The `EMF` of the following cell is observed to be `0.118V` at `25^(@)C:`

If `30mL` of `0.2M NaOH` is added to the negative terminal of the battery, find the `EMF` of the cell.

Explanation

It is concentration cell, therefore, `E^(c-)=0`.
`E_(cell)=-(0.059)/(1) log .([H^(o+)]_(a))/([H^(o+)]_(c))`
`=0.118=-0.059log.([H^(o+)]_(a))/(0.1)`
`:. [H^(o+)]_(a)=10^(-3)M`
`{:(,HA,rarr,H^(o+),+A^(c-)),(Initial,0.1,,0,0),(At equilibrium,(0.1-1.0^(-3)),,10^(-3),10^(-3)),(, ~~0.1, , ,):}`
`K_(a)=([H^(o+)][A^(c-)])/([HA])=((10^(-3))^(2))/(0.1)=10^(-5)`
When `Na OH` is added to `HA`, acidic buffer is formed at negative electrode.
`{:(,HA,+,NaOH,rarr,NaA,+,H_(2)O),(Intial,0.01,,0.006,,-,,-),(Fi nal,(0.01-0.06),, 0,, 0.006,, ),(, =0.004,,,,,,):}`
Buffer equation `: pH=pK_(a)+log.([Sa l t])/([Ac i d])`
or `H^(o+)=K_(a)xx(["Acid" ])/(["Salt"])=10^(-5)xx(0.004)/(0.006)=6.67xx10^(-6)`
`E_(cell)=E^(c-)._(cell)-(0.059)/(1)log.((6.67xx10^(-6)))/(0.1)`
`=0.246V`