InterviewSolution
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InterviewSolution
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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. |
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Answer» (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` |
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