`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)` . Ksp of Zinc phosphate= 9.1 xx 10^(-33)`.
`[Zn^(2+)]` ion in the soil is

Options

A. `2.9 xx 10^(-11)M`
B. `4.0 xx 10^(-10)M`
C. `3.0 xx 10^(-6)M`
D. `9.1 xx 10^(-5)M`

Explanation

To solve the problem step by step, we will follow the process of calculating the concentration of \( \text{Zn}^{2+} \) ions in the soil based on the dissociation of phosphoric acid and the solubility product of zinc phosphate.

### Step 1: Calculate \( K_a \) values from \( pK_a \)

Given:
- \( pK_a1 = 1.12 \)
- \( pK_a2 = 7.21 \)
- \( pK_a3 = 12.32 \)

Using the formula:
\[ K_a = 10^{-pK_a} \]

Calculating each \( K_a \):
\[ K_a1 = 10^{-1.12} \approx 7.76 \times 10^{-2} \]
\[ K_a2 = 10^{-7.21} \approx 6.17 \times 10^{-8} \]
\[ K_a3 = 10^{-12.32} \approx 4.79 \times 10^{-13} \]

### Step 2: Calculate the overall dissociation constant \( K \)

The overall dissociation constant \( K \) for the reaction can be expressed as:
\[ K = K_a1 \times K_a2 \times K_a3 \]
Substituting the values:
\[ K = (7.76 \times 10^{-2}) \times (6.17 \times 10^{-8}) \times (4.79 \times 10^{-13}) \]

Calculating \( K \):
\[ K \approx 2.36 \times 10^{-22} \]

### Step 3: Relate \( K \) to concentrations

The equilibrium expression for the dissociation of \( \text{H}_3\text{PO}_4 \) is:
\[ K = \frac{[H^+]^3 [PO_4^{3-}]}{[H_3PO_4]} \]

Given:
- \( [H_3PO_4] = 10^{-3} \, M \)
- \( pH = 7 \) implies \( [H^+] = 10^{-7} \, M \)

Substituting these values into the expression:
\[ 2.36 \times 10^{-22} = \frac{(10^{-7})^3 [PO_4^{3-}]}{10^{-3}} \]

### Step 4: Solve for \( [PO_4^{3-}] \)

Rearranging the equation:
\[ [PO_4^{3-}] = \frac{2.36 \times 10^{-22} \times 10^{-3}}{(10^{-7})^3} \]
\[ = \frac{2.36 \times 10^{-25}}{10^{-21}} = 2.36 \times 10^{-4} \, M \]

### Step 5: Use \( K_{sp} \) of Zinc Phosphate to find \( [Zn^{2+}] \)

Given:
- \( K_{sp} = 9.1 \times 10^{-33} \)
- The dissociation of zinc phosphate is:
\[ Zn_3(PO_4)_2 \rightleftharpoons 3Zn^{2+} + 2PO_4^{3-} \]

The \( K_{sp} \) expression is:
\[ K_{sp} = [Zn^{2+}]^3 [PO_4^{3-}]^2 \]

Substituting \( [PO_4^{3-}] \):
\[ 9.1 \times 10^{-33} = [Zn^{2+}]^3 (2.36 \times 10^{-4})^2 \]
\[ = [Zn^{2+}]^3 \times 5.57 \times 10^{-8} \]

### Step 6: Solve for \( [Zn^{2+}] \)

Rearranging gives:
\[ [Zn^{2+}]^3 = \frac{9.1 \times 10^{-33}}{5.57 \times 10^{-8}} \]
\[ = 1.63 \times 10^{-25} \]

Taking the cube root:
\[ [Zn^{2+}] = (1.63 \times 10^{-25})^{1/3} \approx 2.5 \times 10^{-9} \, M \]

### Final Answer

Thus, the concentration of \( \text{Zn}^{2+} \) ions in the soil is approximately:
\[ \boxed{2.5 \times 10^{-9} \, M} \]