The titration of a weak base versus weak acid cannot be carried out by using acid-base indicator because

During the titration, a buffer solution of weak acid and salt of its co jugate base is formed. Its pH is given by

$\mathrm{pH}=\mathrm{p}{K}_{\mathrm{a}}^{\circ }+\log \left(\right[\text{ salt }]/[\text{ acid }\left]\right)$       From the given date, we find

$4.14=\mathrm{p}{K}_{\mathrm{a}}^{\circ }+\log [10/(50-10\left)\right]$         This gives $\mathrm{p}{K}_{\mathrm{a}}^{\circ }=4.14-\log (1/4)=4.14+0.60=4.74$ After the addition of 40 mL of NaOH solution, we will have $\mathrm{pH}=4.74+\log [40/(50-40\left)\right]=4.74+0.60=5.34$