`{:("List-I"," ""List-II"),(1("In" H_(3)PO_(4)"solution"),(C=[H_(3)PO_(4)])),((A)[H^(+)],P.sqrt(K_(1)C)),((B)[H_(2)PO_(4)^(-)],Q.K_(2)),((C)[HPO_(4)^(-2)],R.sqrt(K_(1)K_(2))),((D)[PO_(4)^(-3)],S.(K_(2)K_(3))/(sqrt(K_(1).C))):}`

Answer

To solve the problem, we need to match the items in List-I with the corresponding items in List-II based on the dissociation of phosphoric acid (H₃PO₄) and the relationships between the concentrations of the species involved.

### Step-by-Step Solution:

1. **Understanding the Dissociation of H₃PO₄:**
- H₃PO₄ dissociates in water to form H₂PO₄⁻ and H⁺.
- The equilibrium constant for this reaction is denoted as K₁.
- The concentration of H⁺ ions at equilibrium can be expressed as:
\[ [H^+] = \sqrt{K_1 \cdot C} \]
- Since one mole of H₃PO₄ produces one mole of H₂PO₄⁻ and one mole of H⁺, we have:
\[ [H_2PO_4^-] = \sqrt{K_1 \cdot C} \]
- Thus, we can match:
- **(A)** \([H^+]\) with **(P)** \(\sqrt{K_1 C}\)
- **(B)** \([H_2PO_4^-]\) with **(P)** \(\sqrt{K_1 C}\)

2. **Dissociation of H₂PO₄⁻:**
- H₂PO₄⁻ further dissociates to form H⁺ and HPO₄²⁻.
- The equilibrium constant for this reaction is denoted as K₂.
- The concentration of HPO₄²⁻ can be expressed as:
\[ [HPO_4^{2-}] = K_2 \cdot [H^+] \]
- Therefore, we can conclude:
- **(C)** \([HPO_4^{2-}]\) with **(Q)** \(K_2\)

3. **Dissociation of HPO₄²⁻:**
- HPO₄²⁻ dissociates to form H⁺ and PO₄³⁻.
- The equilibrium constant for this reaction is denoted as K₃.
- The concentration of PO₄³⁻ can be expressed as:
\[ [PO_4^{3-}] = \frac{K_3 \cdot [H^+]}{[HPO_4^{2-}]} \]
- Substituting the expression for \([H^+]\) and \([HPO_4^{2-}]\):
\[ [PO_4^{3-}] = \frac{K_3 \cdot \sqrt{K_1 C}}{K_2} \]
- Therefore, we can conclude:
- **(D)** \([PO_4^{3-}]\) with **(S)** \(\frac{K_2 K_3}{\sqrt{K_1 C}}\)

### Final Matching:
- A matches with P: \([H^+] = \sqrt{K_1 C}\)
- B matches with P: \([H_2PO_4^-] = \sqrt{K_1 C}\)
- C matches with Q: \([HPO_4^{2-}] = K_2\)
- D matches with S: \([PO_4^{3-}] = \frac{K_2 K_3}{\sqrt{K_1 C}}\)

### Summary of Matches:
- A → P
- B → P
- C → Q
- D → S