The Concentration Of An Unknown Solution Of Hydrochloric Acid
The aim of this experiment was to calculate the concentration of an unknown solution of Sodium Hydroxide (NaOH) by titrating it with Hydrochloric Acid (HCl) and to identify any possible sources of error. A colour change from pink to yellow signified the end of each individual titre as the NaOH had been neutralised.
Experimental:
Method:
Firstly two solutions were prepared to show the colour of the Phenol Red indicator in acid and alkaline conditions. These colours were then used to determine the end point of each titre. In the experiment 0.100M HCl (acid) was added, from a burette, to a solution of 25cm3 of NaOH (base) and several drops of Phenol red indicator. This was done until the base had been neutralised (signified by a colour change from pink to yellow). A magnetic stirrer was used to ensure even mixing of the acid being added.
Problems: …show more content…
Throughout the experiment several problems were encountered and accordingly dealt with.
Firstly, it took time for the tap on the burette to be closed, so even though the titration was complete more acid was being added to the conical flask, this resulted in the reading on the burette higher than usual. Also each titration was carried out by a different person, the reaction times of different individuals will vary and so the tap would be closed at a faster/slower rate depending on the individual. Additionally, different individuals may spot the colour change at different times. Once this problem was noticed the remaining titrations were carried out by the same person to have more consistency in the experiment. Furthermore parallax became an issue when measurements were taken from the burette, if the reading were took at different angles then the values appeared to be different. To ensure that effects of parallax were minimised readings were taken standing directly in front of the burette and verified by another
person.
Calculations:
At the point of neutralisation the moles of HCl are equal to the moles of NaOH. This means if the moles of HCl are calculated then the moles of NaOH in 25cm3 is also known.
Moles=Concentration ×Volume (1)
Using equation (1) the moles in 20cm3 of 0.5M HCl can be calculated:
Moles=20.(1/(1000〖dm〗^3 ))×0.5
Moles = 0.001 mol
To calculate the average titre from the results of several titration the following formula can be used:
Average Titre=(Sum of all titrations)/(Number of Titrations) (2)
This means that if the results of three titrations are known to be 15.4cm3, 16.2cm3 and 16.4cm3 then the average titre can be calculate using equation (2).
Average Titre=(15.4+16.2+16.4)/3
Average Titre=16.0cm3
Results:
Table 1- showing results of titrations carried out Initial Burette Reading (cm3) (±0.1cm3) Final Burette Reading (cm3) (±0.1cm3) Final Reading – Initial Reading Titre (cm3)
(±0.3cm3) Average titre
Approximation 0.1 11.2 11.1 11.1
Precise Titration 1 11.2 22.2 11.0 11.0
Precise Titration 2 22.2 33.2 11.0 11.0
Precise Titration 3 33.2 44.3 11.1 11.1
Mean Titre 11.0*
*Only precise titration values used to calculate the average titre
At the end point of the reaction a colour change from pink to yellow was witnessed. This colour change showed that the base had been neutralised by the acid that was being added to it. The overall equation for this process is:
NaOH(aq) + HCl(aq) →┴( ) H2O(l) + NaCl(aq) (3)
Equation (3) shows that NaOH and HCl are stoichiometrically equivalent, therefore at the endpoint of the reaction the number of moles of NaOH are equal to the number of moles of HCl. This piece of information can be used to calculate the molarity of NaOH. The moles of HCl can be calculated using equation (1):
MolesHCL = 0.100 M × 11/1000dm3
MolesHCl = 0.0011 mol
The moles of HCl are equal to the moles of NaOH. This means that there are 0.0011 moles of NaOH in 25cm3. The moles and volume of NaOH are known, so equation (1) can be rearranged to calculate the molarity of NaOH.
Concentration=Moles/Volume
Concentration=0.0011mol/(25/1000 dm^3)
Concentration=0.0440 M
Discussion:
There is always an uncertainty in results obtained from laboratory experiments. This is a direct result of errors that occur during the experiment. These errors could be random or systematic. The possible effect of certain errors will be discussed in further detail.
