The molarity of the NaOH solution was 0.425 M and 26.30 Ml of H2SO4 was added to the Erlenmeyer flask .
H2SO4 + 2 NaOH ——->Na2SO4+2 HO
Calculate the molaraty of the H2SO4 solution if 16.45 Ml of NaOH was necessary to reach the end point of titration.?
2018-02-25 9:21 am
回答 (3)
2018-02-25 5:58 pm
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
Mole ratio H₂SO₄ : H₂SO₄ = 1 : 2
Amount of NaOH reacted = (0.425 mmol/ml) × (16.45 ml) = 6.99 mmol
Amount of H₂SO₄ required = (6.99 mol) × (1/2) = 3.50 mmol
Molarity of H₂SO₄ = (3.50 mmol) / (26.30 ml) = 0.133 M
Mole ratio H₂SO₄ : H₂SO₄ = 1 : 2
Amount of NaOH reacted = (0.425 mmol/ml) × (16.45 ml) = 6.99 mmol
Amount of H₂SO₄ required = (6.99 mol) × (1/2) = 3.50 mmol
Molarity of H₂SO₄ = (3.50 mmol) / (26.30 ml) = 0.133 M
2018-02-25 11:16 am
0.425 moles NaOH/liter x 16.45 mL x 1 liter /1000 mL = 6.99 x 10^-3 moles NaOH
H2SO4 + 2 NaOH ——-> Na2SO4 + 2H2O
6.99 x 10^-3 moles NaOH x 1 moles H2SO4/2moles NaOH = 3.495 x 10^-3 moles H2SO4
Molarity of H2SO4 = 3.495 x 10^-3moles H2SO4 /26.30 mL x 1000 mL/liter = 0.133 M
H2SO4 + 2 NaOH ——-> Na2SO4 + 2H2O
6.99 x 10^-3 moles NaOH x 1 moles H2SO4/2moles NaOH = 3.495 x 10^-3 moles H2SO4
Molarity of H2SO4 = 3.495 x 10^-3moles H2SO4 /26.30 mL x 1000 mL/liter = 0.133 M
2018-02-25 9:23 am
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