In the acid-base titration [H3PO4(0.1M)+NaOH(0.1M)] , the e.m.f of the solution is measured by coupling this electrode with a suitable reference electrode. When an alkali is added, the pH of the solution is in accordance with the equation: Ecell=E∘cell+0.059 pH For H3PO4:Ka1=10−3,Ka2=10−8,Ka3=10−13 What is the e.m.f of the cell at the IInd end point of the titration if E∘cell at this stage is 1.3805 V

In the acid-base titration [H3PO4(0.1M)+NaOH(0.1M)] , the e.m.f of the

1.	In the acid-base titration [H3PO4(0.1M)+NaOH(0.1M)] , the e.m.f of the solution is measured by coupling this electrode with a suitable reference electrode. When an alkali is added, the pH of the solution is in accordance with the equation: Ecell=E∘cell+0.059 pH For H3PO4:Ka1=10−3,Ka2=10−8,Ka3=10−13 What is the e.m.f of the cell at the IInd end point of the titration if E∘cell at this stage is 1.3805 V
Answer» In the acid-base titration $[H_{3} P O_{4} (0.1 M) + N a O H (0.1 M)]$ , the e.m.f of the solution is measured by coupling this electrode with a suitable reference electrode. When an alkali is added, the pH of the solution is in accordance with the equation: $E_{cell} = E_{cell}^{\circ} + 0.059 pH$ For $H_{3} P O_{4} : K a_{1} = 10^{- 3}, K a_{2} = 10^{- 8}, K a_{3} = 10^{- 13}$ What is the e.m.f of the cell at the ${II}^{nd}$ end point of the titration if $E_{cell}^{\circ}$ at this stage is $1.3805 V$