Two buffers (X) and (Y) of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA. Both the buffers are 0.50 M in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers ? (K_(HA)=1.0xx10^(-5))

Two buffers (X) and (Y) of pH 4.0 and 6.0 respectively are prepared fr

1.	Two buffers (X) and (Y) of pH 4.0 and 6.0 respectively are prepared from acid HA and the salt NaA. Both the buffers are 0.50 M in HA. What would be the pH of the solution obtained by mixing equal volumes of the two buffers ? (K_(HA)=1.0xx10^(-5))
Answer» Solution :For BUFFER X, `PH=pK_(a) + log. (["Salt"])/(["Acid"]), i.e., 4 = 5 + log. (["Salt"])/(0.5) or log .(["Salt"])/(0.5) = - 1 or (["Salt"])/(0.5) = 10^(-1) or ["salt"]=0.05 M` For buffer Y, `pH = pK_(a) + log. (["Salt"])/(["Acid"]), i.e., 6 = 5 + log .(["Salt"])/(0.5) orlog. (["Salt"])/(0.5) or (["Salt"])/(0.5) = 10 or ["Salt"]=5 M` When EQUAL VOLUMES of both the buffers are mixed, [Acid] = 0.5, [Salt] `=(5+0.05)/(2) = 2.5025` `pH = pK_(a) + log. (["Salt"])/(["Acid"])=5 + log. (2.5025)/(0.5) = 5 + log 5.05 = 5.7 `