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Demonstrating Sulphuric acid is Dibasic

Free essay example:

Sophie Keltie

Candidate Number 2076

Centre Number 17625

Chemistry Practical Plan


The aim of my experiment is to demonstrate that sulphuric acid, H2SO4, is dibasic. Dibasic means that one mole of sulphuric acid can release two moles of hydrogen ions, H+ when it reacts in an aqueous solution. To demonstrate the dibasic nature of sulphuric acid, I shall do two experiments, one involving a titration, the other a gas collection.


In my titration, I shall neutralise an acid with a base. Firstly, I will titrate Hydrochloric Acid against Sodium Hydroxide. The balanced chemical equation for this reaction is as follows.

HCl(aq) + NaOH(aq)→ NaCl(aq)  + H20(l)

Hydrochloric Acid, HCl, is a monoprotic acid, meaning that it will release one Hydrogen ion, H+ when reacted in an aqueous solution. The molar ratio in this reaction is 1:1, and therefore 1 mole of Hydrogen ions, H+, will be needed to react with 1 mole of Hydroxide ions, OH-, to form 1 mole of Water, H20. Therefore, I predict that this will mean that an equal number of moles of Hydrochloric Acid will be needed to neutralise the Sodium Hydroxide.

I shall then titrate Sulphuric Acid against Sodium Hydroxide, in comparison to Hydrochloric Acid. HCl is monoprotic, whereas H2SO4  is diprotic, and will therefore release double the amount of Hydrogen ions, H+. The reaction is as follows:

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

To demonstrate that Sulphuric Acid is dibasic, and that it will release two Hydrogen ions, H+, I predict that only half the number of moles of H2SO4  will be needed to neutralise one mole of NaOH. The molar ratio of acid to alkali is now 1:2, so for every Hydroxide ion released from the Sodium Hydroxide, two Hydrogen ions will be released from the sulphuric Acid, and so only 0.5 mole H2SO4 will be needed to neutralise 1 mole NaOH.

For the gas collection experiment, I shall again react firstly a monoprotic acid, then a diprotic acid and compare the amounts of gas collected. My first reaction will be between Hydrochloric Acid (monoprotic) with Magnesium Carbonate.

2HCl(aq) +  MgCO3(s) →  MgCl2(aq)  +  H2O(l)  +  CO2(g)  

My second reaction will be between Sulphuric Acid (diprotic), and Magnesium Carbonate.

H2SO4(aq) + MgCO3(s) → MgSO4(aq) + H2O(l)  +  CO2(g)

For both reactions I shall collect the gas in a gas syringe, and compare the amounts of gas released. When metal reacts with acid, carbon dioxide is released. I will use acids of equal molarity, and the same mass of magnesium carbonate. I predict that the H2SO4 will produce twice the volume of gas in comparison to HCl, as this will demonstrate the dibasity of Sulphuric Acid.


Sulphuric Acid is corrosive and therefore goggles must we worn at all times. If spillages occur, these should be washed liberally with cold water. Sulphuric Acid can burn skin and eyes, so lab coats and gloves should be worn. In case of contact with eyes, immediately flush eyes with plenty of water for at least 15 minutes. If swallowed drink plenty of water and seek medical help. Also follow standard laboratory safety such as removing lose clothing and tying hair away from face.

Fair Test

Titration – to ensure a fair test in my titration, I will use the same amounts of acid the same in each titration, and measure the variation in alkali needed to neutralise the acid. I will use a pipette and burette because they are the most accurate equipment available, being accurate to +/- 0.05cm3. Before filling the burette and pipette I shall rinse them first with distilled water and then with the chemical to be used. This will allow me to take more accurate readings from the equipment and reduce the possible error margin. I shall rinse the burette and pipette also with the solution to be used to ensure no dilutions in my experiment, which could lead to inaccuracy.

Gas Collection – to ensure a fair test in the gas collection, I shall use equal masses of Magnesium Carbonate in both collections, and measure the varying amount of Carbon Dioxide released. I will use a gas syringe to collect the carbon dioxide because I think that this I more accurate than upwards delivery – carbon dioxide is soluble in water and this could affect my results.

