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.

Explanation

`{:(,NH_(4)OHhArr,NH_(4)^(+)+,OH^(-)),("Before dissociation",1,0,0),("After dissociation", 1-alpha,alpha,alpha):}`
`:. [OH^(-)]=C.alpha`
`=Csqrt((K_(b)//C))=sqrt((K_(b).C))`
Also `K_(b)=Calpha^(2)=0.01xx(0.026)^(2)`
`= 6.76xx10^(-6)`
`:. [OH^(-)]=sqrt([6.76xx10^(-6)xx0.01])`
`=2.6xx10^(-4)M`
`:. [H^(+)]=10^(-14)//2.6xx10^(-4)`
`=3.846xx10^(-11)M`
`:. pH= -log[H^(+)]= - log 3.846xx10^(-11)`
`=10.415`