802 mL of aluminum hydroxide is exactly neutralized with 628 mL 0.320 M sulfurous acid solution. What is the concentration of the base?

Balanced equation for the reaction :
2Al(OH)₃ + 3H₂SO₃ → Al₂(SO₃)₃ + 6H₂O
Mole ratio Al(OH)₃ : H₂SO₃ = 2 : 3

No. of millimoles of H₂SO₃ = (0.320 mmol/mL) × (628 mL) = 201 mmol
No. of millimoles of Al(OH)₃ = (201 mmol) × (2/3) = 134 mmol
Molarity of Al(OH)₃ = (134 mmol) / (802 mL) = 0.167 M = 167 mM