15 cm³ + 15 cm³ ?
I can use the value 58 KJ or 58000 J as it is the value for the heat of neutralization per mole and it is fixed to find the value of energy released using 15 cm³ of alkali with 15 cm³ of acid.
2000 cm³ 58000 J
30 cm³ ?
30 x 58000 / 2000 =
870 J
30 cm³
2000 cm³ 58000 J
(30 cm³ + 30 cm³) 60 cm³ ?
60 x 58000 / 2000 =
1470 J
45cm³
2000 cm³ 58000 J
(45 cm³ + 45 cm³) 90 cm³ ?
90 x 58000 / 2000 =
2610 J
If the results of the experiment match those values then I know that I carried the experiment successfully. I also decided that I plot a graph of the results I am going to find and the graph below shows what I expect my graph to be:
From the graph we could see that if the acid and the alkali were exactly of the same concentration then 15 cm³ of hydrochloric acid would require exactly 15 cm³ of sodium hydroxide to neutralize it. As the acid is added a little at a time more and more heat is evolved. the highest temperature is recorded at the exact and the point. This maybe a little lower or higher. After the neutralization the temperature gradually drops because the excess acid absorbs the heat added.
After finishing finding the heat evolved from each volume I decided to calculate the rise of temperature for each volume by the formulae
Q=mc∆θ
Q = Heat Evolved.
M = for the mass or the volume used.
C = is the specific heat of capacity of water which is the same for the acid and the alkali (4.2 KJ)
∆θ = is the rise in the temperature
and then see if my results are going to match the predicted rise in the temperature.
Below is the calculation that I am going to carry theoretically and then when carrying on the experiment I will see if the results are going to match.
15 cm³
Q = m x c x ∆θ
870 = (15 cm³ + 15cm³) x 4.2 x ∆θ
∆θ = 6.9º C
30 cm³
Q = m x c x ∆θ
1470 = (30 cm³ +30 cm³) x 4.2 x ∆θ
∆θ = 6.9ºC
45 cm³
Q = m x c x ∆θ
2160 = (45 cm³ + 45 cm³) x 4.2 x ∆θ
∆θ = 6.9ºC
from the values above we can observe an interesting feature and that is thought the volume is doubled and tripled the rise in temperature remains the same and this can be explained that even when the volume is doubled or tripled the temperature remains the same because the heat given is absorbed by a larger volume of solution. In another words the temperature is distributed on a larger volume. The representation below would explain what I said above in a simple manner.
When we double that is what happens
30 cm³ + 30 cm³ 2 X J (60 cm³)
60 / 2 = 30 cm³ which is same
The rise in temperature can be represented in two graphs one which is a theoretical graph just for the understanding and the other is the one which is going to result due the obtaining of results of the experiment which is going to show the second graph when plotted.
In the graph, the representation of the rise in temperature means change in temperature – this will also be used in the energy formula that I’ll use to calculate what I expect the values for heat of neutralization and energy to be.
Also from the above we can deduce two observations:
- That the heat evolved is directly proportional to the volume of acid but only up to the point of neutralization.
- After that, the temperature becomes inversely proportional to the volume of acid.
After finishing explaining all of this I am going now the list the apparatus and how I am going to carry on the coursework.
Apparatus
- Burette : Which is going to be used to measure the in the correct amounts as it offers a good reading
- Plastic cup: This is where the reaction is going to take place and the reason I choose it is because plastics are poor conductors of heat thus I have time to record the temperature.
- Measuring cylinder: This is where I am going to measure the alkali used for the reaction
- Thermometer: It is going to be used for recording the initial temperature and the rise in the temperature.
- Stirrer: Is going to be used so that on can mix the mixture and make sure that everything is has reacted.
-
Clamps: I am going to use it for support of the suspended apparatus.
Method
First I am going to show the experiment setup as shown below to make the understanding of how I am going to use it easy:
Carrying on the experiment is not hard, first I am going to measure 15 cm³ of NaOH in the measuring cylinder then pour them in the plastic cup. Then I am going to fill the burette up to the zero mark of it (N.B When carrying on the process of filling and measuring I am going to make sure that lower meniscus is perpendicular to the line of my sight) I am going then to record the temperature of the alkali and note it down. Then I am going to let the acid from the burette 5 cm³ at a time. Thus finding the temperature first when the total is 20 cm³ then I am going to throw the mixture and then I will note the temperature when the total is 25 cm³ and then note the temperature and so on. When adding the acid I am going to stir it with the stirrer to make sure everything reacted. I am going to follow this method for the other two sets of readings which are 30 cm³ and 45 cm³. I am going to obtain 6 sets of readings with difference of 5 cm³ between each of them.
Factors to be kept constant:
The volume of the NaOH is going to be kept constant only the volume of acid which going to be kept constant. Other factors like the concentration, temperature and the pressure are to be kept constant so that the only varying constant is the volume of the acid.
Precautions when carrying on the experiment:
- First of all I am going to clean all the apparatus so as to make sure that there is no undesirable substances in the apparatus.
- Then as I said above I have to make sure the line of my sight is correct.
- I have to make sure that I do not spill any acid or alkali on my hands so that I will not burn them or ware gloves.
I have shown below the table in which I am going to record the observation.
Analysis
After finishing drawing the graphs and carrying out the experiment and tabulating down the results. Now I am going to discuss the graph and the pattern in the graph. I am also going to check and see if the results are going to match what I got are going to match the values I got in the planning section which I got by the theoretical equations. And the trends and the patterns I predicted in the planning section.
