What volume (mL) of 0.13 M K2Cr2O7 would be required to oxidize 30 mL of 0.18 M Na2SO3 in acidic solution, ?
according to the following balanced reaction?
8H+ + Cr2O7–2 + 3SO3–2 → 2Cr+3 + 3SO4–2 + 4H2O
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In the balanced reaction eq'n , 1 mole of dichromate produces 3 moles sulphite.
mol(SO3^2-) = 0.18 x 30 /1000 = 5.4 x 10^-3 mol (equivalent tp '3' molar ratios)
hnce
mol(Cr2O7^2-) = 5.4 x 10^-3 / 3 = 1.8 x 10^-3 mol (equivalent to '1' molar ratio)
Hence
1.8 x 10^-3 = 0.13 x vol(mL) / 1000
Algebraically rearrange
vol(mL) = 1.8 x 10^-3 x 1000 / 0.13 = 13.846 mL is the required volume.
Equation for the reaction:
8H⁺ + Cr₂O₇⁻² + 3SO₃⁻² → 2Cr⁺³ + 3SO₄⁻² + 4H₂O
Mole ratio Cr₂O₇⁻² : 3SO₃⁻² = 1 : 3
Moles of SO₃⁻² = (0.18 mol/L) × (30/1000 L) = 0.0054 mol
Moles of Cr₂O₇⁻² = (0.0054 mol) × (1/3) = 0.0018 mol
Volume of Cr₂O₇⁻² = (0.0018 mol) / (0.13 mol/L) = 0.0138 L = 13.8 mL
Volume of K₂Cr₂O₇ = 13.8 mL
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OR:
(0.18 mol SO₃⁻² / 1000 mL SO₃⁻²) × (30 mL SO₃⁻²) × (1 mol Cr₂O₇⁻² / 3 mol SO₃⁻²) × (1000 mL Cr₂O₇⁻² / 0.13 mol Cr₂O₇⁻²) × (1 mol K₂Cr₂O₇ / 1 mol Cr₂O₇⁻²)
= 13.8 mL K₂Cr₂O₇