Calculate the pH value of a solution of 0.1 M NH_(3) (K_(b)=1.8xx10^(-

1.	Calculate the pH value of a solution of 0.1 M NH_(3) (K_(b)=1.8xx10^(-5))
Answer» Solution :`NH_(3)+H_(2)O hArr NH_(4)^(+)+OH^(-), K_(B) = ([NH_(4)^(+)][OH^(-)]^(2))/([NH_(3)])` `:. [OH^(-)]=sqrt(K_(b)xx[NH_(3)]) `. Hence, `[H^(+)]=(K_(w))/([OH^(-)]), pH = - LOG [H^(+)]` Alternatively,`pOH = (1)/(2) [pK_(b)-log C]=(1)/(2) [-log (1.8xx10^(-5))-log (0.1)]=2.88`

Discussion