sulphuric acid is dibasic

Method: Titration:

1. Pour in 100cm3of sulphuric acid and 100cm3of Sodium hydroxide in two separate beakers and label them with their corresponding names, H2SO4 and NaOH.
2. Rinse burette first with distilled water then with small amount of H2SO4 while tap is still closed. Leave some of the acid and run it through the tap.
3. Close the tap and clamp the burette. Make sure it is secure.
4. Using filter funnel, fill burette with H2SO4.
5. Also, rinse conical flask with distilled water first and then with NaOH.
6. Bind pipette filler with pipette and fill in 25 cm3 of NaOH and read lower meniscus.
7. Hold pipette above conical flask and remove the pipette filler which will allow NaOH to run down into it.
8. Put a whit tile underneath burette.
9. Add few drops of phenolphthalein in the conical flask containing NaOH and place it on the tile.
10. Open the tap fully and allow the acid to flow and at the same time swirl the conical flask.
11. If the colour of solution is going lighter, twist the tap so that only drops of acid flow into the conical flask.
12. Keep adding the drops until one last drop changes the colour of the solution to colourless.
13. Close the tap with that last drop and record the reading of the acid that was used in a table.
14. Repeat the same procedure 3 more times and find the average volume of acid used.
15. Carry out exactly the same procedure from step 1 to 13 but using HCl instead of H2SO4.
16. See the figure below for how to set up the apparatus.

 

Since I have got two reading similar, I can stop titration.

Titration: Equations

H2SO4 (aq) + 2NaOH (aq) →Na2 SO4 (aq) + 2H2O (l)

Concentration (H2SO4) =1.00moldm-3

Volume (H2SO4) = 25.60 cm3

no. of moles of (H2SO4) = 1.00 x (25.60/1000) = 0.025mol

no. of moles of (NaOH) = 0.025 x 2 = 0.050 mol

From this equation we can see that molar ratio of H2SO4: NaOH is 1:2.So, 1 mole of H2SO4 requires two moles of NaOH and produces two moles of water in titration.

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

Concentration of (HCl) = 1.00moldm-3

Volume of (HCl) = 25.0cm3

No. of moles of (HCl) = 1.00 x (25/1000) = 0.025mol

No. of moles of (NaOH) = 0.025mol

From this equation we can see that 1 mole of HCl requires half a mole of NaOH than that required by H2SO4. There is only 1 mole of water produced. Two moles of water has double the number of hydrogen atoms. Therefore, it is proved that H2SO4 is dibasic.

Apparatus:  Collection of gas

Method:

1. Rinse conical flask with distilled water first and then with sulphuric acid.
2. Pour in 4cm3of sulphuric acid into it.
3. Prepare bowl of water and place measuring cylinder into it carefully.
4. Fit delivery tube in bung.
5. Calculate minimum mass of magnesium require from equation and add 1 extra spatula of magnesium powder to make it in excess
6. Make sure all the other apparatus are ready for experiment.
7. Add magnesium powder into the conical flask and start the timer.
8. Record volume of hydrogen gas produced from the measuring cylinder when all of the metal (magnesium powder) has completely reacted with the acid and this is indicated by no more production of bubbles.
9. Repeat step 1-9 using hydrochloric acid.

Collection of gas: Equations

Mg(S) + H2SO4 (aq) → Mg SO4 (aq) + H2 (g)      

Concentration (H2SO4) =1.00moldm-3

Volume (H2SO4) = 4.00 cm3

no. of moles of (H2SO4) = 1.00 x (4.00/1000) = 4.00x10-3 mole

no. of moles of (Mg) = 4.00x10-3 mole

Mass of (Mg) = 0.097g

Since 1 mole of a gas occupies 24dm3 of space.

Therefore, 1mole: 24dm3= 4.00x10-3 moles: x dm3 (where x is the volume of hydrogen gas produced.)

1/ 24dm3 =4.00x10-3 /x

x = 0.096 dm3

x = 96.00cm3

Volume of H2 = 96.00cm3

Mg(S) + 2HCl (aq) → MgCl2 (aq) + H2 (g)      

Concentration of (HCl) = 1.00moldm-3

Volume of (HCl) = 4.00 cm3

No. of moles of ( HCl) = 1.00 x (4.00/1000) = 4.00x10-3 mole

No. of moles of (H2) = 4.00x10-3 /2 = 2.00x10-3mole

Since 1 mole of a gas occupies 24dm3 of space.

Therefore, 1mole: 24dm3= 4.00x10-3 moles: x dm3 (where x is the volume of hydrogen gas produced.)

1/ 24dm3 =2.00x10-3 /x

x = 0.048 dm3

x = 48.00cm3

Volume of H2 = 48.00cm3

1 mole of H2SO4 gives 96.00cm3of H2 gas where as, 1 mole of HCl produces only half a mole of H2 gas (48.00cm3).

Consequently, I have proven that H2SO4 is dibasic.

Analysis:

1. Titration gives an accurate reading because the apparatus used such as pipette and burette are accurately calibrated and so can measure up to 0.10 cm3 than the collection of gas using a measuring cylinder.
2. Titration is more reliable as it is repeated several times and an average volume of titres is found.
3. Since I am using an indicator in titration, I can determine the end point whereas there isn’t an end point for the gas evolved.
4. The results of the gas collection may be inaccurate because some of the gas may have escaped while the metal is added into the acid.

References:

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