A) 0.9542
B) 0.5924
C) 0.9245
D) 0.5294
Correct Answer: A
Solution :
Once NaOH is added, the solution becomes a buffer. \[C{{H}_{3}}COOH+NaOH\xrightarrow{\,}\,C{{H}_{3}}COONa\] \[+{{H}_{2}}O\] 0.1 M 0.5 M 0 When \[\frac{1}{4}\] \[0.1\frac{-0.1\times 1}{4}\] \[\frac{0.1\times 1}{4}\] is neutralized When \[\frac{3}{4}\] \[0.1-\frac{0.1\times 3}{4}\] \[\frac{0.1\times 3}{4}\] Is neutralized \[{{(pH)}_{1/4}}=p{{K}_{a}}+\log \,\frac{[C{{H}_{3}}COONa]}{[C{{H}_{3}}COOH]}\] \[=p{{K}_{a}}+\log \,\frac{\frac{0.1}{4}}{\frac{0.4-0.1}{4}}\] \[=pKa+\log \,\frac{1}{3}\] Similarly, \[{{(pH)}_{3/4}}=p{{K}_{a}}+\log \,\frac{\frac{0.3}{4}}{\frac{0.4-0.3}{4}}\] \[=p{{K}_{a}}+\log \,\frac{3}{1}\] \[\therefore \] \[\Delta pH\] (by Henderson?s equation) \[=\,\log \,\left( \frac{3}{1} \right)\,-\log \,\left( \frac{1}{3} \right)=2\,\log \,3\] \[=2\times 0.4771=0.9542\]
You need to login to perform this action.
You will be redirected in
3 sec
