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Investigating titration
Investigating titration
Investigating titration
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Determining the Concentration of Citric Acid
Introduction
Titrations are performed to calculate the unknown concentration of solutions using standard solutions. A solution of known concentration and volume is added to a solution of known volume and unknown concentration, a burette is used to find the exact amount of the known solution is required for the reaction to come to completion. A pH indicator is used to determine when a reaction has completed.
Aim
To determine the concentration of Citric acid in both fresh and bottled lemon juice.
Materials
Burette Pipette Retort stand Clamp Funnel Beakers Conical Flask White tile Vacuum pump Buchner Flask Distilled water Universal Indicator Phenolphthalein Hydrochloric acid solution
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HCl
C=1M n=CV
V=0.01887L n=(1)(0.01887) n= ?moles n=0.01887 moles
The ratio of HCl to NaOH is 1:1 so they have the same number of moles
NaOH
C= ?M C=n/V
V=0.02L C=0.02/0.01887 n=0.01887 moles C=1.06M
Bottled Lemon Juice + Sodium Hydroxide Volume (mL) Titre Volume Initial Final mL L
Rough 0.0 3.9 3.9 0.0039
Titration 1 3.9 7.6 3.7 0.0037
Titration 2 7.6 13.7 6.1 0.0061
Titration 3 13.7 19.1 5.4 0.0054
Average in Litres (not including rough) 0.00567 L
C6H8O7 +3NaOH 3H2O + Na3C6H5O7
3NaOH
C=1.06M n=CV
V=0.02L n=(1.06)(0.02) n= ?moles n=0.0212 moles
The ratio of sodium hydroxide to citric acid is 3:1 so the number of moles in citric acid is:
0.0212÷3=7.067×〖10〗^(-3)
C6H8O7
C=?M C=n/V
V=0.00567L C=(7.067×〖10〗^(-3))/0.00567 n=7.067×〖10〗^(-3) moles C=1.25M Fresh Lemon Juice + Sodium Hydroxide Volume (mL) Titre Volume Initial Final mL L
Rough
Titration 1 0.0 5.5 5.5 0.0055
Titration 2 4.9 0.0049
Titration 3
Average in Litres (not including rough) 0.0052 L
C6H8O7 +3NaOH 3H2O + Na3C6H5O7
3NaOH
C=1.06M n=CV
V=0.02L n=(1.06)(0.02) n= ?moles n=0.0212 moles
The ratio of sodium hydroxide to citric acid is 3:1 so the number of moles in citric acid is:
0.0212÷3=7.067×〖10〗^(-3)
C6H8O7
C=?M
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When the titration was performed to find the concentration of the sodium hydroxide solution, phenolphthalein was used to determine when the pH reached 7. The disadvantage of using phenolphthalein is that it changes from pink to transparent when the solution has a pH of less than 8, meaning that most of the time the point at which the colour change occurs is not the same as the point at which the solution has become neutral.
Whilst both of the calculated values for the concentration of the lemon juices are quite far from the actual values for the concentration, there are a number of errors that may have occurred while performing the titrations that may have led to these inaccuracies. For the bottled lemon juice titration, universal indicator was used to measure the pH and to determine when the solution had reacted completely and reached a pH of 7, indicating neutrality. This indicator is good to use because the colour change is obvious as it changes from purple to green. The downside and the factor that may have led to errors is that it is best used in solutions that are transparent. As the solution being added to the sodium hydroxide solution was not a transparent solution, the yellow tinge in the lemon juice made it difficult to tell at what point the solution had reacted completely and turned green. This may have led to more or less acid
Record the volume of the sodium thiosulfate solution used in the titration, and repeat the procedure in a duplicate titration.
This lab contains two different procedures to titrating vinegar. One procedure uses phenolphthalein while the other uses a pH meter. Bothe procedures can be found on “An Analysis of a Household Acid: Titrating Vinegar” by the Department of Chemistry at APSU.
4 "That amount of any gas that occupies a volume of 22414 mL in normal conditions is called one mole [eine solche Menge irgendeines Gases, welche das Volum von 22412 ccm im Normalzustand einnimt nennt man ein Mol]"
from both sides, leaving us with ½ V2 = GH. When the above equation is
Number of moles of〖 K_2 CrO〗_4, mol = (4.0 ×〖10〗^(-2) mol L^(-1))(5.0×〖10〗^(-2) L)= 2.0×〖10〗^(-3) mol
In this, the amount of moles in the sodium hydroxide solution after it has been reacted with the aspirin is found using titration, and then compared with the amount of moles it had without the aspirin being added. The difference in moles is the number of moles of sodium hydroxide that reacted with the aspirin, and therefore the number of moles of
Neutralization Experiment AIM:- To investigate how heat is given out in neutralizing sodium hydroxide (NaOH) using different concentrations of Hydrochloric Acid. Background Information:- Substances that neutralize acids are called alkalis. An acid is a substance that forms hydrogen ions (H+ ) when placed in water. It can also be described as a proton donor as it provides H+ ions. An example of an acid is hydrochloric acid (HCl), Sulphuric acid (H2SO4) etc.
I decided to experiment with pHs within the range pH 2 to pH7, as I
Moles Volume HCl Volume Water 2 M 10 cm 3 0 cm 3 1.5 M 7.5 cm 3 2.5 cm 3 1 M 5 cm 3 5 cm 3 0.5 M 2.5 cm 3 7.5 cm 3
NaOH * 1mol HCl/1mol NaOH=.00925mol HCl. .00925mol HCl/.005L =.185 M HCl.
Some improvements to the experiment might be using Na Acetate or Na Citrate as buffers instead of KHPO4. The pH ranges are 4.5-5.5 and 4.7-5.5, respectively. This range falls closer to the ideal pH of 5, then KHPO4 (pH
Despite straining the lemon juice before titration, small amounts of pulp would have affected the titrations, because pulps was in some cases blocking the pipette when the juice is transformed from volumetric to conical flask. For future investigation, instead of straining the lemon juice, it could have been filtered to remove all pulp and thus increase the accuracy of the result
Therefore, the Number of moles × molecular mass = mass of the sugars in grams in 10 mL of distilled water (with uncertaintyof ±0.2 mL on the graduated cylinder)
I hypothesize that by titrating a known amount of a substance in a solution, I can use the data from the experiment to calculate how many moles of the solution of the known concentration were required to completely react with the titrant and then using the formula C=n/V(in liters) to calculate the molarity of NaOH.