The titration of a weak base versus weak acid cannot be carried out by using acid-base indicator because
- there is no variation in \(\mathrm{pH}\) of the solution
- the variation in \(\mathrm{pH}\) is gradual with no steep change in pH near the equivalence point.
- the change in pH near the equivalence point does not encompasses an interval equal to the \(\mathrm{pH}\) transition range of the indicator
- no indicator can be found which changes colour at pH of equivalence point.
Solution
\(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}^{\circ}+\log ([\mathrm{salt}] /[\text { acid }])\)
From the given date, we find
\(4.14=\mathrm{p} K_{\mathrm{a}}^{\circ}+\log [10 /(50-10)]\)
This gives \(\mathrm{p} K_{\mathrm{a}}^{\circ}=4.14-\log (1 / 4)=4.14+0.60=4.74\)
After the addition of \(40 \mathrm{~mL}\) of \(\mathrm{NaOH}\) solution, we will have
\(\mathrm{pH}=4.74+\log [40 /(50-40)]=4.74+0.60=5.34\) ^
Asked in: JEE-TOPICTESTS-CHEMISTRY