Consider a buffer solution containing `0.1mol` each of acetic and sodium acetate in `1.0L` of solution, `0.01mol` of `NaOH` is gradully added to this buffer solution. Calculate the average buffer capcity of the solution and as well as initial and final buffer capcity. `[K_(a) = 2 xx 10^(-5)] pK_(a) = 4.7`

Explanation

`pH` of the solutions is given by
`pH = pK_(a_("initial")) +"log"_(10) (["salt"])/(["acid"]), pH_("initial") = 4.7`
when `0.01M NaOH` is added. `["salt"] = 0.1 + 0.01 = 0.11M` and `["acid"] = 0.1 - 0.01 = 0.09M`
`rArr` Final `pH` of solution:
`pH = pK_(a) + "log" (0.11)/(0.09) = 4.787`
`rArr [H^(o+)] = K_(a) (["acid"])/(["salt"])`
`= 2 xx 10^(-5) xx (0.09)/(0.11) = 1.64 xx 10^(-5)M`
`Delta(pH) = 4.787 - 4.7 = 0.087`
Thus, `beta_(Avg) = (Deltan)/(Delta(pH)) = (0.01)/(0.087) = 0.1148`
`C_("Buffer") = [HA] + [A^(Θ)] = 0.1 + 0.1 = 0.2M`
`pH "initial" = 4.7, [H^(o+)] "initial" = 2 xx 10^(-5), K_(a) = 2 xx 10^(-5)`
`beta_("initial") = 2.303 (C_("Buffer").K_(a)[H^(o+)])/([K_(a) +[H^(o+)]^(2))`
`beta_("initial") = 2.303 (0.2 xx 2xx10^(-5) xx2xx10^(-5))/((2xx10^(-5)+2xx10^(-5))^(2)) = 0.115`
`[H^(o+)]_("Final") = 1.64 xx 10^(-5)M`.
`beta_("Final") = 2.303 (0.2 xx2xx10^(-5)xx1.64xx10^(-5))/((2xx10^(-5)+1.64xx10^(-5))^(2)) = 0.114`
Note: Buffer capcity will decrease as more and more `NaOH` is added into the above solution. check yurself that initially the buffer capcity was maximum. so when `0.01M NaOH` was added, we will see a very slight decrease in the buffer capcity but as more and more `NaOH` is added, buffer capcity will decrease sharply and the difference between the average and maximum buffer capcity will get magnified.
If `0.05M NaOH` is added in teh solution initially, we can calculate the final `pH` as:
`pH = pK_(a) + log (0.15)/(0.05) = 5.177`
Thus, `beta_(Avg.) = (Deltan)/(Delta(pH)) = (0.05)/(0.477) = 0.105`
Calculate `beta_("initial")` and `beta_("Final")` yourself.