Question

To `1.0L` solution containing `0.1mol` each of `NH_(3)` and `NH_(4)C1,0.05mol NaOH` is added. The change in `pH` will be `(pK_(a)` for `CH_(3)COOH = 4.74)`
A. `0.30`
B. `-0.30`
C. `0.48`
D. `-0.48`

Answer 1

Correct Answer - C
Rule `(BBB)`: in basic buffer `(B)`, on addition of `S_(B)(B)`, the concentration of `S_(B)(B)` increases and that of salt decreases.
`pOH_("initial") = pK_(b) + "log"([NH_(4)C1])/([NH_(3)])`
`pOH_("initial") = 4.74 + "log"(0.1)/(0.1)`
`:. pOH_("initial") = 4.74`.
On adding `0.05 M NaOH`:
`[NH_(4)C1] = 0.1 - 0.05 = 0.05M`
`[NH_(3)] = 0.1 + 0.05 = 0.15M`
`pOH_(final) = pK_(b) +"log"(0.15)/(0.15)`
`= pK_(b) - log3`
`= 4.74 - 0.48 = 4.26`
Change in `pOH (DeltapOH) = pOH_(f) - pOH_(i) = 4.26 - 4.74 =- 0.48`
Change in `pH =- ("changes in pOH") =- (-0.48) = 0.48`