The concentration of an unknown sample of sulfuric acid was determined by the me

Question

The concentration of an unknown sample of sulfuric acid was determined by the method used in this experiment, using two sets of titrations. In the first titration the sodium hydroxide was standardized by titrating 0.1355g of oxalic acid dihydrate (molar mass 126.07g/mole) with 25.30mL of sodium hydroxide solution. In the second titration 20.00mL of the unknown sulfuric acid solution was titrated with 22.85mL of the sodium hydroxide solution. What was the concentration of the sulfuric acid? (Show your work!!) 5.

Explanation / Answer

Answer

0.0486M

Explanation

Titration 1

H2C2O4 + 2NaOH - - - - - > Na2C2O4 + 2H2O

1mol Oxalic acid react with 2moles of NaOH

No of moles of oxalic acid = 0.1355g/126.07g/mol = 0.001075

Therefore

No of moles of NaOH consumed = 2× 0.001075mol= 0.00215

Concentration of NaOH = (0.00215mol/25.30ml)×1000ml = 0.08498M

Titration 2

H2SO4 + 2NaOH - - - - - > Na2SO4 + 2H2O

1mole of Sulphuric acid react with 2moles of NaOH

No of moles of NaOH consumed = (0.08498mol/1000ml)× 22.85ml = 0.001942

0.001942moles of NaOH indicate 0.001942/2 = 0.000971moles of H2SO4

Therefore

Concentration of H2SO4 = (0.000971mol/20.00ml)×1000ml = 0.0486M

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