Respuesta :
Answer:
A) 0.801
Explanation:
The reaction that takes place is:
- 2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
First we calculate the H₂SO₄ moles that reacted:
- 0.355 M * 28.2 mL = 10.011 mmol H₂SO₄
Now we convert H₂SO₄ moles to NaOH moles:
- 10.011 mmol H₂SO₄ * [tex]\frac{2mmolNaOH}{1mmolH_2SO_4}[/tex] = 20.022 mmol NaOH
Finally we calculate the molarity of the NaOH solution:
- 20.022 mmol NaOH / 25.0 mL = 0.801 M
So the answer is option A.
The molarity of the NaOH solution used in the neutralization reaction has been 0.801 [tex]\rm \times\;10^-^3[/tex] M. Thus, option A is correct.
The neutralization reaction has resulted in the formation of the salt and water on reacting an acid with a base. In the neutralization reaction, the strength of acid and base can be given as:
[tex]\rm 2\;NaOH\;+\;H_2SO_4\;\rightarrow\;Na_2SO_4\;+\;H_2O[/tex]
For the neutralization of 1 mole of sulfuric acid, 2 moles of NaOH has been required.
The moles of sulfuric acid can be given as:
Moles = Molarity × Volume
Given, the molarity of acid ([tex]\rm H_2SO_4[/tex]) = 0.355 M
Volume of acid = 28.2 ml = 0.0282 L
Volume of base (NaOH) = 25 ml = 0.025 L
Moles of [tex]\rm H_2SO_4[/tex] = 0.355 × 0.0282 mol
Moles of [tex]\rm H_2SO_4[/tex] = 10.011 mol.
Since, 1 mole [tex]\rm H_2SO_4[/tex] = 2 moles NaOH
10.011 moles [tex]\rm H_2SO_4[/tex] = 10.011 × 2 moles NaOH
10.011 moles [tex]\rm H_2SO_4[/tex] = 20.022 moles NaOH.
The neutralization of 0.355 M [tex]\rm H_2SO_4[/tex] requires 20.022 moles NaOH. The strength of the NaOH solution will be:
Molarity = [tex]\rm \dfrac{Moles}{Volume\;(L)}[/tex]
Molarity of NaOH = [tex]\rm \dfrac{20.022}{0.025}[/tex]
Molarity of NaOH = 0.801 [tex]\rm \times\;10^-^3[/tex] M.
The molarity of the NaOH solution used in the neutralization reaction has been 0.801 [tex]\rm \times\;10^-^3[/tex] M. Thus, option A is correct.
For more information about the neutralization reaction, refer to the link:
https://brainly.com/question/4612545