For both experiments, I shall also use the largest values possible as this will reduce the percentage error in my practical to give me more reliable results.

Titration Method

Apparatus: Sodium Hydroxide (1.0 molar), Sulphuric Acid (1.0 molar), Hydrochloric Acid (1.0 molar), 50 cm 3 pipette, burette, phenolphthalein indicator, white tile, clamp stand, boss, distilled water dispensers, pipette filler, 4 × 100cm3 beakers, 2 × 100cm3 conical flasks, plastic filter funnel.

  1. Perform pre-tests to determine the colour change of the indicator at the end point of the titration and the colour of phenolphthalein in acids and alkalis.
  2. Prepare the equipment as follows:
  • Burette – rinse with distilled water followed by the solution to be used, NaOH as not to dilute the solution with water.
  • Run solution through and invert the burette to ensure no air bubbles.
  • Fill the burette with NaOH using a filter funnel, and remove the funnel. Record the volume of solution within the burette to 0.05 ml.
  • Pipette – rinse the pipette several times by sucking and releasing the solution to be used, HCl or H2SO4, using a pipette filler, (suck up solution to above the measured mark)
  • Using pipette filler, fill pipette with until the meniscus is slightly above the mark.
  • Take the pipette out of the solution to ensure no atmospheric pressure and allow the liquid to run out slowly until the meniscus is level with the mark.
  • Touch the side of the flask with the tip of the pipette but allow any residue to remain in the pipette.
  1. Add 3-4 drops of the indicator phenolphthalein to the solution in the conical flask (no more as indicators are weak acids and so can have an effect on the titration)
  2. Use a white tile to help identify the colour change at the end point.
  3. Run the solution from the burette whilst swirling, stop as soon as the indicator colour changes.
  4. Record the volume of the rough titration.
  5. Repeat the titration carefully and drop wise until the colour is about to change, then add half a drop at a time.
  6. When the faintest detectable colour change can be noted, record the final volume to within 0.05 ml.
  7. Repeat the titration to get three accurate titrations within 0.1 ml of each other.

Gas Collection

Apparatus: Hydrochloric Acid (1 molar), Sulphuric Acid (1 molar), Magnesium Carbonate, 6 conical flasks, gas valves and bungs, gas syringe with delivery tube, Clamp stand and boss.

  1. Set up the experiment as shown.
  1. React 0.34g MgCO3 with 100ml HCl for at least 10 minutes or until the reaction is complete.
  2. Record the volume of CO2 produced.
  3. Repeat three times, ensuring that the gas syringe is always set at zero before the experiment is set.
  4. Repeat with H2SO4to allow the comparison of results.



From my research I know that the maximum volume of Sodium Hydroxide will be needed in the titration with Hydrochloric Acid. The burette I will be using can measure accurately up to 50cm3. I from my calculations, this amount will be used to neutralise 50cm3 of acid. For the reaction with Sulphuric Acid, twice the volume of NaOH shall be needed to neutralise the acid. Therefore, I shall use only 25cm3 acid for both titrations, as I know that 50cm3 NaOH is the maximum volume I will need. The advantage of using the largest volumes possible is that the percentage error is reduced, and my experiments will therefore me more accurate.

Gas Collection

H2SO4(aq) + MgCO3(s) → MgSO4(aq) + H2O(l)  +  CO2(g)

This is the reaction that I predicted would release the most carbon dioxide. I therefore had to work out what the maximum amount of product I could use within the practical limits of my apparatus.

Max volume gas released – 100cm3

No. moles = actual volume ÷ 24

= (100÷1000) ÷ 24

= 0.004 moles CO2

Molar ratio MgCO3: CO2 =1:1

Therefore 0.004 moles MgCO3

Mr MgCO3 = 24+12+ (16 x 3)


Actual Mass = number of moles x mass 1 mole

= 0.004 x 84

0.34g Magnesium Carbonate to be used.

The Sulphuric Acid must also be present in excess to ensure the reaction goes to completion, and so I will therefore use 0.1 mole H2SO4.

