In the acid base titration `[H_3PO_4(0.1 M)+NaOH(0.1 M)]` e.m.f of the solution is measured by coupling this electrodes with suitable reference electrode.When alkali is added pH of solution is in acoodance with equation
`E_(Cell)=E_(Cell)^(@)+0.059 pH`
For `H_3PO_(4) Ka_(1)=10^(-3) , Ka_(2)=10^(-8), Ka_(3)=10^(-13)`
What is the cell e.m.f at the lind end point of the titration if `E_(cell)^(@)` at this stage is 1.3805 V.

Explanation

To solve the problem, we need to find the cell e.m.f at the end point of the titration of phosphoric acid (H₃PO₄) with sodium hydroxide (NaOH). We will follow these steps:

### Step 1: Calculate the pH at the end point of the titration
At the end point of the titration of a diprotic acid like H₃PO₄ with NaOH, we can assume that the pH is determined by the average of the pKa values of the acid. The pKa values are derived from the given Ka values.

Given:
- \( K_{a1} = 10^{-3} \)
- \( K_{a2} = 10^{-8} \)
- \( K_{a3} = 10^{-13} \)

Calculating the pKa values:
- \( pK_{a1} = -\log(10^{-3}) = 3 \)
- \( pK_{a2} = -\log(10^{-8}) = 8 \)
- \( pK_{a3} = -\log(10^{-13}) = 13 \)

At the end point, we can use the average of \( pK_{a1} \) and \( pK_{a3} \):
\[ pH = \frac{pK_{a1} + pK_{a3}}{2} = \frac{3 + 13}{2} = \frac{16}{2} = 8 \]

### Step 2: Use the Nernst equation to calculate the cell e.m.f
The Nernst equation is given by:
\[ E_{cell} = E_{cell}^0 + 0.059 \cdot pH \]

We are given:
- \( E_{cell}^0 = 1.3805 \, V \)
- \( pH = 8 \)

Substituting the values into the equation:
\[ E_{cell} = 1.3805 + 0.059 \cdot 8 \]
Calculating \( 0.059 \cdot 8 \):
\[ 0.059 \cdot 8 = 0.472 \]

Now, substituting this back:
\[ E_{cell} = 1.3805 + 0.472 = 1.8525 \, V \]

### Step 3: Round off the answer
Rounding off the value, we can state that the e.m.f at the end point of the titration is approximately:
\[ E_{cell} \approx 1.85 \, V \]

### Final Answer
The cell e.m.f at the end point of the titration is approximately **1.85 V**.

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