The reaction between HNO3 and Ba(OH)2 is given by the equation below; 2HNO3 + Ba(OH)2 = Ba(NO3)2 + 2H2O Moles of Barium hydroxide used; = 0.200 × 0.039 l = 0.0078 Moles The mole ratio of HNO3 and Ba(OH)2 is 2: 1 Therefore; moles of nitric acid used will be; = 0.0078 ×2 = 0.0156 moles But; 0.0156 moles are equal to a volume of 0.10 The concentration of Nitric acid will be; = (0.0156 × 1)/0.1 = 0.156 M