1.

Given at 25^(@)C [Ag(NH_(3))_(2)]^(+)hArrAg^(+)+2NH_(3) will be

Answer»

`10^(-8)`
`10^(-10)`
`10^(-12)`
`10^(-14)`

Solution :`[Ag(NH_(3))_(2)}^(+)+e^(-)toAg+2NH_(3),E^(@)=0.02V`
`underline(""AgtoAg^(+)+e^(-),E^(@)=-0.80V)`
`[Ag(NH_(3))_(2)]^(+))HARR[Ag^(+)]+2NH_(3),E_(cell)^(@)=-0.78V`
`E_(cell)^(@)=(0.0591)/(n)logKtherefore-0.78=(0.0591)/(1)LOGK`
or `logK=(-0.78)/(0.0591)=--13.2`
K=Antilog`overline(14).8=6.3xx10^(-14)`


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