Temperature rise
First of all I have got to calculate the heat rise from the graph and how much acid was needed in the following table, which I have created:
As we can from the above table it is clearly shown that my experiments did not match the expected results that I have shown in the planning where I expected all of them to have the same temperature rise which was 6.9ºC. the only thing which came true was that the volume of HCl needed to neutralize the 15 cm³ of NaOH was the same which is 15 cm³. All the other volumes were the same also. I am also aware that there are different reasons behind this which are listed below:
- The reason due to the difference in the temperature rise is may be because I made mistakes while taking down the temperature from the thermometer or noting down the temperature in the note book
- It may be also due to me waiting for too long to note down the temperature or waiting less time thus not giving the thermometer a chance to adjust to the temperature of the solution
- The solution could have also lost some of the heat to the surrounding while recording the temperature.
- It could have been also due to that my hand raised the temperature as the cup absorbed it.
- A parallax error could have occurred in the form that my line of sight was not perpendicular to the thermometer scale.
- The surrounding environment could have affected the experiment like by switching on the fan or air condition.
I decided to present the above data in the form of a bar graph so that I t would be easier to comprehend the data.
Heat energy evolved:
Now that after finishing investigating the temperature rise I going now to investigate the heat evolved from the graph. I am going to find the values using the formula Q=mc∆θ. I am going to start in order:
15cm³
Q = m x c x ∆θ
Q = (28/1000) x 4.2 x 6
Q = 0.7056 KJ of energy or 705.6 J
30 cm³
Q = m x c x ∆θ
Q = (55.5/1000) x 4.2 x 6
Q = 1.3986 KJ of energy or 1398.6 J
45 cm³
Q = m x c x ∆θ
Q = (81/1000) x 4.2 x 6.5
Q = 2.0412KJ of energy or 2041.2 J
So after finding the values I am going to tabulate the above results in the following table:
In the planning I mentioned that the heat evolved would be directly proportional to the total volume of the solution which means that when a doubling occurs in the volume the evolved should also double and when it is tripled the heat evolved is also tripled and I am going to see if that is correct or no by checking it below:
0.756 x 2 = 1.512 while I got 1.3986 which roughly near to the value
0.756 x 3 = 2.268 while I got 2.0412 which is some what lower but could be considered as tripling.
I decide to draw a bar graph to represent the doubling and the tripling as shown below:
Heat of heat neutralization
After investigating the temperature rise and the heat evolved I am now going to investigate the heat of neutralization which needs the to have the values I calculated above. To achieve that I have got to calculate the number of moles in each solution first by the formula n =cv
15 cm³
n = 1 x 15 /1000
n = 0.015 moles
30 cm³
n = 1 x 30/1000
n = 0.030 moles
45 cm³
n = 1 x 45/1000
n = 0.045 moles
Now I am going to find the heat of neutralization:
0.015 moles 705.6 J
1 mole ?
705.6 / 0.015 = 47040 J
0.030 moles 1398.6J
1 mole ?
1398.6 / 0.030 = 46620 J
0.045 moles 2041.2 J
1 mole ?
2041.2 / 0.045 = 45360 J
I tabulated the above results in the following table:
In the planning section of the coursework I said that the heat of neutralization is going to be the same but it appears to the previous mistakes committed while doing the experiment it also affected the heat of neutralization value and thus we could see one mistake could to lead to other mistakes.
I also decide to draw a bar chart for the results :
Conclusion
Coming to the conclusion part, I might as well say that not all of the predictions I mentioned in the planning section came as accurate as they should have been. Most of the predictions I mentioned came roughly true. At the end we see that the doubling and the tripling of the heat evolved occurred. And that most of the results are satisfying.
Evaluate
After finishing my work and the experiment I am going to evaluate them now and comment on the obstacles that faced me and in what this experiment could have been improved to carry it successfully and obtain accurate results.
Commenting on the method:
On the overall the experiment was easy to carry and the method used was good to obtain accurate result if one is fast in action and careful.
- While doing the experiment one of the problems, which faced us, was the purity and concentration of acid and alkali.
- Not all of the acid and alkali was poured in the plastic cup due to some of the drops sticking to the side of the burette and the measuring cylinder.
- The temperature loss to the surrounding from the plastic cup thus not giving accurate results.
All of these problems could have been solved by the following list and it also contains ways on improving the experiment also:
- The loosing of temperature to the surrounding could have been overcome by lagging the plastic cup by a lagging material so to stop the loss.
- The experiment could have been done twice and the average or the mean taken to improve the results
- A digital thermometer could have been used instead of the analogue one so a more accurate results could be achieved
- A pipette could have been used instead of the measuring cylinder to achieve more accuracy.
- The new technology could also be used in this case where the sensors could be used to measure the temperature where the computer records the temperature at an interval of 10 seconds, which could help the one to achieve a highest possible accuracy that could be achieved.
The experiment could have been expanded in various possible directions as seen below
- The experiment could have been used to investigate different concentration thus having a wider range to improve the results.
- The experiment could have been made using different kind of acids and alkalis. Strong acids and strong alkalis could be used as well as the weak acids and weak alkalis.
- An indicator could have been used so that we can record at what volume the acid neutralizes the alkali and see if it matches with the results of the graph
Obtaining Evidence
I carried now the experiment and below is the results tabulated:
