Question

How much of 0.1 F HC2H3O2 is required to make 600 ml of solution with a pH of 5.0?

Answer 1

PH = -log (M)

Hence M = anti log (-pH)

M = anti log (-5.0)

M$= 1.0×10^{-5}$

Moles $=(1.0×10^{-5})×0.6=6.0×10^{-6} moles$