Question

Question 3 2 pts How many mL of 0.233 M. HNO3(aq) are needed to neutralize 56.90 mL of a 0.1368 M solution of barium hydroxide, Ba(OH)2? 0 33.4 O 100.2 0 66.8 O 16.7 O 53.9

Answer 1

First we will calculate the Normality of both the solutions. Nosmality It is defined as the number of gram equivakuks of Soluti“present in 1lts of a solution - 1 Normality = nfactor X Molarity nfactor - Basicity of acid or acidity of bare for HNO₃ = n factor =/ for Ba@M)= n factor = Nosmality of two = 1X0.233 = 0.233 Normality of Balon) = 2X0.1368 = 0.27736 We will use the dilution fosunula TNV = N2427 0.233X V, = 0.7736x56.90 V = 0.7736756.90 0.233 V = 66.814 med 66.81 ml of HNO is required to neutralize , 0.22;