Respuesta :
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
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
Answer:The correct answer is option a.
The concentration of nitrate ion in the final solution is 0.48056 M.
Explanation:
[tex]Molarity=\frac{\text{Moles of substantiate}}{\text{Volume of the solution(L)}}[/tex]
[tex]HNO_3(aq)\rightarrow H^+(aq)+NO_{3}^-(aq)[/tex]
Moles of nitrate ions in 35.0 ml of 0.255 M nitric acid:
[tex]0.255 M=\frac{Moles}{0.035 L}[/tex]
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:
[tex]Mg(NO_3)_2(aq)\rightarrow Mg^{2+}(aq)+2NO_3^{-}(aq)[/tex]
[tex]0.328 M=\frac{Moles}{0.045 L}[/tex]
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 :
[tex]2\times 0.01476 mol=0.02952 mol[/tex] 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:
[tex]\frac{0.038445 mol}{0.080 L}=0.48056 M[/tex]
