The e.m.f. of cell: `H_(2)(g)` |Buffer| Normal calomal electrode is `0.6885V` at `40^(@)C` when the barometric pressure is `725mm` of Hg. What is the `pH` of the solution `E_("calomal")^(@) = 0.28`.

Answer

6.6

Explanation

`E = E_(calmol)^(@) +E_(H_(2)//H^(+))^(@) -(RT)/(nF) "In" ([H^(+)]^(2))/(P_(H_(2)))`
`0.6885 = 0.28 -(8.314 xx 313)/(2 xx 96500) "In"([H^(+)]^(2))/(725) xx 760`
`[H^(+)] = 2.57 xx 10^(-7)`
`pH = 6.6`