Investigation into the volume of acid needed to neutralise an alkali

In this investigation I am going to investigate the volume of different concentrations of Sulphuric acid (H2SO4) needed to neutralise 25cm� of 0.1M Sodium Hydroxide (NaOH). I will do this by using titration.

Background Knowledge

When a substance/solution is alkaline it contains hydroxide ions, OH .

When a substance/solution is acidic it contains hydrogen ions, H .

Neutralization is when the OH ions neutralize the acid by removing the H ions and turning them into water. The equation for this is:

A salt is always the end product of neutralization. The general equations for this are:

Acid+ metal oxide–>metal salt+ water

Acid+ metal hydroxide–> metal salt + water

Acid+ metal carbonate–> metal salt + water+ carbon dioxide

Acid+ metal hydrogen carbonate–> metal salt+ water + carbon dioxide

The general equation for the reaction of my experiment is:

Acid+ metal hydroxide–> metal salt+ water

And the specific equation is:

2NaOH + H2SO4 –> Na2SO4 + 2H2O

In order to be able to tell the exact point at which the solution becomes neutral, a colour change is needed and so an indicator is used. An indicator is a substance, which has a different colour in acid and alkali. For this experiment, I will be using Methyl Red. Methyl Red is yellow in alkali and bright pink in acid. I’m using this because it gives a good colour change.

To neutralize an acid and alkali you can use a process called titration. Titration enables you to find out the exact volume of acid required to neutralize a certain volume of alkali.

Using your results from titration, you can then use the equation to make a prediction.

no. of moles = concentration (mol/dm�) x volume(v) (cm�)

1000

V to convert cm� into dm�.

1000

The ionic equation of the reaction gives a mole ratio of acid (H ): alkali ( OH ) as 1:1 so at the exact point of neutralization no. moles of acid= no. moles of alkali.

Therefore the equation looks like this:

volume of acid x concentration of acid = volume of alkali x concentration of alkali

1000 1000

Looking at the equation like this enables us to make a prediction. The volume of acid is the output (dependant variable) and is the variable that we are measuring. If you choose to change the volume of alkali then you can predict that as the volume of alkali increases so will the volume of acid. This is also the same for if you choose to change the concentration of the alkali. This is because the volume of alkali and the volume of acid and the concentration of alkali and the volume of acid are on different sides of the equation; therefore if one thing increases the other must also. However if you choose to change the concentration of the acid, which is what I am doing, then as the concentration of the acid increases the volume of acid will decrease because they are on the same side of the equation.

Because I am changing the concentration of the acid I will need to dilute the acid that I am given. The standard concentration given for this experiment is 0.1M so in order to obtain smaller concentrations:

Desired concentration = Volume of acid needed

Original concentration Total volume of solution

E.g. If you want to make 100cm� of 0.08M by diluting 0.1M solution:

V needed = 0.08

100 0.1

V= 0.08 x100 =80cm�

0.1

Therefore the solution will be made up of 80cm� acid and 20cm� water.

Plan

Using this background knowledge, I am going to use titration to neutralize 25cm� of NaOH with 5 different concentrations of H2SO4. I will then work out the volume of acid taken to neutralise the solution by taking the 2nd reading from the burette away from the 1st reading. Five readings is a suitable number because it is enough readings to give me a good graph. The highest concentration I will use is 0.1M and the lowest is 0.02M. I will not record outside these limits because 0.1M is the standard concentration given and 0.02M is a weak solution so a weaker one would take too much time. I will measure the volumes to 2decimal places. The Titration results table means I will do a rough reading and then repeat each accurate reading until I get a result within 1 decimal place. The variables that will be kept the same are:

o The concentration of alkali

o The volume of alkali

The variables that will change are:

o The concentration of the acid

o The volume of acid needed

To make it a fair test I will use the same apparatus, I will do the experiment on the same day so that the conditions are the same and I will also use the same solutions so the concentrations are consistent. I will be using the same volume and concentration of alkali, which will make it a fair test.

