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In the acid base titration [H3PO4(0.1 M)...

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.

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The correct Answer is:
2
`pH=(pKa_2+pKa_3)/2=10.5`
`E_(cell)=E_(cell)^@+0.059 pH=1.3805+0.059xx10.5=2V`
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