Question

a 10.00 ml sample of nitric acid, hno3, requires 0.108 g of barium hydroxide, ba(oh)2, for titration to the equivalence point. what is the concentration of the nitric acid?

Answer 1

To calculate the concentration of nitric acid, use the stoichiometry of the reaction and the amount of barium hydroxide used in the titration.

Step-by-step explanation:

To calculate the concentration of the nitric acid (HNO3), we can use the stoichiometry of the reaction and the information given. From the balanced chemical equation:

1 mol HNO3 reacts with 1 mol Ba(OH)2

We are given that 0.108 g of Ba(OH)2 is required to react with 10.00 ml of HNO3. We can calculate the number of moles of Ba(OH)2:

# moles Ba(OH)2 = 0.108 g / (molar mass of Ba(OH)2)

Then, using the stoichiometry of the reaction, we can calculate the number of moles of HNO3:

moles HNO3 = moles Ba(OH)2

Finally, we can calculate the concentration of the nitric acid:

concentration = moles HNO3 / volume of HNO3 in L

Answer 2

The concentration of nitric acid (HNO₃) in the titration is calculated to be 0.126 M after determining the moles of barium hydroxide (Ba(OH)₂) used and using the stoichiometry of the balanced chemical equation.

Step-by-step explanation:

To calculate the concentration of nitric acid (HNO₃) in the given titration problem, we need to first determine the amount in moles of barium hydroxide (Ba(OH)₂) used. Since mass is given, we can use the molar mass of Ba(OH)₂ (171.34 g/mol) to find the moles.

Moles of Ba(OH)₂ = 0.108 g / 171.34 g/mol = 0.000630 moles.

The balanced chemical equation provides the mole ratio between Ba(OH)₂ and HNO₃:

Ba(OH)₂ + 2HNO₃→ Ba(NO3)2 + 2H₂O

This indicates that 1 mole of Ba(OH)₂ neutralizes 2 moles of HNO₃. Therefore:

Moles of HNO₃= 2 × moles of Ba(OH)₂ = 2 × 0.000630 moles = 0.00126 moles.

To find the molarity of HNO3, we divide the moles of HNO₃ by the volume of the acid in liters (10.00 mL = 0.01000 L):

Molarity of HNO₃ = 0.00126 moles / 0.01000 L = 0.126 M.