determine the ph of a solution prepared by mixing 50 ml of 0.2N hcl with 50ml 0f 0.1N naoh? Share with your friends Share 0 Geetha answered this First find the milli equivalent of acid and baseMilli equivalent of acid = V×N = 50 × 0.2 = 10 meqMilli equivalent of base = V×N = 50 × 0.1 = 5 meqAs acid is neutralised by base, the remaining milliequivalent remaining is = 10 -5 = 5 meqThe total volume is 50 + 50 = 100 mlso the resultant normality = 5100 = 0.05 NNow the pH of the solutionpH= - log H+ = - log (0.05) = - log (5 × 10-2) = 2 log 10 -log 5 = 2 - 0.6990 =1.301 1 View Full Answer