A solution is 0.10 M in `Ag^(+), Ca^(2+), Mg^(2+)` and `Al^(3+)` ions. Which compound will precipitate at the lowest `[PO_(4)^(3-)]` when a solution of `Na_(3)PO_(4)` is added ?

Options

A. `Ag_3PO_4(K_(sp)=1xx10^(-6))`
B. `Ca_3(PO)4)_2(K_(sp)=1xx10^(-33))`
C. `Mg_3(PO_4)_2(K_(sp)=1xx10^(-24))`
D. `AIPO_4(K_(sp)=1xx10^(-20))`

Answer

To determine which compound will precipitate at the lowest concentration of \([PO_4^{3-}]\) when a solution of \(Na_3PO_4\) is added, we need to calculate the concentration of \([PO_4^{3-}]\) required for precipitation for each of the given compounds based on their solubility product constants (\(K_{sp}\)).

### Given Data:
1. **Compounds and their \(K_{sp}\)**:
- \(Ag_3PO_4\): \(K_{sp} = 1 \times 10^{-6}\)
- \(Ca_3(PO_4)_2\): \(K_{sp} = 1 \times 10^{-33}\)
- \(Mg_3(PO_4)_2\): \(K_{sp} = 1 \times 10^{-24}\)
- \(AlPO_4\): \(K_{sp} = 1 \times 10^{-20}\)

2. **Concentration of ions in solution**:
- \([Ag^+] = 0.10 \, M\)
- \([Ca^{2+}] = 0.10 \, M\)
- \([Mg^{2+}] = 0.10 \, M\)
- \([Al^{3+}] = 0.10 \, M\)

### Step-by-Step Calculation:

#### Step 1: Calculate \([PO_4^{3-}]\) for \(Ag_3PO_4\)
The dissociation of \(Ag_3PO_4\) can be represented as:
\[ Ag_3PO_4 \rightleftharpoons 3Ag^+ + PO_4^{3-} \]
The \(K_{sp}\) expression is:
\[ K_{sp} = [Ag^+]^3 [PO_4^{3-}] \]
Rearranging gives:
\[ [PO_4^{3-}] = \frac{K_{sp}}{[Ag^+]^3} = \frac{1 \times 10^{-6}}{(0.1)^3} = \frac{1 \times 10^{-6}}{0.001} = 1 \times 10^{-3} \, M \]

#### Step 2: Calculate \([PO_4^{3-}]\) for \(Ca_3(PO_4)_2\)
The dissociation of \(Ca_3(PO_4)_2\) is:
\[ Ca_3(PO_4)_2 \rightleftharpoons 3Ca^{2+} + 2PO_4^{3-} \]
The \(K_{sp}\) expression is:
\[ K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2 \]
Rearranging gives:
\[ [PO_4^{3-}] = \sqrt{\frac{K_{sp}}{[Ca^{2+}]^3}} = \sqrt{\frac{1 \times 10^{-33}}{(0.1)^3}} = \sqrt{\frac{1 \times 10^{-33}}{0.001}} = \sqrt{1 \times 10^{-30}} = 1 \times 10^{-15} \, M \]

#### Step 3: Calculate \([PO_4^{3-}]\) for \(Mg_3(PO_4)_2\)
The dissociation of \(Mg_3(PO_4)_2\) is:
\[ Mg_3(PO_4)_2 \rightleftharpoons 3Mg^{2+} + 2PO_4^{3-} \]
The \(K_{sp}\) expression is:
\[ K_{sp} = [Mg^{2+}]^3 [PO_4^{3-}]^2 \]
Rearranging gives:
\[ [PO_4^{3-}] = \sqrt{\frac{K_{sp}}{[Mg^{2+}]^3}} = \sqrt{\frac{1 \times 10^{-24}}{(0.1)^3}} = \sqrt{\frac{1 \times 10^{-24}}{0.001}} = \sqrt{1 \times 10^{-21}} = 1 \times 10^{-10.5} \, M \approx 3.16 \times 10^{-11} \, M \]

#### Step 4: Calculate \([PO_4^{3-}]\) for \(AlPO_4\)
The dissociation of \(AlPO_4\) is:
\[ AlPO_4 \rightleftharpoons Al^{3+} + PO_4^{3-} \]
The \(K_{sp}\) expression is:
\[ K_{sp} = [Al^{3+}][PO_4^{3-}] \]
Rearranging gives:
\[ [PO_4^{3-}] = \frac{K_{sp}}{[Al^{3+}]} = \frac{1 \times 10^{-20}}{0.1} = 1 \times 10^{-19} \, M \]

### Summary of Results:
- \([PO_4^{3-}]_{Ag_3PO_4} = 1 \times 10^{-3} \, M\)
- \([PO_4^{3-}]_{Ca_3(PO_4)_2} = 1 \times 10^{-15} \, M\)
- \([PO_4^{3-}]_{Mg_3(PO_4)_2} \approx 3.16 \times 10^{-11} \, M\)
- \([PO_4^{3-}]_{AlPO_4} = 1 \times 10^{-19} \, M\)

### Conclusion:
The compound that will precipitate at the lowest \([PO_4^{3-}]\) is **\(AlPO_4\)**, with a precipitation threshold of \(1 \times 10^{-19} \, M\).

### Answer:
**\(AlPO_4\)** will precipitate at the lowest \([PO_4^{3-}]\).