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Jason Yu (Block C)
Which of the following methods give a value of NaOH concentration closest to the ‘real’ value1?
1. Weighing the solute,
2. Titrating it
1. No literature value available, we need to find by most accurate method
Data Collection: Method A (Weighing NaOH)
Method A: Weighing NaOH
Mass of NaOH pellets (zeroed beaker) = 0.412 ï¿½ 0.001 g
Volume of standard solution = 100.0 ï¿½ 0.5 cm3
On electronic balance: Two white and round NaOH pellets which were dry soon become wet and moist on surface, as seen by it gaining luster at time of measurement, which was a minute or two after taking out of the storing bottle.
In preparing NaOH solution: By crushing with glass rod, white NaOH pellets dissolve in water to form clear solution.
Data Collection: Method B (Titration)
Method B: Titration of NaOH
4 (Vanessa Yu’s)
Indicator (titration by)
Final burette reading (cm)
Initial burette reading (cm)
1 Phn = Phenolphthalein acid-base indicator.
2 Around ï¿½ 9.40 to ï¿½ 9.45
3 Only the titres in bold will be used in calculation
Precision of burette readings = ï¿½ 0.1 cm3
Volume of pipette = 10.00 ï¿½ 0.04 cm3 (class B)
Concentration of acid (HCl) = 0.1036 M ï¿½ 0.1%
The solution in the flask with NaOH and phenolphthalein indicator was initially purple, which turned clear in color after around 9.4 cm3 of HCl was titrated.
Goal: Finding concentration of NaOH by the two experimental methods.
Calculations: Method A (Weighing the NaOH)
The no. of concentration of NaOH = moles of NaOH / volume of solution
Volume of solution prepared = 100.
0 ï¿½ 0.5 cm3 (ï¿½ 0.5%)
= 0.1000 dm3 ï¿½ 0.5%
Mass of NaOH used = 0.412 ï¿½ 0.001 g (ï¿½ 0.2%)
No. of moles (n) of NaOH = 0.412 g / 39.99 g mol-1
= 0.0103 mol ï¿½ 0.2%
So as M = n / V,
The concentration (M) of NaOH = 0.0103 / 0.1000
= 0.103 M ï¿½ 0.7%
Calculations: Method B (Titration)
The equation for the reaction in titration is:
NaOH + HCl ï¿½ NaCl + H2O
The mole ratio of NaOH and HCl is 1:1.
Since concentration = no. of moles / volume (M = n/ V), and the no. of moles of HCl used for titration reaction = no. of moles of NaOH used for titration reaction when the equivalence point is reached, the concentration of the NaOH solution can be found by this equation:
MHCl x VHCl = MNaOH x VNaOH
Where M = Concentration of HCl and V = Volume of HCl
Average Titre Used:
Precision error in burette = ï¿½ 0.1 cm3 (ï¿½ 1%)
Titration data obtained = 9.4, 9.5, 9.4 cm3
Range in titres = 9.5 – 9.4 cm3
= 0.1 cm3
Uncertainty = Range / 2?3 cm3
= 0.1 / 2?3 cm3
= 0.03 cm3
So, average volume of titre (HCl) used = (9.4 + 9.5 + 9.4)/ 3
= 9.43 ï¿½ 0.03 cm3 (ï¿½ 0.3%)
As the data lie between the precision error of the burette, the uncertainty decreases with more trials, so we can take 3 sig. fig.s in our result above.
Volume of NaOH Used:
Volume of NaOH by pipette = 10.00 ï¿½ 0.04 cm3 (ï¿½ 0.4%)
Concentration of HCl Used
Concentration = 0.1036 M ï¿½ 0.1%
So we have, by the equation
MHCl x VHCl = MNaOH x VNaOH,
Concentration of NaOH = MNaOH
= MHCl x VHCl / VNaOH
= 0.1036 x 9.43 / 10.00
= 0.0977 M ï¿½ 0.8%
Presentation: Final Comparison:
Concentration of 100 cm3 NaOH obtained by
Method A (Weighing) (M)
Method B (Titration) (M)
So value of NaOH concentration in Method A > Method B.
1. (Method A) We weigh only the mass of the NaOH pellets and not other substance:
a.) The NaOH pellets did not change weight from their original one during measurement;
b.) The NaOH pellet’s surfaces are clean from impurities;
2. (Method A) No NaOH was lost during transfer of the solution to the volumetric flask.
3. (Method A) The solution put into the volumetric flask is exactly 100 cm3.
4. (Method B) The standard solution in the bottle is well mixed so that we take the same molar of NaOH for each titration from the bottle.
5. (Method B) The apparatus are thoroughly cleansed.
6. (Method B) The equivalence point of the reaction equals the end point of the indicator.
7. (Method B) The amount of HCl used just sufficiently neutralizes the NaOH, no more no less.