From the mean titre it can be seen that it took 11.0cm3, on average, to reach the reaction end point. However there is an uncertainty of ±0.9cm3 in the mean titre. This is due to the fact that there is an uncertainty of ±0.3cm3 in each titration (±0.2cm3 from the two burette readings and ±0.1cm3 from the pipette which was used to measure out 25cm3 of NaOH). The range of the precise titration values in Table 1 is 0.1cm3, there is a significant difference between the range of the precise titrations and the mean uncertainty. This difference could be the result of several errors in the experiment, such as misreading the values on the burette. The repeatability of a measurement is assumed to be half the range. Repeatability can be represented as a percentage of the mean titration:
(0.1/2)/11.00×100=±0.45%=percentage repeatability
Due to the uncertainty in the measurements taken during the titration there will also be an uncertainty in the calculated concentration of NaOH. ‘The percentage repeatability of the NaOH molarity is equal to the percentage repeatability of the titration value’ [1], therefore the percentage repeatability of the NaOH molarity is ±0.45%. The absolute repeatability of NaOH molarity can then be calculated to be ±0.000198 M.
If the HCl solution was actually 0.105M this would lead to a systematic error in all the calculations that involved the concentration of HCl. The moles of HCl would instead be calculated as 0.105 M×11/1000dm3, giving a result of 0.001155 moles. This incorrect value of HCl moles would then be assumed as the number of moles of NaOH and used to calculate NaOH molarity. Molarity=0.001155/(25/1000 dm^3) = 0.0462 M. The size of this systematic error can be calculated as:
(0.0462 M-0.044 M)/(0.044 M)×100=5% error
If the glassware is not calibrated at the proper position then it could be a source of systematic error. If the pipette has an accuracy of ±0.2cm3, then the percentage precision of the initial 25cm3 reading can be calculated.
Precision of pipette = 0.2/25×100=±0.80%
Precision of burette = 0.1/11×100=±0.91%
If the volumes of NaOH and HCl are not precise then the inaccuracy will carried into the value for concentration of NaOH, and so the greatest percentage precision with which the NaOH concentration can be reported is 98.29% or ±1.71%.
It is know that a 2°C will cause a change of 0.002% in the volume of Pyrex equipment. Then the value of HCl in each titration would have an error of ±0.002%, the 25cm3 of NaOH would also have a percentage error of ±0.002%. The maximum error would arise when HCl volume increases by 0.002% and the NaOH volume decreases by 0.002%. So the new mean titre would be 11.033554cm3 ((2(11.0+11.0×0.002%)+(11.1+11.1×0.002%))/3) and the new volume of NaOH would be 24.995cm3. If we use these values to calculate the molarity of NaOH we obtain a result of 0.0441 M. This means the percentage error from a 2°C would be (0.0441-0.0440)/0.0440×100, this comes out to a percentage error of 0.23%. An error of this magnitude cannot be neglected as it is more than 10% of the greatest error being considered in the experiment. The greatest error being considered was ±0.45%, which was the percentage repeatability in the NaOH concentration value.
There were several sources of error in this experiment. Firstly, distilled water from used to clean the some of the glassware and the magnetic stirrer could have diluted the NaOH solution. Another possible error is that over titration could have occurred, this could have happened if the colour change was not spotted immediately. Furthermore, atmospheric CO2 could have been absorbed into the NaOH solution, this would make the solution less basic and so the colour change would occur earlier than it should have. Also if the bottom of the meniscus was in between two markings, the volume of the solution could not be accurately measured.
Improvements could be made to the experiment procedure to minimise errors and so obtain more accurate results. The use of a titration thief [2] could be implemented; this would help counter the problem of over titration. A small volume of the NaOH solution would be removed and then the titration carried out as normal. Once the end point of the titration is reached the removed NaOH solution is reintroduced making the solution basic again. The solution can now be neutralised with extra care, and the effects of over titration are minimised.
Conclusion:
It took an average titre of 11.0cm3 to neutralise the NaOH solution, this was used to calculate the molarity of the NaOH solution to be 0.0440 M. The main sources of error identified were uncertainty from the Pyrex glassware and from systematic errors in the experiment. If the experiment was repeated a lot of the errors could be eliminated by taking extra care and confirming reading from the glassware with other people. However there will always be some uncertainty involved with the experiment, this is because systematic errors are hard to detect. For example the measuring equipment may not be calibrated correctly or the temperature may fluctuate from the calibration temperature required for accurate measurements.
