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 dissociation reactions of phosphoric acid.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Equilibrium Constants:**
- The first dissociation reaction is:
\[ H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^- \quad (K_1) \]
- The second dissociation reaction is:
\[ H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-} \quad (K_2) \]
- The third dissociation reaction is:
\[ HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-} \quad (K_3) \]

2. **Add the Reactions:**
- When we add these three reactions together, we can cancel out the intermediate species:
- From the first reaction, we have \( H_3PO_4 \) producing \( H^+ \) and \( H_2PO_4^- \).
- From the second reaction, \( H_2PO_4^- \) produces \( H^+ \) and \( HPO_4^{2-} \).
- From the third reaction, \( HPO_4^{2-} \) produces \( H^+ \) and \( PO_4^{3-} \).
- Adding these reactions gives:
\[ H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \]

3. **Combine the Equilibrium Constants:**
- According to the principle of equilibrium constants, when we add reactions, the equilibrium constants multiply. Therefore, the equilibrium constant for the overall reaction is:
\[ K = K_1 \times K_2 \times K_3 \]

4. **Final Result:**
- Thus, the equilibrium constant for the reaction \( H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-} \) is:
\[ K = K_1 \times K_2 \times K_3 \]