Search any question & find its solution
Question:
Answered & Verified by Expert
A buffer solution contains $0.1$ mole of sodium acetate dissolved in $1000 \mathrm{~cm}^{3}$ of $0.1 \mathrm{M}$ acetic acid. To the above buffer solution, $0.1$ mole of sodium acetate is further added and dissolved. The $\mathrm{pH}$ of the resulting buffer is
Options:
Solution:
1494 Upvotes
Verified Answer
The correct answer is:
$\mathrm{p} K_{a}+\log 2$
$\mathrm{pH}=\mathrm{p} K_{a}+\log \frac{\text { [salt] }}{\text { [acid] }}$
When extra $0.1$ mole sodium acetate is added
[salt $]=0.2 \mathrm{~mol} \mathrm{dm^{-3 }}$
[acid] $=0.1 \mathrm{~mol} \mathrm{dm}^{-3}$
$\therefore \quad \mathrm{pH}=\mathrm{p} K_{a}+\log \frac{[0.2]}{[0.1]}$
$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log 2$
When extra $0.1$ mole sodium acetate is added
[salt $]=0.2 \mathrm{~mol} \mathrm{dm^{-3 }}$
[acid] $=0.1 \mathrm{~mol} \mathrm{dm}^{-3}$
$\therefore \quad \mathrm{pH}=\mathrm{p} K_{a}+\log \frac{[0.2]}{[0.1]}$
$\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}+\log 2$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.