ACID AND BASE TITRATION
Ms Shailina Khija Chemistry Lab 1
10 September 2018
The very common and simple technique of Titration is demonstrated in this experiment. The main objective of this study was to determine the end point of an acid base titration using a chemical indicator phenolphthalein. And also to gain experience titrating carefully to a visible endpoint. Acid base indicator is mostly used to know the molarity of a solution. This experiment is done to standardize the sodium hydroxide solution. Molarity is the amount of moles of solute per litre of solution. The first experiment the volume of Naoh needed for the equivalent point is 10.0 mL. Nacl(aq) + Hcl(aq) = Nacl(aq) + H2O(l). The result of the first titration was 8.00mL, the second titration was 17.1mL and the third titration was 26.2mL. The total average was 8.7%. And the second experiment the average molarity of Naoh needed to react with 10mL of soda as an unknown concentrated acid is 0.11mL. The first result was 30.7mL, the second was 33.9mL and the third was 36.7mL. The total average was 3.5%.
A titration is a quantitative chemical analysis used to determine the concentration an identified analyte (James R). The purpose of a titration is generally to determine the quantity or concentration of one of the reagents, that of the other being known beforehand. In any titration there must be a rapid quantitative reaction taking place as the titrant is added, and in acid-base titrations this is a acid and base react quantitatively. For example, the of Hydrochloric Acid is equal in value with the amount of Sodium Hydroxide.
NaoH + Hcl NaCl + H2O
Erlenmeyer Flask 125 mLBurette 50 mLVolumetric Flask 100 mLFunnel
Ring Stand ClambBurette ClambMagnetic Stir Plate
deionized water squirt bottle
a. 50.0mL × 0.10m Hcl solution from a 4.0 M.
(0.10) (50.0 mL) V2 (4.0 m)= 1.25 Hclb. 100 mL × 0.10 m NaoH from solid. molarity is defined as the number of moles of solute in a liter of solution (M = mol/L). So 100mL = Mol/1000L = 0.1. Mole =M × L. 0.10 × 0.1 = 0.01 mol NaoH.
0.1 MOL NaoH × 39.98/1 mol NaoH = 0.4g NaoH Onwuta 4
c. Phenolphthalein Solution
d. 10 mL of clear Soda
f. Deionized water
1.25 m HCL solution are put in 50mL volumetric flask with stopper, dilluted with distilled water and mix thoroughly.
0.4g NaoH was also mix thoroughly with a distilled water in a 100 mL volumetric flask with a stopper.
Burette are clean and the inner walls rinse with 10 mL of 0.10 m NaoH solution. Open the stop cock to release 9 mL of solution.
Record the initial volume to the nearest 2 decimal point which is 0.01 mL.Add 10 mL of 0.10m HCL solution and 2 drops of phenolpthalein in a 125mL Erlenmeyer flask. Magnetic Plate and Stir bar are used to spin the solution quickly.
Titrate HCL solution with NaoH from the burette until the colour turns light pink. The point at which the colour change occurs are called end points.
Record the final volume reading on the burette to the nearest 2 decimal point and calculate the used NaoH volume.
The titration is repeated 3 times inorder to get the average of the solution.
Step V and VIII are repeatated by replacing HCL with 10mL of clear Soda.
Titration between NaoH and HCL
Initial point: 0.10
Trial 1: 8.0mL
Trail 2: 17.1mL
Trail 3: 26.2mL
Volume of NaoH needed for the titration
8.0 – 0.10 = 7.9
17.1 – 8.0 = 9.1
26.2 – 17.1= 9.1
The average volume of NaoH solution = 7.9+9.1+9.1 ÷ 3 = 8.7%
Titration between NaoH and clear Soda as an unknown concentration acid
Trial 1: 30.7
Trial 2: 33.9
Trial 3: 36.7
The volume of unknown concentrated acid needed for titration
30.7 – 26.2 = 4.5
33.9 – 30.7 = 3.5
36.7 – 33.9 = 2.8
The average volume of unknown = 4.5+3.5+2.8 ÷ 3 = 3.5%
Chemical equation for reaction between NaoH and HCL
NaoH + Hcl NaCl + H2O
Trial 1 (M1)(V1) (M2)(V2)
M1 = 0.10
V1 = 10.0mL
M2 = ?
V2 = 8.7
(0.10 )( 10.0) M2 (8.7) = 0.11 NaoHTrial 2:
V1 = 10.0mL
M2 = 0.11
V2 = 3.5
M1 (10.0mL) (0.11)(3.5) = 0.0039
The sample in this experiment is Hydrochloric acid and the clear soda solution substance. Sodium hydroxide as base substance. Phenolphthalein was the indicator used in this experiment. Base was used as titrant and acid as solution in the Erlenmeyer flask, the solution in the flask turn to light pink at the end point.
During the experiment, there was some error in the process. More NaoH was added to the solution, making the volume to pass the endpoint. To avoid this problem, titration process should be carefully monitored and slowly open the tap of the burette. As soon as the colour change is noticed, stop the cock of the burette. Another error was the contamination of Erlenmeyer flask that has been used with other solution. To avoid this problem, the flask should be properly cleaned and dry. The error affected some of our result making them not to be precise, mostly in the second experiment. Manual titration is still used where the cost of automation is not justified. This includes short-term or occasional use, or simply lack of money in many countries.
The advantage is that it is fairly fundamental and direct in its measurement, and is easy to check using known standards. In addition, there are a wide range of types of titration, and specific methods have been developed for a vast range of materials.
This experiment is a valuable learning aid for teaching chemistry and chemical engineering, converting theoretical chemistry into a practical operation. It gives practice in balancing equations and the relevant mathematics to convert between volume, mass and mole, which is vital for both disciplines. Effectively it is done on a large scale in industry in batch neutralization, where one chemical is reacted with another. For example a waste acid may be neutralized with alkali to give a neutral precipitate which is then filtered. The solid and liquid are then safe for disposal or other use.