Question

An unknown sample of sulfuric acid was determined by the method used in this experiment. The sodium hydroxide was standardized by titration of 0.0855g of oxalic acid dihydrate with 25.28mL of sodium hydroxide. 10.00ml of the unknown sulfuric acid was then titrated to endpoint with 32.34mL of the same sodium hydroxide solution. What was the concentration of the sulfuric acid? (Show your work!!)

Answer 1

In this method, we are finding the concentration of sulfuric acid by volumetry.

For that the first standardization of the titrant. Since sulfuric acid is a strong acid it is better to use a strong base. Thus the given base is NaOH.

The whole titration works under the principle acidimetry and alkalimetry

The reaction between NaOH and oxalic acid is

H₂C₂O_4(aq) + 2 NaOH _(aq) --> Na₂C₂O_4(aq) + 2 H₂O _(l)

That is one oxalic acid consumes 2 moles of NaOH

Given that 0.0855 g of oxalic acid took 25.28 mL of Na OH for neutralization.

Molecular weight of oxalic acid is 90.03 g/mol

Thus 25.28 mL of NaOH took 0.0855/90.03 moles of oxalic acid

25.28 mL of NaOH took 9.46 x 10^-4moles of oxalic acid. Which means 25.28 mL contains twice 9.46 x 10^-4moles of NaOH (since one oxalic acid consumes two NaOH molecules)

Thus 25.28 mL contains 0.001899 moles. For converting into molarity (number of moles per liter) 0.001899 x 1000/25.28 =0.075 M (molarity and normality are same for NaOH since it is monoprotic base)

Standardization of NaOH is over. Now we have to estimate the unknown sulfuric acid. Reaction between sulfuric acid and NaOH is

H₂SO_4(aq) + 2 NaOH _(aq) --> Na₂SO_4(aq) + 2 H₂O _(l)

In this case also, one-mole sulfuric acid consumes 2 moles NaOH.

Given that 10 mL (volume _(H2SO4)) of this sulfuric acid consumes 32.34 mL of NaOH (volume _(NaOH))

For finding concentration

Normality _(NaOH)x volume _(NaOH)= Normality _(H2SO4) x volume _(H2SO4)

thus normality of sulfuric acid = 32.34 x 0.075/10 = 0.2425 N.

Since sulfuric acid is diprotic acid (it can react with 2 base molecules) = normality/2 = 0.2425/2

= 0.1212 M