References:
1) Christensen, P.A. (2014). CME 1020 Stage 1 Chemistry and Chemical engineering laboratory manual.p56. Newcastle University
2) Kolthoff, I. and Belcher, R. (1957). Volumetric Analysis. New York: Interscience
Appendix:
Approximate Titration Value
Burette reading after the solution colour has changed (step 4) = 11.2cm3
Burette reading initially = 0.1cm3
Approximate titration value = 11.1cm3 Precise Titration Value
Precise titration 1
Burette reading after the solution colour has changed = 22.2cm3
Uncertainty in burette reading = ± 0.1cm3
Burette reading initially = 11.2cm3
Uncertainty in initial burette reading = ± 0.1cm3
Difference in burette readings (Titration value 1) = 11.0cm3
Uncertainty in difference between set 1 burette readings = ± 0.2cm3 Precise titration 2
Burette reading after the solution colour has changed = 33.2cm3
Uncertainty in burette reading = ± 0.1cm3
Burette reading initially = 22.2cm3
Uncertainty in initial burette reading = ± 0.1 cm3
Difference in burette readings (Titration value 2) = 11.0cm3
Uncertainty in difference between set 2 burette readings = ± 0.2cm3 Precise titration 3
Burette reading after the solution colour has changed = 44.3cm3
Uncertainty in burette reading = ±0.1cm3
Burette reading initially = 33.2cm3
Uncertainty in initial burette reading = ± 0.1cm3
Difference in burette readings (Titration value 3) = 11.1cm3
Uncertainty in difference between set 3 burette readings = ± 0.2cm3
Mean of the three titration values and the mean uncertainty (Mean ± mean uncertainty) = 11.0cm3 ± 0.9cm3
Range of precise titration values (highest - lowest) = 0.1cm3
Questions: Absolute error is the difference between a measured value and the known correct value. It is just the physical difference between the measurement and the correct value. For example, if the length of a wire is measured to be 5cm but the actual length is 3cm, then the absolute error will be 2cm.
Relative error allows the error in two different measurements to be compared. Relative error relies upon absolute error. This is because it is a ratio between the absolute error and the actual value. As it is a ratio there are no units and so the relative error for any two measurements can be compared. Relative error = absolute error/actual value. Phenol red is used as an indicator as it has a transition point of approximately 7.5. The target pH in a titration is 7 and the transition point of Phenol red is close to this so it can be used for the experiment. At its transition point Phenol red will undergo a colour change, which will signify the end of the titration. Phenol red is pink in basic conditions but yellow in acidic conditions, this colour change can be easily distinguished making Phenol red a good choice as an indicator for the experiment. The burette is not filled at above eye level for safety reasons; if it is filled above eye level the liquid could easily splash onto the users face and eyes. A lot of the solution used in titrations are irritants and so it is best to avoid a situation in which a person’s skin comes into contact with them. Also if the burette is filled above eye level then it is difficult to properly judge what level the liquid is at. If the magnetic stirrer is set at a high level it would result in heating the solution up. The movement of the stirrer would increase the energy of the molecules in the solution and so heat it up. This additional heat could have an effect on the experiment which would may not be accounted for, leading to more uncertainty in the experiment. Also the increase in temperature could affect the glassware as previously mentioned. If the stirrer is set at a high level then it would cause excessive splashing of the solution in the conical flask. This means that some of The NaOH solution will remain on the side of the glassware and not be involved in the titration. Therefore a lower volume of acid will be needed to neutralise the base, causing inaccuracy in the results. The end point of a titration is when the base or acid in the conical flask below the burette has been neutralised. An indicator is usually used in titration to show when the end point has been reached. The chosen indicator solution will be a different colour in acidic conditions than in basic conditions. So when the end point of the titration is approached the indicator will change in colour to show this. Determining the exact end point from the colour change is difficult, this is because the initial colour change may only be faint and so easily missed. So instead a digital pH monitor could be used in conjunction with the Phenol red indicator. The pH monitor would show the exact end point and the colour change of the Phenol red indicator would confirm this.
