moles of HCl = 50*0.1 = 5 moles of NaOH = 20.5*0.2 = 4.1 so moles of H+ left after titration = 5-4.1 = 0.9 so [H+] = 0.9/20.5+50 = 0.0127 so pH = -log([H+]) = - log(0.0127) =1.896 ANSWER Solution moles of HCl = 50*0.1 = 5 moles of NaOH = 20.5*0.2 = 4.1 so moles of H+ left after titration = 5-4.1 = 0.9 so [H+] = 0.9/20.5+50 = 0.0127 so pH = -log([H+]) = - log(0.0127) =1.896 ANSWER.