A titration of an unknown concentration of sulfuric acid needs .0836 L of a 0.12 M lithium hydroxide solution. There is 34.3 mL of the unknown concentration sample. What is the concentration of the unknown sulfuric acid sample? (Hint: you need to know the reaction.
a) 0.17 M
b) 0.20 M
c) 0.36 M
d) 0.34 M
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2016-06-16 3:53 pm
回答 (2)
2016-06-16 4:57 pm
✔ 最佳答案
Equation: 2LiOH(aq) + H2SO4(aq) → Li2SO4(aq) + 2H2O(l)
2mol LiOH react with 1 mol H2SO4
Mol LiOH in .0836 L of a 0.12 M = 0.0836*0.12 = 0.01003 mol LiOH
This will react with 0.01003/2 = 0.005016 mol H2SO4
There are 0.005016 mol H2SO4 in 34.3 mL solution = 0.0343L
Mol /L H2SO4 = 0.005016/0.0343 = 0.146mol/L
To 2 significant figures [H2SO4] = 0.15M
This does not match any option - check that you have submitted the correct data.
2016-06-16 4:08 pm
H₂SO₄ + 2LiOH → Li₂SO₄ + 2H₂O
OR: Mole ratio H₂SO₄ : LiOH = 1 : 2
No. of moles of LiOH = (0.12 mol/L) × (0.0836 L) = 0.010 mol
No. of moles of H₂SO₄ = (0.010 mol) × (1/2) = 0.0050 mol
Concentration of H₂SO₄ = (0.0050 mol) / (34.3/1000 L) = 0.15 M
The answer is: None of the 4 options
OR: Mole ratio H₂SO₄ : LiOH = 1 : 2
No. of moles of LiOH = (0.12 mol/L) × (0.0836 L) = 0.010 mol
No. of moles of H₂SO₄ = (0.010 mol) × (1/2) = 0.0050 mol
Concentration of H₂SO₄ = (0.0050 mol) / (34.3/1000 L) = 0.15 M
The answer is: None of the 4 options
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