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For NH(3), K(b)=1.8xx10^(-5). K(a) for N...

For `NH_(3)`, `K_(b)=1.8xx10^(-5)`. `K_(a)` for `NH_(4)^(+)` would be

Text Solution

Verified by Experts

The reaction are:
at anode: `1/2 H_(2) to H_(a)^(+) +e^(-)`
at cathode `H_(c)^(+) +e^(-) to 1/2 H_(2)`
=overall reaction `: H_(c)^(+) to H_(a)`
`E=E^(0)-0.059 log ([H_(a)^(+)])/([H_(c)^(+)]`
At anode:
From `CH_(3)COOH Leftrightarrow CH_(3)COO^(-) +H_(a)^(+)`
`[H_(a)^(+)]=C alpha =C sqrt((K_(a))/(C))=sqrt(K_(a)C)`
`[H_(a)^(+)]=sqrt(K_(a) xx [CH_(3)COOH])=sqrt(1.8 xx 10^(-5) xx 0.1)`
`=1.34 xx 10^(-3)M`
At cathode
From
`NH_(4)OH Leftrightarrow NH_(4)^(+) + OH^(-)`
`[OH^(-)]= C alpha =sqrt((K_(b))/(C))=sqrt(K_(b)C)`
`sqrt((K_(a))/(C))=sqrt(K_(a)C)=(K_(w))/([OH^(-)])=(K_(w))/(sqrt(K_(b)))=(K_(w))/(sqrt(K_(b) xx [NH_(4) OH])`
`=(10^(-14))/sqrt(1.8 xx 10^(-5) xx 0.01) =2.359 xx 10^(-11)M`
`E=0-0.591 log (1.34 xx 10^(-3))/(2.359 xx 10^(-11))`
=-0.4575V
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