Nitric acid + mg (no3) 35.0 ml of 0.255 m nitric acid is added to 45.0 ml of 0.328 m mg (no3) 2. what is the concentration of nitrate ion in the final solution?

a. 0.481 m

b. 0.296 m

c. 0.854 m

d. 1.10 m

e. 0.0295 m

HNO₃ → H⁺ + NO₃⁻

v₁=35.0 mL

c₁=0.255 mmol/mL

n₁ (NO₃⁻) = v₁c₁

Mg (NO₃) ₂ → Mg²⁺ + 2NO₃⁻

v₂=45.0 mL

c₂=0.328 mmol/mL

n₂ (NO₃⁻) = 2c₂v₂

c₃={n₁+n₂} / (v₁+v₂) = {c₁v₁ + 2c₂v₂} / (v₁+v₂)

c₃={35.0*0.255+2*0.328*45.0} / (35.0+45.0) ≈0.481 mmol/mL

a. 0.481 m