Question

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

Answer 1

Answer:The correct answer is option a.

The concentration of nitrate ion in the final solution is 0.48056 M.

Explanation:

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

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

Moles of nitrate ions in 35.0 ml of 0.255 M nitric acid:

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

Moles of nitric acid = 0.008925 mol

1 mol of nitric acid gives 1 mol of nitrate ions,

Then 0.008925 moles of nitric acid will give 0.008925 moles of nitrate ions.

Moles of nitrate ions in 35.00 mL of solution = 0.008925 mole ..(1)

Moles of nitrate ions in 45.0 ml of 0.328 M magnesium nitrate:

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

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

Moles of magnesium nitrate = 0.01476 mol

1 mol of magnesium nitrate gives 2 mol of nitrate ions,

Then 0.01476 moles of magnesium nitrate will give :

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

Moles of nitrate ions in 45.00 mL of solution = 0.02952 mol...(2)

Total number of moles of nitarte 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

The concentration of nitrate ion in the final solution:

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

Answer 2

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