Chemical Reaction: Finding the Concentration of NaOH Used

What is the concentration of NaOH that is used?

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

c₁=2.00 mol/L
v₁=0.25 L
v₂=2.00 L
c₂-?

n(NaOH)=c₂v₂
n(H₂SO₄)=c₁v₁
n(NaOH)=2n(H₂SO₄)

c₂v₂=2c₁v₁

c₂=2c₁v₁/v₂

c₂=2*2.00*0.25/2.00=0.5 mol/L

0.5 M NaOH


Answer:

0.5 M is the concentration of NaOH used.

Explanation:

Considering:

Molarity= Moles of solute / Volume of the solution

Or,

Moles = Molarity x Volume of the solution

Given :

For H₂SO₄ :

Molarity = 2.00 M

Volume = 0.25 L

Thus, moles of H₂SO₄ :

Moles = 2.00 x 0.25 moles

Moles of H₂SO₄ = 0.5 moles

According to the given reaction:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

1 mole of H₂SO₄ reacts with 2 moles of NaOH

0.5 mole of H₂SO₄ reacts with 2 x 0.5 moles of NaOH

Moles of NaOH = 1.0 moles

Given that volume of NaOH reacted = 2.00 L

So,

Molarity= Moles of solute / Volume of the solution

Molarity= 1.0 / 2.00 M = 0.5 M

0.5 M is the concentration of NaOH used.

← Fastest movement of na ions across a permeable membrane Understanding reduction half reaction in chemistry →