Calculate the amount of (NH_(4))_(2)SO_(4) in g which must be added to 500 mL of 0.2 M NH_(3) to yield a solution of pH = 9.35, K_(b) for NH_(3) = 1.78xx10^(-5).

Calculate the amount of (NH_(4))_(2)SO_(4) in g which must be added to

1.	Calculate the amount of (NH_(4))_(2)SO_(4) in g which must be added to 500 mL of 0.2 M NH_(3) to yield a solution of pH = 9.35, K_(b) for NH_(3) = 1.78xx10^(-5).
Answer» Solution :As it is a BASIC buffer, `pOH=pK_(b) + LOG. (["Salt"])/(["Base"])=-log K_(b) + log. ([NH_(4)^(+)])/([NH_(4)OH])` As `pH = 9.35, :. pOH = 14 - 9.35 = 4.65` Millimoles of `NH_(4)OH` in solution `= 0.2xx500 = 100` Suppose millimoles of `NH_(4)^(+)` to be added = X `:. 4.65 = - log (1.78xx10^(-5))+ log. (x//500)/(100//500)= (5 - 0.2504) +log .(x)/(100)` or `log. (x)/(100) = - 0.0996 = bar(1).0004 ~= 0.1 or log x = 2.1 or x = 125.9` `:.` Millimoles of `(NH_(4))_(2)SO_(4)` to be added `=(125.9)/(2) = 62.95` (`:. 1 "millimole of " (NH_(4))_(2)SO_(4)-=2 "millimoles of " NH_(4)^(+)`) `:.` Mass of `(NH_(4))_(2)SO_(4)` to be added `=(62.95xx10^(-3) "MOLES" ) (132 G "mol"^(-1))=8.3094 g`.