`H_(3)PO_(4)` is a tribasic acid with `pK_(a_(1)), pK_(a_(2))` and `pK_(a_(3)) 1.12, 7.21`, and `12.32`, respectively.`10^(-3)M H_(3)PO_(4)(pH = 7)` is used in fertilizers as an aqueous soil digesting. Plants can absorb zinc in water soluble from only. Zinc phosphate is the source of zinc and `PO_(4)^(3-`) ions in the soil. `K_(sp)` of zinc phosphate `= 9.1 xx 10^(-33)`.
`[PO_(4)^(3-)]` ion in the soil with `pH = 7`, is

Options

A. `10^(-3)M`
B. `1.2 xx 10^(-4)M`
C. `2.2 xx 10^(-4)M`
D. `1.1xx10^(-10)M`

Explanation

To solve the problem, we need to find the concentration of the phosphate ion \([PO_4^{3-}]\) in the soil at a pH of 7, given that \(H_3PO_4\) is a tribasic acid with known \(pK_a\) values and a specific concentration.

### Step-by-Step Solution:

1. **Identify the dissociation of phosphoric acid**:
Phosphoric acid (\(H_3PO_4\)) dissociates in three steps:
\[ H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \]
\[ H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \]
\[ HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-} \]

2. **Determine the \(K_a\) values from \(pK_a\)**:
Using the given \(pK_a\) values:
- \(pK_{a1} = 1.12 \Rightarrow K_{a1} = 10^{-1.12} \approx 7.76 \times 10^{-2}\)
- \(pK_{a2} = 7.21 \Rightarrow K_{a2} = 10^{-7.21} \approx 6.17 \times 10^{-8}\)
- \(pK_{a3} = 12.32 \Rightarrow K_{a3} = 10^{-12.32} \approx 4.79 \times 10^{-13}\)

3. **Calculate the overall dissociation constant (\(K\))**:
The overall dissociation constant \(K\) for the complete dissociation of \(H_3PO_4\) can be expressed as:
\[ K = K_{a1} \times K_{a2} \times K_{a3} \]

4. **Calculate \(H^+\) concentration**:
At pH = 7:
\[ [H^+] = 10^{-7} \, M \]

5. **Concentration of \(H_3PO_4\)**:
The initial concentration of \(H_3PO_4\) is given as \(10^{-3} \, M\). Since it is a weak acid, we can assume that its dissociation is negligible.

6. **Set up the equilibrium expression**:
The equilibrium expression for the dissociation of \(H_3PO_4\) can be written as:
\[ K = \frac{[H^+]^3 \cdot [PO_4^{3-}]}{[H_3PO_4]} \]

7. **Substitute known values into the equilibrium expression**:
Substitute \(K\), \([H^+]\), and \([H_3PO_4]\) into the equilibrium expression:
\[ K = \frac{(10^{-7})^3 \cdot [PO_4^{3-}]}{10^{-3}} \]

8. **Solve for \([PO_4^{3-}]\)**:
Rearranging the equation:
\[ [PO_4^{3-}] = K \cdot 10^{-3} / (10^{-7})^3 \]
\[ [PO_4^{3-}] = K \cdot 10^{-3} / 10^{-21} \]
\[ [PO_4^{3-}] = K \cdot 10^{18} \]

9. **Calculate \(K\)**:
Calculate \(K\) using the values of \(K_{a1}\), \(K_{a2}\), and \(K_{a3}\):
\[ K \approx (7.76 \times 10^{-2}) \times (6.17 \times 10^{-8}) \times (4.79 \times 10^{-13}) \approx 2.36 \times 10^{-22} \]

10. **Final Calculation**:
\[ [PO_4^{3-}] = (2.36 \times 10^{-22}) \cdot 10^{18} \approx 2.36 \times 10^{-4} \, M \]

### Conclusion:
The concentration of \([PO_4^{3-}]\) ions in the soil at pH = 7 is approximately \(2.36 \times 10^{-4} \, M\).