Produce two different metal salts (NaHSO4 and Na2SO4) through an application of specific stoichiometric ratios in a reaction of diprotic sulfuric acid (H2SO4) with sodium hydroxide (NaOH).

Authors Avatar

IB Chemistry – Practical Report

Name: Chris Bolton

Partner: Jamie Gearing

Date(s) Conducted: 20/08/04

Topic: Acids and Bases

Experiment:         Neutralisation of Polyprotic Acids

Aim:

To produce two different metal salts (NaHSO4 and Na2SO4) through an application of specific stoichiometric ratios in a reaction of diprotic sulfuric acid (H2SO4) with sodium hydroxide (NaOH).

Apparatus:

Refer to attached worksheet for given apparatus and materials

Method:

Refer to attached worksheet for given method

        AMMENDMENTS/ADDITIONS/SAFETY PRECAUTIONS:

  • Avoid adding excess solution to evaporating basins whilst hot, as this may cause damage to the basins
  • Evaporate only a small volume of the evolved acids to reduce time required to produce the stock solution
  • Note: as acids evaporate small quantities of the salt may splatter – ensure that safety glasses are on securely
  • Do not add distilled water (for the stock solution) directly to the basins, rather wait momentarily for some cooling to occur
  • Agitate evolved solutions prior to evaporation to ensure homogeneity
  • To test the ≈pH of NaHCO3, add a [small] sample to a [small] volume of distilled water and test the resulting solution with litmus

Preliminary Calculations:

Prior to commencement of the experiment, several preliminary calculations were necessary (stipulated by steps 2 and 3 in method).  First it is necessary to find the number of moles in each 20.00 cm3 aliquot of sodium hydroxide using the known concentration.

        Given concentration is obviously necessary;

                [NaOH]        = 1.0010 mol.dm-3 (accuracy must be assumed)

Using 20.00 cm3 bulb pipette to transfer the sodium hydroxide solution into each conical flask, so;

                          V        = 20.00 cm3 ± 0.01 cm3

                        = 0.02000 dm3 ± 1 x 10-5 dm3 (± 0.05 %)

Now simply rearrange the general molarity equation;

                          C        = n/V

                           n        = C.V

        And solve;

             n(NaOH)        = (1.001mol.dm-3)(2.000x10-2 dm3)

                     n(NaOH)        = 2.002x10-2 mol ± 0.05 %

Of cardinal importance, however, are the appropriate volumes of H2SO4 required to satisfy the specific stoichiometric ratios to produce the two possible salts.  First, the different ways in which diprotic sulfuric acid can react with sodium hydroxide must be explicated.

        In the first instance there is a 1 : 1 ratio between the acid and the base;

        (I)        NaOH(aq) + H2SO4(aq) → NaHSO4(aq) + H2O(l)

        Since this is a 1:1 ratio;

n(H2SO4) = n(NaOH) = 2.002x10-2 mol

        The concentration of the sulfuric acid is given as;

                [H2SO4]        = 1.0707 mol.dm-3 (accuracy must be assumed)        

Using the molarity equation again;

                               V = n/C

               V = (2.002x10-2mol)/(1.0707mol.dm-3)

                               V = 0.01870 dm3 ± 0.05 %

                V1(H2SO4) = 18.70 cm3 ± 0.05 %

This is the volume of sulfuric acid required to induce the reaction in (I), producing NaHSO4, as it will result in the presence of an equivalent number of moles for each reactant.

        Another reaction is possible, and it involves a 2:1 ratio to sodium hydroxide to sulfuric acid; 

        (II)        2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

Join now!

Since this is a 2:1 ratio;

n(H2SO4) = ½n(NaOH) = ½(2.002x10-2 mol)

          = 1.001x10-2 mol

Using the molarity equation (again);

                               V = n/C

                               V = (1.001x10-2mol)/(1.0707mol.dm-3)

                               V = 0.009349 dm3 ± 0.05 %

                V2(H2SO4) = 9.349 cm3 ± 0.05 %

This is the volume of sulfuric acid required to induce the reaction in (II), producing Na2SO4, it will result in the presence of half the number of mole NaOH in H2SO4.  Note that as would be ...

This is a preview of the whole essay