An aqueous solution of nitric acid is standardized by titration with a 0.174 M solution of barium hydroxide. If 13.9 mL of base are required to neutralize 28.7 mL of the acid, what is the molarity of the nitric acid solution?

Explanation:

2HNO₃ + Ba( OH)₂ = Ba( NO₃ )₂ + 2H₂O .

2 mole      1 mole

13.9 mL of .174M barium hydroxide contains .0139 x .174 moles

= 2.42 x 10⁻³ moles .

1 mole of barium hydroxide reacts with 2 moles of nitric acid

2.42 x 10⁻³ moles of barium hydroxide will react with 2 x 2.42 x 10⁻³ moles of nitric acid

= 4.84 x 10⁻³ moles

4.84 x 10⁻³ moles of nitric acid is contained in 28.7 mL

molarity of nitric acid = 4.84 x 10⁻³ / 28.7 x 10⁻³

= .169 M .

the correct answer would be d. hope this !

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.