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 nitrat

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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 nitrat

e 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

Chemistry

2 answers:

Tems11 [2.7K] · Answer 1 · 2026-02-27T17:07:15+03:00

Answer: The accurate option is a.

The nitrate ion concentration in the final mixture is approximately 0.48056 M.

Explanation:

$Molarity=\frac{\text{Moles of substantiate}}{\text{Volume of the solution(L)}}$

$HNO_3(aq)\rightarrow H^+(aq)+NO_{3}^-(aq)$

Calculating nitrate ions from 35.0 mL of 0.255 M nitric acid:

$0.255 M=\frac{Moles}{0.035 L}$

Calculated moles of nitric acid = 0.008925 mol

1 mole of nitric acid yields 1 mole of nitrate ions,

Therefore, 0.008925 moles of nitric acid will produce 0.008925 moles of nitrate ions.

Resulting in moles of nitrate in 35.00 mL of the solution = 0.008925 mole..(1)

Calculating nitrate ions from 45.0 mL of 0.328 M magnesium nitrate:

$Mg(NO_3)_2(aq)\rightarrow Mg^{2+}(aq)+2NO_3^{-}(aq)$

$0.328 M=\frac{Moles}{0.045 L}$

Calculated moles of magnesium nitrate = 0.01476 mol

1 mole of magnesium nitrate generates 2 moles of nitrate ions,

Hence, 0.01476 moles of magnesium nitrate results in:

$2\times 0.01476 mol=0.02952 mol$ of nitrate ions

Thus, moles of nitrate ions in 45.00 mL of solution = 0.02952 mol...(2)

Total moles of nitrate ions =

0.008925 mol+0.02952 mol = 0.038445 mol

Total volume after mixing = 35.00 mL+ 45.00 mL =

80.00 mL = 0.080 L

Final solution's nitrate ion concentration:

$\frac{0.038445 mol}{0.080 L}=0.48056 M$

lorasvet [2.7K] · Answer 2 · 2026-02-27T17:05:18+03:00

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

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 nitrat

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