A 25.0 mL solution of HNO₃ is neutralized with 17.6 mL of 0.250 M Ba(OH)₂. What is the concentration of the original HNO₃ solution?

Answer 1

ALWAYS write a correctly balanced equation before beginning any calculations:

2 HNO₃ + Ba(OH)₂ ==> Ba(NO₃)₂ + 2 H₂O

moles Ba(OH)₂ used in the titration = 17.6 ml x 1 L/1000 ml x 0.250 mol/L = 0.0044 moles Ba(OH)₂

moles HNO₃ present = 0.0044 mol Ba(OH)₂ x 2 mol HNO₃ / 1 mol Ba(NO₃)₂ = 0.0088 moles HNO₃ present

Concentration of HNO₃ = moles/liter. We have 0.0088 moles and that is in 25.0 ml. Converting 25.0 mls to liters, we have 25 ml x 1 L/1000 ml = 0.025 L

Concentration of HNO₃ = 0.0088 moles / 0.025 L = 0.352 M (3 sig. figs.)