]
The ion product [tex]\( Q \)[/tex] is given by the expression:
[tex]\[ Q = [\text{Zn}^{2+}] \times [\text{F}^{-}]^2 \][/tex]

Substitute the final concentrations:
[tex]\[ Q = (0.09230769230769231) \times (0.15384615384615385)^2 \approx 0.002184797451069641 \][/tex]

### Step 4: Compare the Ion Product (Q) with the Solubility Product Constant ([tex]\( K_{sp} \)[/tex])
The solubility product constant for [tex]\( \text{ZnF}_2 \)[/tex] is given as [tex]\( K_{sp} = 0.030 \)[/tex].

To determine if a precipitate forms, we compare [tex]\( Q \)[/tex] with [tex]\( K_{sp} \)[/tex]:
- If [tex]\( Q > K_{sp} \)[/tex], a precipitate forms.
- If [tex]\( Q \leq K_{sp} \)[/tex], no precipitate forms.

In this case:
[tex]\[ 0.002184797451069641 < 0.030 \][/tex]

Since [tex]\( Q \)[/tex] is less than [tex]\( K_{sp} \)[/tex], no precipitate will form.

### Conclusion
In this mixture, you would observe:
B. No reaction taking place.