If it requires 23.4 milliliters of 0.65 molar barium hydroxide to neutralize 42.5 milliliters of nitric acid, solve for the molarity of the nitric acid solution. Show all of the work used to solve this problem.

Unbalanced equation: Ba(OH)2 + HNO3 -> Ba(NO3)2 + H2O

The numbers may be slightly different but the process is the same. http://www.jiskha.com/display.cgi?id=1458010856

To solve for the molarity of the nitric acid solution, you need to use the concept of stoichiometry. Stoichiometry allows us to determine the amount of one substance needed to react completely with another substance based on the balanced chemical equation.

1. Begin by writing out the balanced chemical equation for the reaction:
Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O

2. Determine the number of moles of barium hydroxide (Ba(OH)2) using the given volume and molarity:
moles of Ba(OH)2 = volume (in L) x molarity
moles of Ba(OH)2 = 23.4 mL x (1 L / 1000 mL) x 0.65 mol/L

3. Use stoichiometry to determine the number of moles of nitric acid (HNO3):
From the balanced equation, we see that 2 moles of HNO3 react with 1 mole of Ba(OH)2. Therefore,
moles of HNO3 = 2 x moles of Ba(OH)2

4. Determine the molarity of the nitric acid solution using the given volume and the moles of HNO3:
molarity of HNO3 = moles of HNO3 / volume (in L)
molarity of HNO3 = (2 x moles of Ba(OH)2) / (42.5 mL x (1 L / 1000 mL))

Plug in the values to calculate the molarity of the nitric acid solution. Make sure to convert the volume from milliliters to liters in order to maintain consistent units throughout the calculation.