A 25.00 mL sample of NaOH is titrated with 17.13 mL of 0.3150 M sulfuric acid. What is the concentration of the NaOH solution?
回答 (2)
2017-12-17 6:59 pm
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
Mole ratio H₂SO₄ : NaOH = 1 : 2
No. of milli-moles of H₂SO₄ reacted = (0.3150 mmol/mL) × (17.13 mL) = 5.396 mmol
No. of milli-moles of NaOH reacted = (5.396 mmol) × 2 = 10.792 mmol
Concentration of NaOH solution = (10.792 mmol) / (25.00 mL) = 0.4317 M
Mole ratio H₂SO₄ : NaOH = 1 : 2
No. of milli-moles of H₂SO₄ reacted = (0.3150 mmol/mL) × (17.13 mL) = 5.396 mmol
No. of milli-moles of NaOH reacted = (5.396 mmol) × 2 = 10.792 mmol
Concentration of NaOH solution = (10.792 mmol) / (25.00 mL) = 0.4317 M
2017-12-17 2:13 pm
assuming both H from the acid react
0.3150 X 2 / 25.00 <<< use a calculator
0.3150 X 2 / 25.00 <<< use a calculator
收錄日期: 2021-04-18 18:03:03
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