The `emf` of the following cell is observed to be `0.118V` at `25^(@)C`.
`[Pt,H_(2)(1atm)|HA(100mL 0.1M||H^(o+)(0.1M)|H_(2)(1atm)|Pt]`
a. If `30mL` of `0.2M NaOH` is added to the negative terminal of battery, find the emf of the cell.
b. If `50mL` of `0.2 M NaOH` is added to the negative terminal of battery, find the emf of teh cell.

Explanation

(a = anode, c = cathode)
`E_(H_(2)) = - 0.059 (pH_(c) - pH_(a)) [[[H^(o+)]_(c)=,0.1M],[pH=1,]]`
`0.118 =- 0.059 (1-pH_(a))`
`- (0.118)/(0.059) = 1 - pH_(a)`
`:. pH_(a) = 1+(0.118)/(0.059) = 3`
`[H^(o+)]_(a) = 10^(-3)M`
Since `HA` is a `W_(A)`
`:. pH_(W_(A)) = (1)/(2) (pK_(a) - logC)`
`3 = (1)/(2) (pK_(a) - log 0.1)`
`:. pK_(a) = 5`.
a. When `30mL` of `0.2M NaOH`
`(=30 xx 0.2 = 6mmol)` is added, acidic buffer is formed.
`{:(,HA +,NaOHrarr,NaA+,H_(2)O),("Initial",100xx0.1,6mmol,0,0),(,=10mmol,,,),("Final",10-6=4,0,6,-):}`
`:. ["Salt"] = (6mmol)/(130mL)`
`["Acid"]_("left") = 10 - 6 = (4mmol)/(130mL)`
`:. pH_(a) = pK_(a) + log.((6//130)/(4//130))`
`= 5 + log .(6)/(4) = 5 + log 3 - log2`
`= 5 + 0.48 - 0.3 = 5.18`
`E =- 0.059 (pH_(c) - pH_(a))`
`=- 0.059 (1-5.18)`
`= 0.246V`
b. When `50mL` of `0.2M NaOH(=50 xx 0.2 = 10 mmol)` is added, salt of `W_(A)//S_(B)` is formed.
`{:(,HA +,NaOHrarr,NaA+,H_(2)O),(Initial,rArr10mmol,10mmol,-,-),("Final",rArr0,0,10mmol,-):}`
`:. ["Salt"] = 10//150 = 0.06M`
`:. pH_(a) = (1)/(2) (pK_(w) + pK_(A) + logC)`
`= (1)/(2) [(14 +5 + log (6 xx 10^(-2))]`
`= (1)/(2) (19 + log 3 + log 2 - 2)`
`= (1)/(2) (19 + 0.48 + 0.3 - 2) = 8.89`