InterviewSolution
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InterviewSolution
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The `EMF` of the following cell is `0.180 V` at `30^(@)C`. Find `EMF_(cell)` when `a. 40 mL of 0.2M NaOH` is added to the negative terminal of the battery . `b.` `50 mL` of `0.2 M NaOH` is added to the negative terminal of the battery . `c.` `50mL` OF `0.2M NaOH` is added to `100 mL` of `H_(2)SO_(4)` at the positive terminal of the battery. |
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Answer» 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` |
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