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 Na2SO4 + 2H2O What is the concentration of NaOH that is used?

Answer 1

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

v(H₂SO₄)=0.25 L
c(H₂SO₄)=2.00 mol/L
v(NaOH)=2.00 L

n(H₂SO₄)=c(H₂SO₄)v(H₂SO₄)
n(NaOH)=2n(H₂SO₄)=2c(H₂SO₄)v(H₂SO₄)

c(NaOH)=n(NaOH)/v(NaOH)=2c(H₂SO₄)v(H₂SO₄)/v(NaOH)

c(NaOH)=2*2.00*0.25/2.00=0.5 mol/L

the concentration of solution of NaOH is 0.5 mol/L

Answer 2

Answer: The molarity of the NaOH solution is 0.5 M.

Explanation:

Moles of $H_2SO_4$ in 0.25 L of 2.00 M solution:

$Molarity=\frac{\text{Moles of }H_2SO_4}{\text{Volume in Liters of}H_2SO_4}$

$2.00 M=\frac{\text{Moles of }H_2SO_4}{0.25 L}$

Moles of $H_2SO_4$ = 2.00 mol/L × 0.25 L = 0.5 moles

Moles of $NaOH$ in 2.00 L of an unknown Molarity :

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

1 mole $H_2SO_4$ gives 2 moles of $H^+$ ions.

Then 0.5 mole $H_2SO_4$ will give: 0.5 mol × 2 = 1.0 mol

1.0 mol of $H^+$ will neutralize the 1.0 mol of $OH^-$ ions.

Moles of $H^+$ = Moles of $OH^-$ = 1.0 moles

1 mol of $OH^-$are produced by 1 mol of NaOH

The 1 mole of $OH^-$will be produced from = 1 mol of NaOH

$Molarity=\frac{\text{Moles of }NaOH}{\text{Volume in Liters of NaOH}}$

$M=\frac{1 mol}{2.00L}=0.5 mol/L=0.5M$

The molarity of the NaOH solution is 0.5 M.