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Acetic acid has `K_a=1.8xx10^(-5)` while formic acid has `K_(a)=2.1xx10^(-4)`.What would be the magnitude of the emf of the cell `Pt(H_(2))|0.1 M` acetic acid+ 0.1 M sodium acetate ||0.1 M formic acid + 0.1 M sodium formate|`Pt(H_(2))` at `25^(@)C` ?A. 0.032 voltB. 0.063 voltC. 0.0456 voltD. 0.055 volt |
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Answer» Correct Answer - B Reaction is `H_C^+ overset(1e^(-))to H_A^+` `therefore E=(0.059)/1 "log" (2.1xx10^(-4))/(1.8xx10^(-5))=0.0629 V=0.063 V` |
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