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))`

Options

A. `4.7033`
B. `5.7033`
C. `6.7033`
D. `8.7033`

Answer

B

Explanation

`pH` of buffer is given by :
`pH = - "log" K_(a) + "log" (["Salt"])/(["Acid"])`
Case I : `4 = - "log" 1.0 xx 10^(-5) + "log" (["Salt"])/((0.5))`
or [Salt] `= 0.1 xx 0.5 = 0.05 M`
Case II : `6 = - "log" 1.0 xx 10^(-5) + "log" (["Salt"])/((0.5))`
`"log" (["Salt"])/((0.05)) = 1`
`["Salt"] = 10 xx 0.5 = 5 M`
Now the two buffer `[(I.NaA = 0.05 M` and `HA = 0.5 M`) and (II. `NaA = 5M` and `HA = 0.5 M`)] are mixed in equal proportion.
Thus, new conc. of NaA in mixed buffer
`= (0.5 xx V + 5 xx V)/(2V) = 0.5 M`
Thus, `pH = - log(1.0 xx 10^(-3)) + "log" ([0.05//2])/([0.5])`
`pH = 5 + 0.7033 = 5.7033`