Find out the concentrations of 5 acids by carrying out volumetric analysis, by which molarity of acids can be determined when they titrate an alkali.

Safety:

• Be careful to make sure the acid doesn’t spill on your hands, and in case of the acid spilling on your hands wash immediately with luke warm water
• Do not spill any of the substances on the work surface and wipe away immediately if dropped
• Handle the glass equipment with care
• Make sure the beakers are labelled accurately

Apparatus:

• Burette
• Conical flask
• Clamp
• Retort stand
• 2 beakers
• White tile
• Pipette
• Filler
• Funnel
• Sticky labels
• Distilled water
• Hydrochloric acid (HCl)
• Sulphuric acid (H2SO4)
• Nitric acid (HNO3)
• Ethanoic acid (CH3COOH)
• Citric acid (C6H8O7)
• Sodium Hydroxide 1M
• Indicator- Phenolphthalein

Method:

• Collect all the equipment and place it on a clean, uncluttered surface
• Wash the burette, Conical flask, and beakers with tap water
• Wash this equipment again with distilled water
• Fill one beaker with 1Molar solution of NaOH and label it and the other with H2SO4 again putting on a label
• Rinse the burette with a small amount of H2SO4
• Fill the burette, using a funnel, with H2SO4 just above the 0 mark
• Let the acid run up till the 0 mark into the beaker marked H2SO4, by opening the tap
• Hold the burette to the retort stand and tighten the clamp
• Take the temperature of the acid and record it
• Attach the filler to the pipette and pull in 25cm³ of NaOH, this is marked on the pipette, again make sure you read the lower meniscus
• Release the NaOH into the conical flask by detaching the filler from the pipette
• Put few drops of the indicator Methyl Orange in the conical flask
• Place the white tile underneath the burette and place the conical flask containing the NaOH on the tile so that the tap is partway in the flask but not touching the solution
• Open the tap of the burette fully and let the acid flow all the time swirling the flask
• When you see a subtle change in colour from yellow to darker orangey colour twist the tap so that only droplets of acid fall into the flask
• When the first signs of red colour show in the solution stop and swirl the flask, if the solution has not turned completely red then add another drop and stop
• Carry on doing this until the solution has turned red but make sure you stop as soon as it does
• Take the reading from the burette and record it making sure to read from the lower meniscus
• Repeat the experiment for H2SO4 two more times
• Wash all the equipment again in the same way as before but this time rinse the burette with the next acid
• Carry out the experiment for HCl, HNO3, CH3COOH, C6H8O7  following the procedure and then record all the results
• Make sure you change the label of the beaker containing the acid each time you change the acid
• Lastly make sure each experiment is done three times

Prediction:

I predict that when the mol of alkali used is greater than the mol of acid used, the volume of acid needed to titrate the same volume of NaOH will be less, than when the mol.s are equal or the mol of acid is greater.

Below is the method of predicting how much of acid will be used in the neutralisation of 25cm³ of NaOH.

H2SO4 + 2NaOH        Na2SO4 + 2H2O

    1       :      2

  12.5    :     25

So… 25/2 =12.5, because 1mol of H2SO4 is needed to neutralise 2 mols of NaOH. 2 mols = 25, so 1 mol equals half of 2 mols.

HNO3 + NaOH         NaNO3 + H2O

    1      :     1

   25     :    25

HCl + NaOH         NaCl + H2O

   1   :     1

  25  :    25

CH3COOH + NaOH         CH3COONa + H2O

         1        :      1

        25       :     25

C6H8O7 + 3NaOH        C6H5O7Na3 + 3H2O

      1      :       3

     8.3    :      25

I predict that 12.5cm³ of H2SO4 will be needed to titrate 25cm³ of NaOH

I predict that 25cm³ of  HNO3 will be needed to titrate 25cm³ of NaOH

I predict that 25cm³ of  HCl will be needed to titrate 25cm³ of NaOH

I predict that 25cm³ of  CH3COOH will be needed to titrate 25cm³ of NaOH

I predict that 8.3cm³ of C6H8O7 will be needed to titrate 25cm³ of NaOH

This is a general prediction, for 1M of each acid. The volume used will depend on how many moles of acid there are. So if the concentration of HCl is 2M then the volume used will be two times the volume for 1M.

An acid is a solution that contains H+ ion when dissolved in H2O.

• A strong acid is an acid that produces H+ ion completely when reacted with H2O. For example, H2SO4, HCl, HNO3, etc.
• A weak acid is an acid that DOES NOT produce H+ ion completely when reacts with H2O. For example, H2CO3, CH3COOH, etc

The maximum number of H+ ions produced by one mole of pure acid is called the basicity (the number of hydrogen atoms available to react with a base) of the acid. Generally, they are Monobasic, Dibasic, and Tribasic.  

This means that with acids such as sulphuric acid, only ½ the amount of it will be used when it reacts with the base sodium hydroxide, as it is Dibasic. There is a similar case with citric acid, as it is Tribasic, only 1/3 of it will be used when it reacts with the base NaOH.

Obtaining Evidence:

Analysis:

From my results I can see that the least volume of acid used for neutralising 25cm³ of 1M NaOH, was 8.9cm³, which was citric acid. The next was sulphuric acid and then ethanoic, nitric and then Hydrochloric.

Using the volume of each acid used I can determine what the molarity of each one is. The formula to work out molarity is

No. of moles = Volume x Concentration

Sulphuric acid-

H2SO4 + 2NaOH        Na2SO4 + 2H2O

Amount of NaOH = Vol. x Conc.

