The equilibrium constants for the reactions are
`H_3PO_(4) overset(K_1)( `H_2PO_(4) overset(K_2)( `HPO_(4)^(2-) overset(K_2)( The equilibrium constant for `H_3PO_4

Options

A. `K_1//K_2//K_3`
B. `K_1 xx K_2 xx K_3`
C. `K_2//K_1K_3`
D. `K_1 + K_2 + K_3`

Answer

To find the equilibrium constant for the reaction \( H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \), we can use the equilibrium constants of the individual reactions provided.

### Step-by-step Solution:

1. **Identify the Reactions and Their Equilibrium Constants**:
- The first reaction is:
\[ H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \]
with equilibrium constant \( K_1 \).

- The second reaction is:
\[ H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \]
with equilibrium constant \( K_2 \).

- The third reaction is:
\[ HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-} \]
with equilibrium constant \( K_3 \).

2. **Combine the Reactions**:
To find the overall reaction \( H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \), we can add the three reactions together:
- From the first reaction, we have:
\[ H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \]
- From the second reaction, we have:
\[ H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \]
- From the third reaction, we have:
\[ HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-} \]

When we add these reactions, the intermediate species \( H_2PO_4^- \) and \( HPO_4^{2-} \) will cancel out:
\[ H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \]

3. **Calculate the Overall Equilibrium Constant**:
When we add reactions, the equilibrium constants multiply. Therefore, the overall equilibrium constant \( K \) for the reaction \( H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \) is given by:
\[ K = K_1 \times K_2 \times K_3 \]

### Final Answer:
The equilibrium constant for the reaction \( H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \) is:
\[ K = K_1 \times K_2 \times K_3 \]