Equilibrium Question 750
Question: When 10 ml of 0.1 M acetic acid ( $ pK_{a} $ . = 5.0) is titrated against 10ml of 0.l Mammonia solution ( $ pK_{b} $ =5.0), the equivalence point occurs at pH
Options:
A) 5.0
B) 6.0
C) 7.0
D) 9.0
Show Answer
Answer:
Correct Answer: C
Solution:
$ pK_{a}=-\log K_{a}p{K_{a,}}pK_{b}=-\log K_{b} $
$ pH=-\frac{1}{2}[logK_{a}+logK_{w}-logK_{b}] $
$ =-\frac{1}{2}[-5+log(1\times {10^{-14}})-(-5)] $
$ =-\frac{1}{2}[-5-14+5]=-\frac{1}{2}(-14)=7 $
