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

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