Answer:
[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
