An NH_(4)^(+) buffer is supposed to keep the pH of the solution constant within0.3 pH unit during the reaction CH_(3)COOH_(3)(aq)+2H_(2)O(aq)rarrCH_(3)COO^(-)(aq)+H_(3)O^(+)(aq)+CH_(3)OH(aq) If this solution had initial concentration :[NH_(4)^(+)]_(10)=0.1 M ,[NH_(3)]_(0)=0.06 M ,[CH_(3)COOH_(3)]_(0)=0.02Mdetermine the magnitude of pH change as a result of reaction. Multiply the magnitude by 10 & add 1 if it is a satisfactory buffer, otherwise subtract 1.Report the answer rounding it off to the nearest whole number. [K_(b)NH_(3)=1.8xx10^(-6)]

An NH_(4)^(+) buffer is supposed to keep the pH of the solution consta

1.	An NH_(4)^(+) buffer is supposed to keep the pH of the solution constant within0.3 pH unit during the reaction CH_(3)COOH_(3)(aq)+2H_(2)O(aq)rarrCH_(3)COO^(-)(aq)+H_(3)O^(+)(aq)+CH_(3)OH(aq) If this solution had initial concentration :[NH_(4)^(+)]_(10)=0.1 M ,[NH_(3)]_(0)=0.06 M ,[CH_(3)COOH_(3)]_(0)=0.02Mdetermine the magnitude of pH change as a result of reaction. Multiply the magnitude by 10 & add 1 if it is a satisfactory buffer, otherwise subtract 1.Report the answer rounding it off to the nearest whole number. [K_(b)NH_(3)=1.8xx10^(-6)]
Answer» Solution :`K_(B)(NH_(3)) = 1.8 xx 10^(-5)` `CH_(3)COOCH_(3) (aq)+2H_(2)O (aq) rarr CH_(3)COO^(-) (aq) + H_(3)O^(+) (aq) + CH_(3)OH(aq)` `[NH_(4)^(+)]_(0) = 0.1M, [NH_(3)] = 0.06M, [CH_(3)COOCH_(3)]_(0) = 0.02 M` `POH=pK_(b)+log(([NH_(4)^(+)])/([NH_(3)]))=4.74+log((0.1)/(0.06))` `(pOH)_("initial")=4.74+0.22=0.22=4.96:'(pH)_("intial")=9.04` `{:(NH_(3)(aq),+,H^(+)(aq.),rarr,NH_(4)^(+)(aq),),(0.06,,0.2,,0.1,"mole"),(0.04,,-,,0.12,"mole"):}` `(PoH)_("final")=4.74+log((0.12)/(0.04))=4.74+(log)((0.12)/(0.04))=4.74+log3=4.74+0.48=5.22` `:.(pH)_("final")=8.78` `DeltapH=9.04-8.76=0.26` YES this is satisfactory buffer.