For `H_(3)PO_(4)` ,
`H_(3)PO_(4) rarr H_(2)PO_(4)^(-)+H^(+)(K_(1)), H_(2)PO_(4)^- rarr HPO_(4)^(-)+H^(+) (K_2), `
`HPO_(4)^(2-) rarr PO_(4)^(3-) + H^(+) (K_(3))` then

Options

A. `K_(1) gt K_(2) gt K_(3)`
B. `K_(1) lt K_(2)lt K_(3)`
C. ` K_(1) K_(2) lt K_(3)`
D. `K_(1).K_(2).K_(3)=K_(w)`

Answer

To determine the relationship between the equilibrium constants \( K_1 \), \( K_2 \), and \( K_3 \) for the dissociation of phosphoric acid \( H_3PO_4 \), we can analyze the ionization steps and the nature of the species involved.

### Step-by-Step Solution:

1. **Identify the Ionization Steps**:
- The first ionization of phosphoric acid is:
\[ H_3PO_4 \rightleftharpoons H_2PO_4^- + H^+ \quad (K_1) \]
- The second ionization is:
\[ H_2PO_4^- \rightleftharpoons HPO_4^{2-} + H^+ \quad (K_2) \]
- The third ionization is:
\[ HPO_4^{2-} \rightleftharpoons PO_4^{3-} + H^+ \quad (K_3) \]

2. **Understanding the Nature of Each Ionization**:
- In the first ionization, \( H_3PO_4 \) is a neutral molecule, and it loses one proton to form \( H_2PO_4^- \). This process is relatively easy because it involves a neutral species.
- In the second ionization, \( H_2PO_4^- \) is already negatively charged, and losing another proton to form \( HPO_4^{2-} \) is more difficult than the first ionization.
- In the third ionization, \( HPO_4^{2-} \) is even more negatively charged, making the loss of a proton to form \( PO_4^{3-} \) the most difficult of the three steps.

3. **Comparing the Equilibrium Constants**:
- Since the first ionization involves a neutral molecule, it will have the highest equilibrium constant \( K_1 \).
- The second ionization, involving a singly charged anion, will have a lower equilibrium constant \( K_2 \).
- The third ionization, involving a doubly charged anion, will have the lowest equilibrium constant \( K_3 \).
- Therefore, we can conclude that:
\[ K_1 > K_2 > K_3 \]

4. **Final Conclusion**:
- The relationship between the equilibrium constants for the ionization of phosphoric acid is:
\[ K_1 > K_2 > K_3 \]