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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)`. |
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Answer» 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`. |
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