It is an electrolytic concentration cell.
`[H_(2)SO_(4)]0.05M=0.05xx(n` factor`)`
`=0.1N=10^(-1)N`
`pH_(c)=-log(10^(-1)N)=1`
`pH_(a)=?`
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(1-pH_(a))`
`0.180V=-0.06(1-Ph_(a))`
`pH_(a)=(0.180)/(0.06)+1=4`
`:. [H^(o+)]_(a)=10^(-4)M `
Since `HA` is a weak acid `(W_(A))`,
`:. pH_(a)=pH_(wA)=(1)/(2)(pK_(a)-logC)`
`4=(1)/(2)(pK_(a)-log10^(-1))`
`:. pK_(a)=7`
`a.` When` 40mL` of `0.2 M NaOH(=40xx0.2=8 mmol)` is added to a weak acid, acidic buffer is formed.
`HAA+NaOH rarrNaA+H_(2)O`
Total volume `=100+40=140mL`
`[Sa l t] =[NaA]=(8mmol)/(140mL)`
`[W_(A)]_(l e f t)=[HA]_(l eft)=(2mmol)/(140mL)`
`pH_(a)=pH_(aci dic buffter)=pK_(a)+log .([Sa l t])/([Ac i d])`
`=7+log.((8//140)/(2//140))`
`=7+log2^(2)`
`=7+2xx0.3=7.6`
Thus,
`E_(cell)=-0.06(pH_(c)-pH_(a))`
`=-0.06(1-7.6)`
`=-0.06xx-6.6=0.396V`
`b.` When `50 mL` of `0.2M NaOH(=50xx0.2=10mmol)`
is added to weak acid `(HA)`, salt of `W_(A)//S_(B)(NaA)` is formed.
`:.[NaA]=(10mmol)/(150mL)=(1)/(15)M`
Thus, `pH` of a salt of `W_(A)//S_(B)` is given by `:`
`pH_(a)=pH_(sal tW_(A)//S_(B))=(1)/(2)(pK_(w)+pK_(a)+log C)`
`[pK_(a)` of `HA=7` from part `(a)` above `]`
`pH_(a)=(1)/(2)(14+7+log.(1)/(15))`
`=(1)/(2)(21+log1-log15)`
`=(1)/(2)(21+0-log3-log5)`
`=(1)/(2)(21-0.48-0.7)=9.91`
Thus,
`E_(cell)=-0.06(pH_(c)-pH_(a))`
`=-0.06xx(1-0.01)`
`-0.06xx(-8.91)=0.5346V`
`c.` When `50mL` of `0.2M NaOH(=50xx0.2=10mmol=10mEq)` is added to `100mL ` of `0.05 M H_(2)SO_(4)` to the positive terminal of battery , then
`mEq` of `NaOH=10`
`mEq` of `H_(2)SO_(4)=100mLxx0.05xx2(n` factor `)`
`=10`
When `10mEq` of `H_(2)SO_(4)` reacts with `10mEq` of `NaOH,` a salt `(Na_(2)SO_(4))` of `S_(A)//S_(B)` is formed, which does not hydrolyze . So `pH` of such solution is `=7`.
`pH_(c)=7`
`pH_(a)=?`
`HA` is a weak acid whose `pK_(a)=7(` determined in part `(a)` above `)` and `[HA]=0.1M=10^(-1)M`.
Thus, `pH_(wA)=(1)/(2)(pK_(a)-logC)`
`=(1)/(2)(7-log10^(-1))=4`
Hence,
`E_(cell)=-0.06(pH_(c)-pH_(a))`
`=-0.06(7-4)=-0.06xx3=-0.18V`
