Question

A chemist uses 0.25 L of 2.00 M H2SO4 to completely neutralize a 2.00 L of solution of NaOH. The balanced chemical equation of the reaction is given below. 2NaOH + H2SO4 mc012-1.jpg Na2SO4 + 2H2O What is the concentration of NaOH that is used?

Answer 1

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 2

Answer:

0.5 M is the concentration of NaOH used.

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$  

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $H_2SO_4$ :

Molarity = 2.00 M

Volume = 0.25 L

Thus, moles of $H_2SO_4$ :

$Moles=2.00 \times 0.25\ moles$

Moles of $H_2SO_4$ = 0.5 moles

According to the given reaction:

$2NaOH + H_2SO_4\rightarrow Na_2SO_4 + 2H_2O$

1 mole of $H_2SO_4$ reacts with 2 moles of $NaOH$

0.5 mole of $H_2SO_4$ reacts with 2*0.5 moles of $NaOH$

Moles of $NaOH$ = 1.0 moles

Given that volume of NaOH reacted = 2.00 L

So,

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$  

$Molarity=\frac{1.0}{2.00}\ M=0.5\ M$  

0.5 M is the concentration of NaOH used.