Equinormal solutions of two weak acids `HA(pK_(a)=3)` and `HB(pK_(a)=5)` are each placed in contact with standard hydrogen electrode at `25^(@)C(T=298

Question

Equinormal solutions of two weak acids `HA(pK_(a)=3)` and `HB(pK_(a)=5)` are each placed in contact with standard hydrogen electrode at `25^(@)C(T=298

Equinormal solutions of two weak acids `HA(pK_(a)=3)` and `HB(pK_(a)=5)` are each placed in contact with standard hydrogen electrode at `25^(@)C(T=298K)` when a cell is constructed by interconnecting them through a salt bridge find the e.m.f. of the cell.

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - `E=0.059V`
`Pt//H_(2)^(@)//H^(+)(HA)//H^(+)(HB)//H_(2)//Pt`
`H_(2)to2H^(+)+2e^(-)`
`2H^(+)+2e^(-)toH_(2)`
`2H_(HB)^(+)to2H_(HA)^(+)" "E^(@)=0`
`E=0-(0.0591)/(2)log(([H^(+)]_(HA)^(2))/([H^(+)]_(HB)^(2)))`
But `Ka=([H^(+)]^(2))/(c)`
`=-(0.0591)/(2)log((10^(-3)xxC)/(10^(-5)xxC))=-0.0591`
the cell is constructed in reversed direction.
`E_(cell)=0.0591"volt"`