A 0.105 L sample of an unknown HNO 3 solution required 45.5 mL of 0.150 M Ba ( OH ) 2 for complete neutralization. What is the concentration of the HNO 3 solution?

[HNO₃] = 0.0325 M

Explanation:

Formula for neutralization is:

Normality of the acid . Acid volume = Normality of the base . Base volume

We verify dissociations of base and acid

HNO₃ → H⁺ + NO₃⁻      (Since we have 1 H⁺, N = M)

Ba(OH)₂ → Ba²⁺  + 2OH⁻   (We have 2 OH⁻, the N = M/2)

[Ba(OH)₂] = 0.150M / 2 =0.075 N

We replace data:

(First of all, we convert the volume of base from mL to L)

45.5 mL . 1L / 1000mL = 0.0455 L

0.105 L . Normality of the acid = 0.0455L . 0.075N

Normality of the acid = (0.0455L . 0.075N) / 0.105L → 0.0325 N

N = M → [HNO₃] = 0.0325 M

it takes on an electronical energy