Two buffer, (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))`

Explanation

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) or log. (["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 `