I am going to use the following apparatus and set it up as shown:

o 50cm burette

o 25cm pipette

o White tile

o Methyl red indicator

o Conical Flask

o 1 beaker for sulphuric acid, 1 beaker for sodium hydroxide

o Clamp stand

o Mat

o 50cm Measuring cylinder

I am using the burette to measure the acid and the pipette to measure the alkali and because these are the most accurate pieces of apparatus. I am using the white tile because it shows the colour change clearly when neutralization occurs and I am using the methyl red indicator to show the colour change. The conical flask is used to store the alkali and titrate the acid into. The beakers are to store the solutions. The clamp stand is to hold the burette and the mat is where all the equipment is placed. I am using a measuring cylinder to measure the volumes of acid and water to dilute the acid. The measuring cylinder is accurate enough for its purpose and if I used anything more accurate (burette or pipette) I would get confused because I’m already using a burette and pipette. I will make the method safe by wearing goggles so no corrosive substances splash in my eyes and I will wear my hair tied back as a standard safety regulation in a science lab. I will wipe spills to prevent risk to other people as sulphuric acid and sodium hydroxide are both irritant. When pouring acid into the burette, I will bring it down to waist level so the acid doesn’t splash in my face.

Prediction

Using my background knowledge, I predict that as the concentration of the sulphuric acid increases, the volume of sulphuric acid needed to neutralize the sodium hydroxide will decrease (inverse proportion). This is because the dependant variable is on the same side of the equation as the independent variable (see background knowledge). I also predict this because the higher the concentration the more moles there are per dm� and so you don’t need as many moles to neutralize the solution. Therefore I predict that 0.02M of sulphuric acid will need 5 times the volume of sulphuric acid to neutralize the alkali than 0.1M will need because 0.02M is a 1/5th of 0.1M and will therefore have a 1/5th the number of moles per dm�.

I will display my results on a graph, which I expect to look like this:

I expect the graph to look like this because this shows that as the concentration of acid increases the volume of acid needed decreases (inverse

proportion) which is what I predicted.

The no. moles of alkali is constant because I’m keeping the volume and concentration of alkali the same. Therefore the no. moles of acid will also be constant because no. moles of alkali= no. moles of acid.

Because the no. moles of acid= volume of acid x concentration of acid

1000

the volume of acid x concentration of acid will also be constant so I could plot this against the concentration of acid added and predict it would be a straight line (constant). I would therefore expect the graph to look like this:

The unit for volume x concentration of acid is mol x 1000 because:

cm� x mol so mol x 1000

dm�

Another prediction I could make would be to predict the volume of acid needed to neutralize the solution and check the accuracy of my experiment.

I can do this because I know the volume and concentration of alkali (which equals the number of moles) and so I could work out the number of moles for the acid. That would be 1/2 the number of moles of alkali because

2NaOH: H2SO4

2:1

The volume of the alkali is 25cm� and the concentration is 0.1M and so I worked out the volume of acid needed to neutralize the solution:

No. moles of alkali

No. moles of acid

Concentration of acid mol/dm�

Volume of acid needed/ cm�

25×0.1 = 2.5x 10��

2.5x 10�� = 1.25 x10��

0.1

1.25x 10�� =v x 0.1

12.50

25×0.1 = 2.5x 10��

2.5x 10�� = 1.25 x10��

0.08

1.25x 10�� =v x 0.08

15.63

25×0.1 = 2.5x 10��

2.5x 10�� = 1.25 x10��

0.06

1.25x 10�� = v x 0.06

20.83

25×0.1 = 2.5x 10��

2.5x 10�� = 1.25 x10��

0.04

1.25x 10 �� = v x 0.04

31.25

25×0.1 = 2.5x 10��

2.5x 10�� = 1.25 x10��

0.02

1.25×10 �� = v x 0.02

62.50

To help with my method I carried out some preliminary experiments. I titrated the highest and lowest concentrations to get an idea of the range of concentrations I wanted to use. I did this by:

Method

1. Rinse the burette with a small amount of acid, then close the tap and fill above the 0.00mark. Run out the solution until there are no air bubbles and close the tap. Put the burette in the clamp stand and read the volume to 2decimal places. Record this reading as ‘1st burette reading’ in the results table.

