A chemistry student weighs out 0.103g of sulfurous acid (H2SO3) , a diprotic acid, into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.0800M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.
Let us write the Balanced equation
H_{2}SO_{3} +
2 NaOH ====> Na_{2}SO_{3} + 2
H_{2}O
Reaction type: double replacement.
Mass of sulfurous acid = 0.103 gm
Molar mass of sulfurous acid = 82.079 g/mol
Moles of sulfurous acid = 0.103 gm / 82.079 g/mol = 0.00125488 Moles
Moles of NaOH required = 0.00250977 Moles
Concentration of NaOH = 0.08 M
Volume of NaOH = 0.08 / 0.00250977 = 31.87 ml
Hence 31.87 ml of NaOH is required to neutralize 0.103 g of H2SO3.
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