Experimental:
Method:
Firstly two solutions were prepared to show the colour of the Phenol Red indicator in acid and alkaline conditions. These colours were then used to determine the end point of each titre. In the experiment 0.100M HCl (acid) was added, from a burette, to a solution of 25cm3 of NaOH (base) and several drops of Phenol red indicator. This was done until the base had been neutralised (signified by a colour change from pink to yellow). A magnetic stirrer was used to ensure even mixing of the acid being added.
Problems: …show more content…
Throughout the experiment several problems were encountered and accordingly dealt with.
Firstly, it took time for the tap on the burette to be closed, so even though the titration was complete more acid was being added to the conical flask, this resulted in the reading on the burette higher than usual. Also each titration was carried out by a different person, the reaction times of different individuals will vary and so the tap would be closed at a faster/slower rate depending on the individual. Additionally, different individuals may spot the colour change at different times. Once this problem was noticed the remaining titrations were carried out by the same person to have more consistency in the experiment. Furthermore parallax became an issue when measurements were taken from the burette, if the reading were took at different angles then the values appeared to be different. To ensure that effects of parallax were minimised readings were taken standing directly in front of the burette and verified by another
person.
Calculations:
At the point of neutralisation the moles of HCl are equal to the moles of NaOH. This means if the moles of HCl are calculated then the moles of NaOH in 25cm3 is also known.
Moles=Concentration ×Volume (1)
Using equation (1) the moles in 20cm3 of 0.5M HCl can be calculated:
Moles=20.(1/(1000〖dm〗^3 ))×0.5
Moles = 0.001 mol
To calculate the average titre from the results of several titration the following formula can be used:
Average Titre=(Sum of all titrations)/(Number of Titrations) (2)
This means that if the results of three titrations are known to be 15.4cm3, 16.2cm3 and 16.4cm3 then the average titre can be calculate using equation (2).
Average Titre=(15.4+16.2+16.4)/3
Average Titre=16.0cm3
Results:
Table 1- showing results of titrations carried out Initial Burette Reading (cm3) (±0.1cm3) Final Burette Reading (cm3) (±0.1cm3) Final Reading – Initial Reading Titre (cm3)
(±0.3cm3) Average titre
Approximation 0.1 11.2 11.1 11.1
Precise Titration 1 11.2 22.2 11.0 11.0
Precise Titration 2 22.2 33.2 11.0 11.0
Precise Titration 3 33.2 44.3 11.1 11.1
Mean Titre 11.0*
*Only precise titration values used to calculate the average titre
At the end point of the reaction a colour change from pink to yellow was witnessed. This colour change showed that the base had been neutralised by the acid that was being added to it. The overall equation for this process is:
NaOH(aq) + HCl(aq) →┴( ) H2O(l) + NaCl(aq) (3)
Equation (3) shows that NaOH and HCl are stoichiometrically equivalent, therefore at the endpoint of the reaction the number of moles of NaOH are equal to the number of moles of HCl. This piece of information can be used to calculate the molarity of NaOH. The moles of HCl can be calculated using equation (1):
MolesHCL = 0.100 M × 11/1000dm3
MolesHCl = 0.0011 mol
The moles of HCl are equal to the moles of NaOH. This means that there are 0.0011 moles of NaOH in 25cm3. The moles and volume of NaOH are known, so equation (1) can be rearranged to calculate the molarity of NaOH.
Concentration=Moles/Volume
Concentration=0.0011mol/(25/1000 dm^3)
Concentration=0.0440 M
Discussion:
There is always an uncertainty in results obtained from laboratory experiments. This is a direct result of errors that occur during the experiment. These errors could be random or systematic. The possible effect of certain errors will be discussed in further detail.
From the mean titre it can be seen that it took 11.0cm3, on average, to reach the reaction end point. However there is an uncertainty of ±0.9cm3 in the mean titre. This is due to the fact that there is an uncertainty of ±0.3cm3 in each titration (±0.2cm3 from the two burette readings and ±0.1cm3 from the pipette which was used to measure out 25cm3 of NaOH). The range of the precise titration values in Table 1 is 0.1cm3, there is a significant difference between the range of the precise titrations and the mean uncertainty. This difference could be the result of several errors in the experiment, such as misreading the values on the burette. The repeatability of a measurement is assumed to be half the range. Repeatability can be represented as a percentage of the mean titration:
(0.1/2)/11.00×100=±0.45%=percentage repeatability
Due to the uncertainty in the measurements taken during the titration there will also be an uncertainty in the calculated concentration of NaOH. ‘The percentage repeatability of the NaOH molarity is equal to the percentage repeatability of the titration value’ [1], therefore the percentage repeatability of the NaOH molarity is ±0.45%. The absolute repeatability of NaOH molarity can then be calculated to be ±0.000198 M.
If the HCl solution was actually 0.105M this would lead to a systematic error in all the calculations that involved the concentration of HCl. The moles of HCl would instead be calculated as 0.105 M×11/1000dm3, giving a result of 0.001155 moles. This incorrect value of HCl moles would then be assumed as the number of moles of NaOH and used to calculate NaOH molarity. Molarity=0.001155/(25/1000 dm^3) = 0.0462 M. The size of this systematic error can be calculated as:
(0.0462 M-0.044 M)/(0.044 M)×100=5% error
If the glassware is not calibrated at the proper position then it could be a source of systematic error. If the pipette has an accuracy of ±0.2cm3, then the percentage precision of the initial 25cm3 reading can be calculated.
Precision of pipette = 0.2/25×100=±0.80%
Precision of burette = 0.1/11×100=±0.91%
If the volumes of NaOH and HCl are not precise then the inaccuracy will carried into the value for concentration of NaOH, and so the greatest percentage precision with which the NaOH concentration can be reported is 98.29% or ±1.71%.
It is know that a 2°C will cause a change of 0.002% in the volume of Pyrex equipment. Then the value of HCl in each titration would have an error of ±0.002%, the 25cm3 of NaOH would also have a percentage error of ±0.002%. The maximum error would arise when HCl volume increases by 0.002% and the NaOH volume decreases by 0.002%. So the new mean titre would be 11.033554cm3 ((2(11.0+11.0×0.002%)+(11.1+11.1×0.002%))/3) and the new volume of NaOH would be 24.995cm3. If we use these values to calculate the molarity of NaOH we obtain a result of 0.0441 M. This means the percentage error from a 2°C would be (0.0441-0.0440)/0.0440×100, this comes out to a percentage error of 0.23%. An error of this magnitude cannot be neglected as it is more than 10% of the greatest error being considered in the experiment. The greatest error being considered was ±0.45%, which was the percentage repeatability in the NaOH concentration value.
There were several sources of error in this experiment. Firstly, distilled water from used to clean the some of the glassware and the magnetic stirrer could have diluted the NaOH solution. Another possible error is that over titration could have occurred, this could have happened if the colour change was not spotted immediately. Furthermore, atmospheric CO2 could have been absorbed into the NaOH solution, this would make the solution less basic and so the colour change would occur earlier than it should have. Also if the bottom of the meniscus was in between two markings, the volume of the solution could not be accurately measured.
Improvements could be made to the experiment procedure to minimise errors and so obtain more accurate results. The use of a titration thief [2] could be implemented; this would help counter the problem of over titration. A small volume of the NaOH solution would be removed and then the titration carried out as normal. Once the end point of the titration is reached the removed NaOH solution is reintroduced making the solution basic again. The solution can now be neutralised with extra care, and the effects of over titration are minimised.
Conclusion:
It took an average titre of 11.0cm3 to neutralise the NaOH solution, this was used to calculate the molarity of the NaOH solution to be 0.0440 M. The main sources of error identified were uncertainty from the Pyrex glassware and from systematic errors in the experiment. If the experiment was repeated a lot of the errors could be eliminated by taking extra care and confirming reading from the glassware with other people. However there will always be some uncertainty involved with the experiment, this is because systematic errors are hard to detect. For example the measuring equipment may not be calibrated correctly or the temperature may fluctuate from the calibration temperature required for accurate measurements.