2. Rinse the pipette with a small amount of alkali. Then, depending on the pipette filler, press and hold 1 or A and press the bulb to expel the air, then let go. Press and hold 2 or S to suck up the alkali up to above the graduation mark. Now press 3 or E to expel the alkali until the bottom of the meniscus is on the graduation mark. Hold the pipette in the conical flask and press 3 or E to expel the alkali and then add 2-3 drops of methyl red so the solution turns yellow.

3. Carry out a rough titration. Open the tap of the burette so the acid comes out into the conical flask and when the solution turns pink close the tap. Record the second reading from the burette to decimal places and then subtract the 1st reading from the second reading to get the volume of acid needed to neutralize the solution.

4. Then repeat 1-3 until you obtain 2 results within 0.1cm�. However instead of rough titration, carry out accurate titrations and use the burette to run in 1cm� less than the rough titration. Then add the acid drop by drop until the solution turns pink. Record these results.

5. Repeat 1-4 for each concentration but for concentrations other than 0.1M, dilute them before putting them in the burette. To dilute them you use the equation seen in the background knowledge and then measure the desired volumes using a measuring cylinder.

Results from Prelims

Concentration of acid

0.1M

Rough/cm�

Accurate 1/cm�

Accurate 2/cm�

2nd reading

3.00

12.75

12.85

1st reading

0.40

0.20

0.30

Volume of acid/ cm�

12.60

12.55

12.55

Concentration of acid

0.02M

Rough/cm�

Accurate 1/cm�

Accurate 2/cm�

2nd reading

75.00

78.90

77.55

1st reading

0.50

1.60

0.30

Volume of acid/ cm�

74.50

77.30

77.25

These experiments helped me to plan my method because they proved that the apparatus I’m using is suitable and that my original range of concentrations will work. I will not make any changes to my experiment as a result of these preliminaries.

Concentration of acid/ M

0.1

Rough/cm�

Accurate 1/cm�

Accurate 2/cm�

2nd reading

14.60

13.50

13.35

1st reading

2.00

0.60

0.35

Volume of acid/ cm�

12.60

12.90

13.00

Concentration of acid/ M

0.08

Rough/cm�

Accurate 1/cm�

Accurate 2/cm�

2nd reading

16.65

16.10

16.15

1st reading

0.60

0.20

0.45

Volume of acid/ cm�

16.05

15.90

15.70

Concentration of acid/ M

0.06

Rough/cm�

Accurate 1/cm�

Accurate 2/cm�

2nd reading

22.40

21.75

21.40

1st reading

0.55

0.75

0.50

Volume of acid/ cm�

21.85

21.00

20.90

Concentration of acid/ M

0.04

Rough/cm�

Accurate 1/cm�

Accurate 2/cm�

2nd reading

33.70

33.00

32.70

1st reading

0.40

0.50

0.20

Volume of acid/ cm�

33.30

32.50

32.50

Concentration of acid/ M

0.02

Rough/cm�

Accurate 1/cm�

Accurate 2/cm�

2nd reading

64.70

63.70

64.10

1st reading

0.70

0.20

0.50

Volume of acid/ cm�

64.00

63.50

63.65

Results for experiment

To find the volume of acid, I subtracted the 1st reading from the 2nd reading.

To calculate the volumes of acid and water I would need for different concentrations of acid (see background knowledge- diluting)

Concentration of acid mol/dm�

Volume of water/ cm�

Volume of sulphuric acid/ cm�

0.1

0

100

0.08

20

80

0.06

40

60

0.04

60

40

0.02

80

20

Conclusion

From my 1st graph, I can conclude that as the acid concentration increases, the volume of acid needed to neutralize the solution decreases, which is what I predicted. This also matches my scientific background knowledge. The volume of acid needed to neutralize the solution has decreased as the acid concentration has increased because there are more moles per dm� and so you don’t need as many moles to neutralize the solution. Also, it matches my background knowledge in that the volume of acid and the concentration of acid are on the same side of the equation so as one increases the other will decrease.

