pH is the negative logarithm of `H^(+)`
` pH=log[H^(+)]`
`HCl lt H^(+)+Cl^(-)`
`H_(2)OhArrH^(+)+OH^(-)`
` K_(W)=[H^(+)][OH^(-)]`
`K_(W)` depend on the temperatue. With rise in temperature `K_(W)` increases.
At 298 K, pH of pure water = 7
The pH at first equivalance of `H_(3)PO_(4)` vs NaOH will be

Options

A. 7
B. `gt7`
C. `lt7`
D. Depend on the concentration of titrant

Answer

To find the pH at the first equivalence point of the titration of phosphoric acid (H₃PO₄) with sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Understand the dissociation of H₃PO₄
Phosphoric acid (H₃PO₄) is a triprotic acid, meaning it can donate three protons (H⁺). The first dissociation can be represented as:
\[ H₃PO₄ \rightleftharpoons H₂PO₄^- + H^+ \]
At the first equivalence point, one mole of NaOH will neutralize one mole of H⁺ from H₃PO₄.

### Step 2: Determine the species present at the first equivalence point
At the first equivalence point, the reaction can be summarized as:
\[ H₃PO₄ + NaOH \rightarrow NaH₂PO₄ + H₂O \]
At this point, we have the salt sodium dihydrogen phosphate (NaH₂PO₄) in solution.

### Step 3: Identify the nature of NaH₂PO₄ in solution
NaH₂PO₄ is a weak acid because it can still donate a proton:
\[ NaH₂PO₄ \rightleftharpoons H^+ + HPO₄^{2-} \]
This means that the solution will still have acidic properties, and thus the pH will be less than 7.

### Step 4: Calculate the pH
To find the pH at the first equivalence point, we need to consider the dissociation of NaH₂PO₄. The equilibrium expression for the dissociation can be written as:
\[ K_a = \frac{[H^+][HPO₄^{2-}]}{[NaH₂PO₄]} \]
Since we know that the solution is weakly acidic, we can use the \(K_a\) value for the dissociation of NaH₂PO₄ to find the concentration of H⁺ ions.

### Step 5: Use the \(K_a\) value
The \(K_a\) value for the dissociation of NaH₂PO₄ can be found in tables. For the first dissociation of phosphoric acid, it is approximately \(7.5 \times 10^{-3}\). We can set up an ICE table to find the concentration of H⁺ at equilibrium.

Assuming that the initial concentration of NaH₂PO₄ is \(C\):
- Initial: [NaH₂PO₄] = C, [H⁺] = 0, [HPO₄^{2-}] = 0
- Change: [NaH₂PO₄] = C-x, [H⁺] = x, [HPO₄^{2-}] = x
- Equilibrium: [NaH₂PO₄] = C-x, [H⁺] = x, [HPO₄^{2-}] = x

Substituting into the \(K_a\) expression:
\[ K_a = \frac{x^2}{C-x} \]
Assuming \(C\) is much larger than \(x\), we can simplify to:
\[ K_a \approx \frac{x^2}{C} \]
From this, we can solve for \(x\) (which represents [H⁺]).

### Step 6: Calculate pH
Once we have the concentration of H⁺ ions, we can calculate the pH using:
\[ pH = -\log[H^+] \]

### Final Answer
The pH at the first equivalence point of H₃PO₄ with NaOH will be less than 7 due to the presence of the weak acid NaH₂PO₄.