                              = 25/1000 x 1.0

  2NaOH                = 0.025

2 x Amount of acid = 2NaOH = 0.025

So if c= concentration-

       0.025      = 2 (12.6/1000) x c

0.025 x 1000 = 2 x 12.6 x c

         25         = 25.2 x c

         25/25.2 = c

         0.99      = c

So the concentration of H2SO4 is 0.99M.

Nitric acid-

HNO3 + NaOH         NaNO3 + H2O

0.025 = 25.5/1000 x c

(0.025 x1000)/25.5 = c

0.98 = c

So the concentration of HNO3 is 0.98M.

Hydrochloric acid-

HCl + NaOH         NaCl + H2O

0.025 = 28.6/1000 x c

(0.025 x 1000)/28.6 = c

0.91 = c

So the concentration of HCl is 0.91M.

Ethanoic acid-

CH3COOH + NaOH         CH3COONa + H2O

0.025 = 24.3/1000 x c

(0.025 x 1000)/24.3 = c

1.03 = c

So the concentration of CH3COOH is 1.03M.

Citric acid-

C6H8O7 + 3NaOH        C6H5O7Na3 + 3H2O

0.025 = 3 x 8.9/1000 x c

(0.025 x 1000)/26.7 = c

0.94 = c

So the concentration of CH3COOH is 0.94M.

All the concentrations are close to 1M, and in my prediction when I did calculations for 1M of acids I got similar volumes to that in the experiment. This leads me to conclude that we were probably given 1M concentration of all acids for this titration experiment.

What has taken place in this reaction is a neutralisation of the alkali NaOH. An Acid or alkali is neutralised when the opposite is added to it to reach the neutral pH 7. For this to happen you have to have equal number of alkali and acid molecules.

From these results you can say something about each of these acids. 12.6 cm³ of H2SO4 has the same number of molecules as 25cm³ of NaOH. Similarly 25.5cm³ of nitric acid, 28.6 of hydrochloric acid, 24.3 of ethanoic acid and 8.9 of citric acid has the same no. of molecules as 25cm³ of NaOH.

My results match with my prediction. Because I had predicted the volumes for 1M concentration of acids, and the acids were in fact of 1M concentration, you can see that the volumes I recorded are very similar to my prediction, hence I can say that my prediction was correct. As I had mentioned in my prediction, the reason for different volumes of acids used is because of its basicity. Depending on whether an acid is monobasic, dibasic or tribasic, the volume used is determined. The maximum number of H+ ions produced by one mole of pure acid is called the basicity (the number of hydrogen atoms available to react with a base)  of the acid. Generally, they are Monobasic, Dibasic, and Tribasic.

A monobasic acid produces H+ ion. A dibasic acid produces 2+ ions. A tribasic acid produces 3+ ions. The base NaOH is H-, this means it can join with 1 H+ ion to produce a neutral substance, but if an acid reacting with this base is either Dibasic or Tribasic, then only ½ or1/3, of the acid will be needed because if its dibasic then it has 2 H+ ions and if its tribasic it has 3 H+ ions. Using half the amount means only half of the ions, which will mean it will be H+ instead of 2 H+, and so it will be able to join with the H+ base. It is the same case with the tribasic acid, which instead of half the number of ions, will be a third of the number of ions to react with H+ base.

To put in theory, when we are carrying out the titration, and adding in the acid, with a dibasic acid by the time you put in half the volume of it in as compared to the base NaOH, the H- ion has joined with 1 of the 2 H+ ion’s of the acid and the solution is now neutral and so the adding of the acid is stopped. Similarly when we are using a tribasic acid, by the time we put a third of the volume of the acid as compared to the volume of the base NaOH, the end-point (neutral stage) is reached, because 1 of the 3 H+ ions joins with the H- ion of the base and so the adding of the acid is stopped. This is why, although the concentrations might be the same, it doesn’t mean that the volume used will be the same because the volume used is dependant on the basicity of the acids used.

Evaluation:

In this investigation I carried out volumetric analysis using titration. Titration is a measured way in which one substance is added to another, of set volume, until the end-point is reached. I was given 5 acids of which I had to work out their concentrations. After doing preliminary research, I found out that this can be done by a process called titration. This way of finding out the molarity is called Volumetric analysis or titrimetric analysis.

In order to do my experiment, I first had to do a pre-test which would give some kind of idea of the end-point of the titration. After this I carried out my experiment having predicted the volumes likely to be used for 1mol concentration of the acids. The volumes I obtained were very close to the predicted results, and so I know that my experiment was quite accurate.

I think that titration can be quite a hard process sometimes, especially because you have to be very careful about adding the titrant when the reaction is at the end point. It also gives way to inaccurate results because an indicator like phenolphthalein which goes colourless as soon as the neutral stage is reached and carries on being colourless no matter how acidic the solution is. This can prove to be difficult in working out the exact end-point.

The other chance of inaccuracy is human error. For the time it takes for the human eye to detect colour change or if the tap is turned so that more liquid than needed goes into the conical flask, can cause missing of the exact end-point, when there is a colour change to indicate neutral pH. This can lead to inaccurate results hence inaccurate calculations.

I think for the experiment to be a lot more accurate, instead of using indicators, it would be better to use a pH sensor. This will mean there are less chances of human error, in noticing the colour change of the solution with the indicator at its exact point of change, although the adding of the acid will still remain in the hands of the person carrying out the experiment, who can make the experiment as accurate as they possibly can.

If I were to do this investigation again, I would carry out the same method but instead of using an indicator I would use a pH sensor.