Which of the following is most soluble in water?

Options

A. `Ba_(3)(PO_(4))_(2)(K_(sp)=6xx10^(-39))`
B. `ZnS(K_(sp)=7xx10^(-16))`
C. `Fe(OH)_(3)(K_(sp)=6xx10^(-38))`
D. `Ag_(3)(PO_(4))(K_(sp)=1.8xx10^(-18))`

Answer

To determine which compound is most soluble in water, we will calculate the solubility of each compound based on their solubility product constants (Ksp). The compound with the highest solubility will be the most soluble in water.

### Step-by-Step Solution:

1. **Identify the Compounds and Their Dissociation:**
- **Barium Phosphate (Ba3(PO4)2)**: Dissociates into \(3 \text{Ba}^{2+} + 2 \text{PO}_4^{3-}\)
- **Zinc Sulfide (ZnS)**: Dissociates into \(\text{Zn}^{2+} + \text{S}^{2-}\)
- **Ferric Hydroxide (Fe(OH)3)**: Dissociates into \(\text{Fe}^{3+} + 3 \text{OH}^{-}\)
- **Silver Phosphate (Ag3PO4)**: Dissociates into \(3 \text{Ag}^{+} + \text{PO}_4^{3-}\)

2. **Calculate the Solubility for Each Compound:**

**For Barium Phosphate:**
\[ K_{sp} = [\text{Ba}^{2+}]^3 [\text{PO}_4^{3-}]^2 \]
Let the solubility be \(S\):
\[ K_{sp} = (3S)^3 (2S)^2 = 27S^3 \cdot 4S^2 = 108S^5 \]
Given \(K_{sp} = 6 \times 10^{-39}\):
\[ 6 \times 10^{-39} = 108S^5 \implies S^5 = \frac{6 \times 10^{-39}}{108} \implies S = \left(\frac{6}{108}\right)^{1/5} \times 10^{-39/5} \]
Solving gives \(S \approx 0.8 \times 10^{-8}\).

**For Zinc Sulfide:**
\[ K_{sp} = [\text{Zn}^{2+}][\text{S}^{2-}] = S \cdot S = S^2 \]
Given \(K_{sp} = 7 \times 10^{-16}\):
\[ S^2 = 7 \times 10^{-16} \implies S = \sqrt{7 \times 10^{-16}} \approx 2.6 \times 10^{-8} \]

**For Ferric Hydroxide:**
\[ K_{sp} = [\text{Fe}^{3+}][\text{OH}^{-}]^3 = S \cdot (3S)^3 = 27S^4 \]
Given \(K_{sp} = 6 \times 10^{-38}\):
\[ 6 \times 10^{-38} = 27S^4 \implies S^4 = \frac{6 \times 10^{-38}}{27} \implies S = \left(\frac{6}{27}\right)^{1/4} \times 10^{-38/4} \]
Solving gives \(S \approx 2.16 \times 10^{-10}\).

**For Silver Phosphate:**
\[ K_{sp} = [\text{Ag}^{+}]^3[\text{PO}_4^{3-}] = (3S)^3 \cdot S = 27S^4 \]
Given \(K_{sp} = 1.8 \times 10^{-18}\):
\[ 1.8 \times 10^{-18} = 27S^4 \implies S^4 = \frac{1.8 \times 10^{-18}}{27} \implies S = \left(\frac{1.8}{27}\right)^{1/4} \times 10^{-18/4} \]
Solving gives \(S \approx 1.6 \times 10^{-5}\).

3. **Compare the Solubility Values:**
- Barium Phosphate: \(0.8 \times 10^{-8}\)
- Zinc Sulfide: \(2.6 \times 10^{-8}\)
- Ferric Hydroxide: \(2.16 \times 10^{-10}\)
- Silver Phosphate: \(1.6 \times 10^{-5}\)

4. **Conclusion:**
The compound with the highest solubility is **Silver Phosphate (Ag3PO4)**, with a solubility of \(1.6 \times 10^{-5}\).

### Final Answer:
**Silver Phosphate (Ag3PO4) is the most soluble in water.**