8. (Method B) The reading drop in burette equals the volume of HCl really used for neutralization.
From our qualitative data, as we will see that in Method A a major error occurs, where NaOH absorbs water on the balance, displaying a greater weight than it should be, while Method B has no major error except for minor manipulative errors which have been as avoided as possible. Therefore Method B is more accurate than Method A.
We have assumed that the weight recorded of NaOH on the balance agrees with the real value of the pellets, i.e. there is constant mass of NaOH during the measuring of weight. However, it is known that NaOH is a hygroscopic substance, i.e. it will readily absorb water from the air, which explains why we see the NaOH become moist and show watery luster on the balance. This makes the value of mass obtained larger than it is without the water absorbed, thus making the mass recorded inaccurately showing a larger value. This will give us a larger calculated concentration (as M = (mass / molar mass) / V), which agrees with our final data, where Method A gives a concentration of our standard solution to 0.103 M, which is larger to the concentration found by the second method, which gives 0.0977 M. Thus method A is more inaccurate by this large systematic error.
The probable accuracy of Method B is higher, as its main errors are due to manipulation error, and also small random errors which both errors do not contribute much to affect the results in our experiment, as I have been very careful with a major source of error which could affect the titration. The major source of error would be to miss the end point of titration by titrating too fast – if more acid was used to titrate the NaOH. It was eliminated in our experiment as for the last few drops we stopped the burette tap, shook the volumetric flask thoroughly, and then let another drop come down, and closed it again, repeatedly until the solution just turned from purple to clear – so that our titration reaction is correct to one drop (0.1 cm cubes). Since phenolphthalein has an equal end point with the equivalence point of the neutralization (all NaOH is neutralized while no excess HCl is put in) in strong acid and strong base (i.e. NaOH and HCl), and we stopped almost exactly at the end point, we can say that the equivalence point is reached. If we hadn’t been carefully controlling the tap, the end point would be passed, meaning that a larger titre (which is VHCl) would be used than needed, thus by the equation MHCl x VHCl = MNaOH x VNaOH, would make the calculated value of concentration of NaOH larger than the ‘real’ one.
We also assume that the measurements taken from the graduation marks are accurate, but still a slight error of no more than 0.1 cm cubes would exist, but that is within the uncertainty of error. We also made sure that no chemicals were left in the volumetric flask after use by washing with water. Also, if the titration is long, some acid might have been evaporated from the burette, thus making the volume of HCl used for reaction recorded higher than the real value, as acid is lost from the burette to atmosphere in addition to being lost to the NaOH solution, thus making the concentration calculated larger. All these challenge the assumption that the recorded drop in reading in burette equals the volume of HCl really used for neutralization, but they are only very minor errors.
Both methods give a higher concentration value of the actual one by the systematic errors – 1. NaOH absorbing water. 2. Missing the end point with the burette. Since the by Method A the resulting value is 0.103 M, which is greater than Method B’s 0.0977 M, this implies the systematic error in Method A is greater than that in B, and thus B is more accurate.
However, Method A will give a more precise measurement than that of B. It is because we used the electronic balance (ï¿½ 0.2%) in Method A which is more precise than the burette (ï¿½ 1% [by 0.1 cm3 / 9.4 cm3 x 100%]) in Method B. But when we used more trials and repeated our experiment in Method B to reduce the uncertainty, from the calculations above we see that the precision become similar: Method A is ï¿½0.7% while Method B is ï¿½0.8%. So Method B’s precision is improved.
Therefore we can see that Method B is more accurate than Method A, and that the precision of both methods is similar after we repeat the titrations in B. The other assumptions stated in the assumption section are irrelevant in contributing to any other major error.
Since Method B gives a more accurate result, the way to minimize the errors most is to NOT use Method A in finding concentration of a solution, as it is very difficult to not let the NaOH absorb water. We could of course do the experiment in an environment absent of water vapor if we have to, but it would be had to control the water vapor level in environments.
As for Method B, we should use a burette which has markings on front and back so you are know you are reading at eye level, thus reducing measurement inaccuracy. We could increase the precision by using more NaOH (more than 10 cm3) to get 3 sig. fig.s, as we are getting 9.4 now, which is just a 2 sig fig. value, in case even trial repetition does not increase the number of sig figs. We should titrate fast until the volume used reaches something close to the rough value we did, and titrate drop by drop, to make sure the time we used in the titration was not too long and that the end point was not passed. At the same time of controlling the tap, we should shake the flask thoroughly.
Some drops of HCl may not reach the NaOH but instead be on the sides of the flask instead. Make sure no drops fell to the outer parts of the flask or the mouth of it. Make sure that the tip of the burette hangs a drop before and after the experiment so that it is not counted as it didn’t go into the flask. Ensure that there are no air bubbles trapped inside before doing the titrations. Do more trials to improve both accuracy and precision of the titration.
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