Volume = number of moles ÷ concentration

= 0.01 ÷1 = 0.01 dm-3, or 100 cm-3

2HCl(aq) +  MgCO3(s) →  MgCl2(aq)  +  H2O(l)  +  CO2(g)  

For my second reaction, I shall use the same mass of Magnesium Carbonate, and provided the acid is still in excess, the same volume of Hydrochloric Acid, to ensure a fair test.

Mass MgCO3 to be used = 0.34g

Molar ratio HCl: CO2 = 2:1

Therefore 0.004 x 2 moles HCl to be used.

To allow the acid to be present in excess, I shall use 0.01 mole of Hydrochloric Acid.

Therefore, volume = number of moles x concentration

=0.01 × 1


Specimen Calculations

Titration using Sodium Hydroxide and Hydrochloric Acid

HCl(aq) + NaOH(aq)→ NaCl(aq)  + H20(l)

Volume HCl used: 25 ml or 0.025 dm-3

Number of moles HCl = concentration (mol dm-3) × volume (dm-3)

= 1m× 0.025 dm-3

= 0.025 moles HCl

Molar ratio NaOH: HCl = 1: 1

Therefore 0.025×1= 0.025 moles NaOH

Volume NaOH = number of moles × concentration (mol dm-3)

= 0.025 × 1

=0.025 dm-3

This shows that an equal volume of Sodium Hydroxide is needed to neutralise the Hydrochloric acid, indicating that for every OH- ion in the NaOH, one H+ ion was released from the Hydrochloric Acid to form H2O, a neutral molecule. This shows that Hydrochloric Acid is monoprotic, as it releases one Hydrogen ion when it is reacted in an aqueous solution.

Titration using Sodium Hydroxide and Sulphuric Acid

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

Volume H2SO4 used: 25 ml or 0.025 dm-3

Number of moles H2SO4 = concentration (mol dm-3) × volume (dm-3)

= 1m× 0.025 dm-3

= 0.025 moles H2SO4

Molar ration NaOH: H2SO4 = 2: 1

Therefore 0.025÷2 = 0.0125 moles NaOH

Volume NaOH = number of moles × concentration (mol dm-3)

=0.0125× 1 = 0.0125 dm-3

This shows that only half the volume of Sodium Hydroxide was needed to neutralise the Sulphuric Acid, indicating that for every OH- ion in the NaOH, two H+ were released from the Sulphuric Acid, demonstrating its dibasity, as for every mole of Sulphuric Acid reacting in solution; two moles of Hydrogen ions were released.

Gas Collection

2HCl(aq) +  MgCO3(s) →  MgCl2(aq)  +  H2O(l)  +  CO2(g)  

Number of moles MgCO3 = actual mass ÷ mass 1 mole

Mr MgCO3 = 84

= 0.34 ÷ 84

= 0.004 moles

Molar ration HCl: CO2 = 2: 1

Therefore 0.004 ÷ 2 = 0.002 moles CO2

Volume = number of moles x 24

= 0.002 × 24

= 0.048 dm-3 or 48cm3

H2SO4(aq) + MgCO3(s) → MgSO4(aq) + H2O(l)  +  CO2(g)

Number of moles MgCO3 = actual mass ÷ mass 1 mole

Mr MgCO3 = 84

= 0.34 ÷ 84

= 0.004 moles

Molar ratio H2SO4: CO2 = 1: 1

Therefore 0.004 moles CO2

Volume = number of moles × 24

= 0.004 × 24

=0.096 dm-3 or 96cm3

When a metal carbonate reacts with acid, carbon dioxide is released. These results show that when Magnesium Carbonate reacts with Sulphuric Acid, twice the volume of Carbon Dioxide is produced in comparison to its reaction with Hydrochloric Acid. This shows that, as we know HCl to be a monobasic acid, Sulphuric Acid is dibasic, as it produced double the volume of gas as a result of its two Hyrdogen ions that are released.


  • ‘Chemistry’ – Chris Conoley and Phil Hills (p. 332)

Collins Educational 1998

  • Steven Doherty – Atoms, Molecules and Stoichiometry  
  • www.catalogue.fisher.co.uk/scripts
  • Cambridge University Press 2000

‘Chemistry’ Brian Ratcliff

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