Respuesta :
the reaction between NaOH and H₂SO₄ is as follows;
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
Therefore molarity of NaOH - 0.352 M
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
NaOH is a strong acid and H₂SO₄ is a strong acid, therefore complete ionization into their respective ions takes place.
number of acid moles reacted - 0.112 M / 1000 mL/L x 39.1 mL = 0.0044 mol
the number of base moles required for neutralisation = 0.0044 x 2 = 0.0088 mol
Number of NaOH moles in 25.0 mL - 0.0088 mol
Therefore in 1000 mL - 0.0088 mol/ 25.0 mL x 1000 mL/L = 0.352 mol/L
Therefore molarity of NaOH - 0.352 M
Taking into account the reaction stoichiometry, 25 mL of a 0.352 [tex]\frac{moles}{liter}[/tex] NaoH solution is required to neutralize 39.1 mL of a 0.112 M H₂SO₄ solution.
Reaction stoichiometry
In first place, the balanced reaction is:
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- H₂SO₄: 1 mole
- NaOH: 2 moles
- Na₂SO₄: 1 mole
- H₂O: 2 moles
Definition of molarity
Molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.
The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:
[tex]Molarity=\frac{number of moles}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
Molarity of NaOH solution
In this case, for H₂SO₄ solution you know:
- Molarity= 0.112 M
- number of moles= ?
- Volume= 39.1 mL= 0.0391 L (being 1000 mL= 1 L)
Replacing in the definition of molarity:
[tex]0.112 M=\frac{number of moles}{0.0391 L}[/tex]
Solving:
number of moles= 0.112 M× 0.0391 L
number of moles= 0.0044 moles
So, 0.0044 moles of H₂SO₄ reacts in the reaction.
Now, the following rule of three can be applied: If by reaction stoichiometry 1 mole of H₂SO₄ react with 2 moles of NaOH, 0.0044 moles of H₂SO₄ react with how many moles of NaOH?
[tex]moles of NaOH=\frac{0.0044 moles of H_{2} SO_{4}x 2 moles of NaOH }{1 mole of H_{2} SO_{4}}[/tex]
moles of NaOH= 0.0088 moles
Then, 0.0088 moles of NaOH reacts in the reaction.
So, for NaOH solution you know:
- Molarity= ?
- number of moles= 0.0088 moles
- Volume= 25 mL= 0.025 L (being 1000 mL= 1 L)
Replacing in the definition of molarity:
[tex]Molarity=\frac{0.0088 moles}{0.025 L}[/tex]
Solving:
Molarity= 0.352 [tex]\frac{moles}{liter}[/tex]
Finally, 25 mL of a 0.352 [tex]\frac{moles}{liter}[/tex] NaoH solution is required to neutralize 39.1 mL of a 0.112 M H₂SO₄ solution.
Learn more about
the reaction stoichiometry:
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molarity:
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