Equilibrium Question 411
Question: When 10 ml of 0.1 M acetic acid $ (pK_{a}=5.0) $ is titrated against 10 ml of 0.1M ammonia solution $ (pK_{b}=5.0) $ , the equivalence point occurs at pH [AIIMS 2005]
Options:
A) 5.0
B) 6.0
C) 7.0
D) 9.0
Show Answer
Answer:
Correct Answer: C
Solution:
pKa = - log Ku, pKb = - log Kb pH = $ -\frac{1}{2}[\log K_{a}+\log K_{w}-\log K_{b}] $
$ =-\frac{1}{2}[-5+\log (1\times {10^{-14}})-(-5)] $
$ =-\frac{1}{2}[-5-14+5]=-\frac{1}{2}(-14)=7 $
