The Volume of 0.272 M potassium hydroxide solution would just neutralize 76.1 ml of 0.308 M H₂SO₄ is 40.4mL.
The balanced chemical equation is given as :
2KOH + H₂SO₄ -----> K₂SO₄ + 2H₂O
given that :
molarity of H₂SO₄ = 0.308 M
volume = 76.1 mL = 0.0761 L
moles of H₂SO₄ = 0.308 × 0.0761
= 0.0234 mol
1 mole of H₂SO₄ react with 2 mole of KOH.
Moles of KOH = 0.0234 / 2 = 0.011 mol
the molarity is given as :
concentration of KOH = moles / volume
volume = moles / concentration
= 0.011 / 0.272
= 0.0404 L = 40.4 mL
Thus, The Volume of 0.272 M potassium hydroxide solution would just neutralize 76.1 ml of 0.308 M H₂SO₄ is 40.4mL.
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