Question

# A chemistry student weighs out 0.103g of sulfurous acid (H2SO3) , a diprotic acid, into a...

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

H2SO3 + 2 NaOH ====> Na2SO3 + 2 H2O
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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