Chemistry Problem Solving: Calculating Molarity of a Nitric Acid Solution

How to calculate the molarity of a solution of nitric acid?

Given the data that 0.283 g of barium hydroxide is required to neutralize 20.00 mL of nitric acid, what is the molarity of the nitric acid solution?

Answer:

The molarity of the nitric acid solution is 6.77 × 10⁻² M

To calculate the molarity of a solution of nitric acid, we need to use the given data and apply the concept of molarity and stoichiometry. In this case, the reaction between barium hydroxide (Ba(OH)₂) and nitric acid (HNO₃) is considered.

The chemical equation for the reaction is: Ba(OH)₂ + HNO₃ = Ba(NO₃)₂ + H₂O

One mole of Ba(OH)₂ reacts with one mole of nitric acid to form one mole of barium nitrate and one mole of water. By calculating the number of moles of barium hydroxide (Ba(OH)₂), we can determine the molarity of the nitric acid solution.

The molar mass of barium hydroxide is 171.34 g/mol. Therefore, 0.232 g of barium hydroxide contains 0.00135 moles or 1.35 × 10⁻³ moles of barium hydroxide. Since the moles of Ba(OH)₂ and HNO₃ are equal in the reaction, the quantity of moles of nitric acid present in 20.00 mL solution is 1.35 × 10⁻³ moles.

Then, the molarity of the nitric acid solution is calculated by dividing the number of moles in one liter of the solution by the volume in liters: (1.35 × 10⁻³ moles)/(20.00 L/1000) = 0.068 = 6.77 × 10⁻² M/L.

Therefore, the molarity of the nitric acid solution is 6.77 × 10⁻² M, indicating the concentration of nitric acid in the solution.