The pH of a buffer solution made by mixing 25 mL of 0.02 M `NH_(4)OH` and 25 mL of 0.2 M `NH_(4)CI` at `25 ^(circ)` is (`p^(K_(b)) of NH_(4)OH = 4.8`)

Options

A. 5.8
B. 8.2
C. 4.8
D. 3.8

Answer

B

Explanation

pH of the mixture of `NH_(4)OH`(base) and `NH_(4)Cl`(salt of strong acid) is greater than 7 since it acts as a basic buffer.
(or)
`pOH = PK_(b) = "log"[Sal t]/[base] = 4.8 + "log" 0.2M/0.02M = 4.8 + "log"_(10) = 5.8`
`:. pH = 14 - 5.8 = 8.2`