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A solution of 0.01M concentration of NH(...

A solution of `0.01M` concentration of `NH_(4)OH` is `2.6%` dissociated. Calculate `[H^(+)], [OH^(-)],[NH_(4)^(+)], [NH_(4)OH]` and pH of solution.

Text Solution

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`NH_(4)OH) Leftrightarrow NH_(4)^(+) +OH^(-)) `
`{:(,"Before dissociation",1,0,0),(,"After dissociation",1-alpha,alpha,alpha):}`
`therefore [OH^(-)]=C.alpha= Csqrt((K_b//C))=sqrt((K_(b).C))`
`"Also "K_(b)=Calpha^(2)=0.01 xx (0.026)^(2)=6.76 xx 10^(-6)`
`therefore [OH^(-)]=sqrt([6.76xx10^(-6)xx 0.01]) =2.6 xx 10^(-4)M`
`therefore [H^(+)]=10^(-14)//2.6 xx 10^(-4)`
`=3.846 xx 10^(-11)M`
`therefore pH=-log [H^(+)]=-log 3.846 xx 10^(-11)`
=10.415
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