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.

A solution of `0.01M` concentration of `NH_(4)OH` is `2.6%` dissociate

1.	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.
Answer» `{:(,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`

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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.

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