References:
1) Christensen, P.A. (2014). CME 1020 Stage 1 Chemistry and Chemical engineering laboratory manual.p56. Newcastle University
2) Kolthoff, I. and Belcher, R. (1957). Volumetric Analysis. New York: Interscience
Appendix:
Approximate Titration Value
Burette reading after the solution colour has changed (step 4) = 11.2cm3
Burette reading initially = 0.1cm3
Approximate titration value = 11.1cm3 Precise Titration Value
Precise titration 1
Burette reading after the solution colour has changed = 22.2cm3
Uncertainty in burette reading = ± 0.1cm3
Burette reading initially = 11.2cm3
Uncertainty in initial burette reading = ± 0.1cm3
Difference in burette readings (Titration value 1) = 11.0cm3
Uncertainty in difference between set 1 burette readings = ± 0.2cm3 Precise titration 2
Burette reading after the solution colour has changed = 33.2cm3
Uncertainty in burette reading = ± 0.1cm3
Burette reading initially = 22.2cm3
Uncertainty in initial burette reading = ± 0.1 cm3
Difference in burette readings (Titration value 2) = 11.0cm3
Uncertainty in difference between set 2 burette readings = ± 0.2cm3 Precise titration 3
Burette reading after the solution colour has changed = 44.3cm3
Uncertainty in burette reading = ±0.1cm3
Burette reading initially = 33.2cm3
Uncertainty in initial burette reading = ± 0.1cm3
Difference in burette readings (Titration value 3) = 11.1cm3
Uncertainty in difference between set 3 burette readings = ± 0.2cm3
Mean of the three titration values and the mean uncertainty (Mean ± mean uncertainty) = 11.0cm3 ± 0.9cm3
Range of precise titration values (highest - lowest) = 0.1cm3
Questions: Absolute error is the difference between a measured value and the known correct value. It is just the physical difference between the measurement and the correct value. For example, if the length of a wire is measured to be 5cm but the actual length is 3cm, then the absolute error will be 2cm.
Relative error allows the error in two different measurements to be compared. Relative error relies upon absolute error. This is because it is a ratio between the absolute error and the actual value. As it is a ratio there are no units and so the relative error for any two measurements can be compared. Relative error = absolute error/actual value. Phenol red is used as an indicator as it has a transition point of approximately 7.5. The target pH in a titration is 7 and the transition point of Phenol red is close to this so it can be used for the experiment. At its transition point Phenol red will undergo a colour change, which will signify the end of the titration. Phenol red is pink in basic conditions but yellow in acidic conditions, this colour change can be easily distinguished making Phenol red a good choice as an indicator for the experiment. The burette is not filled at above eye level for safety reasons; if it is filled above eye level the liquid could easily splash onto the users face and eyes. A lot of the solution used in titrations are irritants and so it is best to avoid a situation in which a person’s skin comes into contact with them. Also if the burette is filled above eye level then it is difficult to properly judge what level the liquid is at. If the magnetic stirrer is set at a high level it would result in heating the solution up. The movement of the stirrer would increase the energy of the molecules in the solution and so heat it up. This additional heat could have an effect on the experiment which would may not be accounted for, leading to more uncertainty in the experiment. Also the increase in temperature could affect the glassware as previously mentioned. If the stirrer is set at a high level then it would cause excessive splashing of the solution in the conical flask. This means that some of The NaOH solution will remain on the side of the glassware and not be involved in the titration. Therefore a lower volume of acid will be needed to neutralise the base, causing inaccuracy in the results. The end point of a titration is when the base or acid in the conical flask below the burette has been neutralised. An indicator is usually used in titration to show when the end point has been reached. The chosen indicator solution will be a different colour in acidic conditions than in basic conditions. So when the end point of the titration is approached the indicator will change in colour to show this. Determining the exact end point from the colour change is difficult, this is because the initial colour change may only be faint and so easily missed. So instead a digital pH monitor could be used in conjunction with the Phenol red indicator. The pH monitor would show the exact end point and the colour change of the Phenol red indicator would confirm this.