A buffer solution contains `0.25M NH_(4)OH` and `0.3 NH_(4)C1`.
a. Calculate the `pH` of the solution.
b. How much `NaOH` should be added to `1L` of the solution to change `pH` by `0.6.K_(b) =2xx10^(-5)`.

Answer

A, B

Explanation

i. Since it is basic buffer. Thus,
`K_(b) = 2 xx 10^(-5)`
`pK_(b) =- log (2 xx 10^(-5)) =- 0.3 +5 = 4.7`
`pOH = pK_(a) + "log"(["Salt"])/(["Base"])`
`= 4.7 + log ((0.3)/(0.25))`
`= 4.7 + log (1.2)`
`= 4.7 + log (12 xx 10^(-1))`
`= 4.7 + log 2^(2) + log 3 - 1`
`= 3.7 + 0.6 + 0.48 = 4.78`
`pH = 14 - 4.78 = 9.22`
ii. Rule `B B B:`
In basic buffer (B),on addition of `S_(B)(B)`, the concentration of `W_(B)(B)` increase and that of salt decreases. On adding base, `pH` will increase and `pOH` will decrease.
Let `xM NaOH` is added.
`pH_("initial") = 9.22`,
`pH_("new") = 9.22 + 0.6 = 9.86`
`pOH_("new") = 14 - 9.86 = 4.14`
`["Base"]_("new") = (0.25 +x)`
`["Salt"]_("new") = (0.3 - x)`
`:. pOH = pK_(b) + log ((0.3 -x)/(0.25 +x))`
`4.14 = 4.74 + log ((0.3 -x)/(0.25 +x))`
`4.14 - 4.74 = log ((0.3 - x)/(0.25+x))`
`- 0.6 = log ((0.3 -x)/(0.25 +x)) "or log" ((0.25 +x)/(0.3-x)) = 0.6`
`:. (0.25+x)/(0.3 -x) = "Antilog" (0.6) = 4`
Solve for `x :`
`x = 0.19` mol in `1L` of solution