The correct option is D 4.76
Amount of CH3COOH in 100 cm3 of the solution =(0.50 mol)(1000 cm3)(100 cm3)=0.05 mol
Amount of CH3COONa in 100 cm3 of the solution =(0.50 mol)(1000 cm3)(100 cm3)=0.05 mol
When alkali is added, the acid is converted into salt. Thus, we have
Amount of the acid after the addition of 0.001 mol of the alkali = 0.049 mol
Amount of the salt=0.051 mol
Since pH = pK∘a+log((salt)(acid))
therefore,
pH=−log(1.8×10−5)+log(0.0510.049)=4.7447+0.0174≈4.76