My graph shows inverse proportion and is a curve, which is what I also predicted.

In my prediction I said that the volume of 0.02M sulphuric acid needed to neutralize the alkali would be 5 times the amount needed when using 0.1M of sulphuric acid. The accurate volumes for 0.02M were 63.50cm� and 63.65cm� and the accurate volumes for 0.1M were 12.90cm� and 13.00cm�, which are roughly 1/5th of 63.50 and 63.65. Therefore the prediction I made was very close.

I predicted that the volume of acid x the concentration of acid would be constant and therefore the graph would be a straight line. My 2nd graph shows a straight line and so my prediction was correct. There are only 2 anomalous results on this graph.

I also predicted the volumes of acid needed to neutralize the solution before the experiment. I predicted 12.50cm� for 0.1M and the results were 12.90cm�and 13.00cm�, 15.63 for 0.08M and the results were 15.90cm� and 15.70 cm�, 20.83cm� for 0.06M and the results were 21.00cm� and 20.90 cm�, 31.25cm� for 0.04M and the results were 32.50cm� and finally 62.50 for 0.02M and the results were 63.50cm� and 63.65cm�. These results are very close to the predictions with the results for 0.08M only being out by 0.07cm�.

Evaluation

My experiment went well as I only had 2 anomalous results. By anomalous results I mean results that don’t fit in with the other readings and aren’t as close to the line of best fit.

Those results were 0.1M and 0.04M on the 2nd graph. These anomalies could be due to using a measuring cylinder for diluting acids, which might not have been accurate enough, me reading the burette incorrectly- not from the bottom of the meniscus each time, therefore making it an unfair test. I could also have noted the volume of acid added incorrectly- written down before the exact end-pint was reached and the indicator hadn’t completely turned pink.

However, even if I had done all of the above accurately, I may still have got errors because if the volume of acid added is small, there is a larger percentage of error than larger volumes of acid.

For example, when you do titration you repeat the experiment until you get two results within +/- 0.2cm�. Using my first anomalous result, the percentage of error will be 0.2 divided by the volume of acid added:

0.2 x 100 = 1.6%

12.90

This is the smaller volume. For the larger volume the percentage of error would be for example:

0.2 x 100 = 0.31%

63.50

Therefore I am more likely to get anomalous results for the smaller volumes of acid added because there is a larger percentage of error.

My results were mostly accurate due to using the same solutions, which meant that the concentrations were consistent and using the same apparatus, which meant my experiment, was a fair test.

My results on my 1st graph show the accuracy of the experiment because they are very close to the line of best fit.

I thought my procedure was suitable. If I did the experiment again I would do repeat readings for the anomalous results I got. I would use a pipette or burette instead of a measuring cylinder to measure volumes when making up concentrations of acid because they would be more accurate and so the results would be more accurate. I would also give myself more time to do the experiment so I wouldn’t feel like I was rushing and therefore make silly mistakes, which could contribute to anomalous results. I might use a different indicator to see if this made a difference to my results. Therefore I would change the equipment, not the method. These changes would make my results more reliable and accurate and therefore make my experiment better.

To get additional evidence for my conclusion I could use smaller acid concentration intervals e.g. 0.1M, 0.95M, 0.9M. This would mean I had more points to plot on the graph and so the curve would be more definite and therefore emphasise my conclusion that as the acid concentration increases, the volume needed to neutralize the solution decreases. I could also do more concentrations (0.07M, 0.05M, 0.03M) to make the curve more definite or even higher concentrations to see if the curve leveled off on the graph e.g. 2M, 3M, 4M. I could use another acid, for example nitric acid, to see if my ionic equation for neutralization is correct.