What volume of a .271 M HNO3 solution is required to neutralize 23.4 mL of a .112 M Ba(Oh)2 solution?

Question

What volume of a .271 M HNO3 solution is required to neutralize 23.4 mL of a .112 M Ba(Oh)2 solution?

Answer 1

0.271 M HNO3 solution will ionize into .271 moles of hydrogen ions in 1 liter. 0.112 M Ba(OH)2 solution will ionize into (0.112)(2) = 0.224 moles of hydroxide ions per liter because there are 2 hydroxides for every barium hydroxide. 0.224 moles/liter (0.0234 L) = 0.0052416 moles of hydroxide. So use the equation M1V1=M2V2. We already have the right side of the equation. 0.271(V1) = 0.0052416. V1 = 0.0193 L or 19.3 mL of nitric acid.