1.

calculate the emf of the following cell: Pt(H_(2)" 1 atm")|CH_(3)CH_(2)COOH(0.15 M) ||0.01 M NH_(4)OH|H_(2)" (1 atm)"Pt K_(a) for CH_(3)CH_(2)COOH=1.4xx10^(-5) K_(b) for NH_(4)OH =1.8xx10^(-5)

Answer»


Solution :`[H^(+)]` in `CH_(3)CH_(2)COOH=sqrt(CxxK_(a))=sqrt(0.15xx1.4xx10^(-5))`
`=1.449xx10^(-3)`
`[OH^(-)]` in `NH_(4)OH=sqrt(C xxK_(b))=sqrt(0.01xx1.8xx10^(-5))=0.4242xx10^(-3)`
`[H^(+)]` in `NH_(4)OH=10^(-14)/(0.4242 XX 10^(-3))=2.3573xx10^(-11)`
`E_(cell)=0.0591 "log"([H^(+)]_(RHS))/([H^(+)]_(LHS))`


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