A 37 mL sample of a solution of sulfuric acid
is neutralized by 42 mL of a 0.113 M sodium
hydroxide solution. Calculate the molarity of
the sulfuric acid solution.
Answer in units of mol/L
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
mols NaOH = M x L = ?
Convert mols NaOH to mols H2SO4.
M H2SO4 = mols H2SO4/L H2SO4.
To calculate the molarity of the sulfuric acid solution, we need to use the balanced chemical equation for the neutralization reaction between sulfuric acid (H₂SO₄) and sodium hydroxide (NaOH). The balanced equation is:
H₂SO₄ + 2NaOH -> Na₂SO₄ + 2H₂O
From the equation, we can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide. Thus, the number of moles of sodium hydroxide in the solution can be determined by multiplying the volume of the sodium hydroxide solution (in liters) by its molarity (in mol/L):
moles of NaOH = volume of NaOH solution (L) x molarity of NaOH (mol/L)
moles of NaOH = 0.042 L x 0.113 mol/L
moles of NaOH = 0.004746 mol
Since sulfuric acid and sodium hydroxide react in a 1:2 ratio, the number of moles of sulfuric acid is twice the number of moles of sodium hydroxide:
moles of H₂SO₄ = 2 x moles of NaOH
moles of H₂SO₄ = 2 x 0.004746 mol
moles of H₂SO₄ = 0.009492 mol
Now, we can calculate the molarity (M) of the sulfuric acid solution by dividing the moles of sulfuric acid by the volume of the sulfuric acid solution (in liters):
molarity of H₂SO₄ = moles of H₂SO₄ / volume of H₂SO₄ solution (L)
molarity of H₂SO₄ = 0.009492 mol / 0.037 L
molarity of H₂SO₄ ≈ 0.256 M
Therefore, the molarity of the sulfuric acid solution is approximately 0.256 mol/L.