What is the molarity of a solution of nitric acid if 0.283 g of barium hydroxide is required to neutralize 20.00 mL of nitric acid?

The answer to the question is

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

Explanation:

The chemical equation for the reaction is Search Results

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

We are required to find the molarity of a solution of nitric acid, to do tis, we require to know the number of moles of barium hydrxide present thus

The molar mass of barium hydroxide is 171.34 g/mol

Therefore 0.232 g of barium hydroxide contains (0.232 g) / (171.34 g/mol) = 0.00135 moles or 1.35 * 10⁻³ moles of barium hydroxide

However since in the eaction the number of noles of Ba (OH) ₂ and HNO₃ are equal, we have

the quantity in moles of nitric acid present in 20.00 mL solution is = 1.35 * 10⁻³ moles

Therefore, the number of moles in one liter of the nitric acid solution, which is the molarity is given by

(1.35 * 10⁻³ moles) / (20.00 L/1000) = 0.068 = 6.77 * 10⁻² M/L

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