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

`K_(HA)=1.0xx10^(-5)`
`therefore pK_a=-log (1.0xx10^(-5))`=5.0
Determination of conc. Of salt in buffer X
`pH=pK_a+"log" "[Salt]"/"[Acid]"`
4.0=5.0 +log `"[Salt]"/0.5`
log `"[Salt]"/0.5=-1`
`"[Salt]"/0.5=10^(-1)`
[Salt]=`0.5xx10^(-1)` =0.05 M
Determination of conc. of salt in buffer Y
6.0=5.0+log `"[Salt]"/0.5`
log `"[Salt]"/0.5=1.0`
`"[Salt]"/0.5=10`
`therefore` [Salt]=10 x 0.5 =5.0 M
Conc. of salt in the mixture ,
`"[Salt]"_"mix"=(0.05+5.0)/2`=2.525 M
pH=`pK_a+.log" "[Salt]"/"[Acid]"`
=5.0+log `2.525/0.5`
=5.0+0.70=5.7