Question

When $10\, mL$ of $0.1\, M$ acetic acid $(pK_{a} = 5.0)$ is titrated against $10\, mL$ of $0.1\, M$ ammonia solution $(pK_{b} = 5.0)$, the equivalence point occurs at $pH$

• A. 5
• B. 6
• C. 7
• D. 9

Answer 1

Answer: (C)

7

Answer 2

$p K_{a} =-\log K_{a} \text { and } p K_{b}=-\log K_{b}$ $pH =-\frac{1}{2}\left[\log K_{a}+\log K_{w}-\log K_{b}\right]$ $=-\frac{1}{2}\left[-5+\log 10^{-14}-(-5)\right]$ $=-\frac{1}{2}[-5-14+5]$ $=7$ Thus, the end point or equivalence point is obtained at $pH =7$ in neutral medium.