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

pH of buffer is given by : `pH =- log K_(a)+log (["Salt"])/(["Acid"])`
For I: `4=-log 1.0 xx 10^(-5) +log (["Salt"])/((0.5))`
`therefore log (["Salt"])/(0.5)=-1`
`or ["Salt"] =0.1 xx 0.5=0.05M`
For II: `6=-log 1.0 xx 10^(-5)+log (["Salt"])/(0.5)`
`therefore log (["Salt"])/(0.5)=1`
`therefore ["Salt"]=10 xx 0.5 =5M`
Now the two buffer [(1 NaA=0.05 M and HA =0.5 M) and (II NaA=5M and HA=0.5M)] mixed in equal proportion.
Thus, new conc. Of NaA is mixsed bufer `=(0.05 xx V +5 xx V)/(2V)=(5.05)/(2)`
New con. Of HA is mixed buffer `=(0.5 xx V+5 xx V)/(2V)=5.05`
Thus, `pH=-log 1.0 xx 10^(-5) +log ([5.05//2])/([0.5])`
pH =5+0.7033=5.7033