702 mL of 0.235 M H₂SO₄ solution are needed to neutralize the NaOH
Explanation:
When sodium hydroxide (NaOH) is neutralized by the sulphuric acid (H₂SO₄) the following chemical reaction occur:
2 NaOH + H₂SO₄ → Na₂SO₄ + 2 H₂O
number of moles = mass / molecular weight
number of moles of NaOH = 13.20 / 40 = 0.33 moles
Taking in account the chemical reaction, we devise the following reasoning:
if 2 moles of NaOH are reacting with 1 mole of H₂SO₄
then 0.33 moles of NaOH are reacting with X moles of H₂SO₄
X = (0.33 × 1) / 2 = 0.165 moles of H₂SO₄
molar concentration = number of moles / volume (L)
volume = number of moles / molar concentration
volume of H₂SO₄ solution = 0.165 / 0.235 = 0.702 L = 702 mL
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molar concentration
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