Chem. Lab. Acid-Base Titration
XV. GIMNAZIJA
International Baccalaureate Department
Group 4 – Chemistry SL
Lab no.2: Acid-base titration
Student: Caterina Rende Dominis
Teacher: Zrinka Toplićan
Date: 19 November 2012
Data Collection and Processing (DCP)
Aspect 1: Recording raw data
Table 1 Table showing raw data collected from titration
Known measurements 25 mL of diluted acid 0,100 M of NaOH solution
Measurement Number | V of alkali needed to neutralize acid /mL/ (±0.01 mL) | 1 | 26.4 | 2 | 26.1 | 3 | 26.1 | 4 | 26.0 | 5 | 26.1 | 6 | 26.1 | 7 | 26.0 | 8 | 26.4 | 9 | 26.9 | 10 | 26.9 | Average | 26.3 |
…. Relevant data (Aspect 3: Presenting Raw Data)
Aspect 2: Processing raw data
Task 1. Calculating the concentration of ethanoic acid in the diluted vinegar
In the equation of the balanced chemical equation between the ethanoic acid (CH3COOH solution) and sodium hydroxide (NaOH solution),
CH3COOH + NaOH → CH3COONa + H2O
We can determine that 1 mol of ethanoic acid is needed to neutralize 1 mol of sodium hydroxide, meaning that the amount of moles ethanoic acid used in the reaction is equal to the amount of moles of sodium hydroxide used, their ration being 1:1.
Figure 1 Calculation of acid concentration by inserting the measured and previously known data.
Amount of moles in CH3COOH = Amount of moles in NaOH
Figure 1 Calculation of acid concentration by inserting the measured and previously known data.
Amount of moles in CH3COOH = Amount of moles in NaOH
Task 2. Calculate the mass percentage of ethanoic acid
Known data Density of vinegar is 1.01 g/cm3
Version 1** Figure 2 Calculating mass percentage of ethanoic acid
Mass % = C x M x 100/1000 = 0.1052 x 60 x 100/1000 = 0,63 %
Figure 2 Calculating mass percentage of ethanoic acid
Mass % = C x M x 100/1000 = 0.1052 x
International Baccalaureate Department
Group 4 – Chemistry SL
Lab no.2: Acid-base titration
Student: Caterina Rende Dominis
Teacher: Zrinka Toplićan
Date: 19 November 2012
Data Collection and Processing (DCP)
Aspect 1: Recording raw data
Table 1 Table showing raw data collected from titration
Known measurements 25 mL of diluted acid 0,100 M of NaOH solution
Measurement Number | V of alkali needed to neutralize acid /mL/ (±0.01 mL) | 1 | 26.4 | 2 | 26.1 | 3 | 26.1 | 4 | 26.0 | 5 | 26.1 | 6 | 26.1 | 7 | 26.0 | 8 | 26.4 | 9 | 26.9 | 10 | 26.9 | Average | 26.3 |
…. Relevant data (Aspect 3: Presenting Raw Data)
Aspect 2: Processing raw data
Task 1. Calculating the concentration of ethanoic acid in the diluted vinegar
In the equation of the balanced chemical equation between the ethanoic acid (CH3COOH solution) and sodium hydroxide (NaOH solution),
CH3COOH + NaOH → CH3COONa + H2O
We can determine that 1 mol of ethanoic acid is needed to neutralize 1 mol of sodium hydroxide, meaning that the amount of moles ethanoic acid used in the reaction is equal to the amount of moles of sodium hydroxide used, their ration being 1:1.
Figure 1 Calculation of acid concentration by inserting the measured and previously known data.
Amount of moles in CH3COOH = Amount of moles in NaOH
Figure 1 Calculation of acid concentration by inserting the measured and previously known data.
Amount of moles in CH3COOH = Amount of moles in NaOH
Task 2. Calculate the mass percentage of ethanoic acid
Known data Density of vinegar is 1.01 g/cm3
Version 1** Figure 2 Calculating mass percentage of ethanoic acid
Mass % = C x M x 100/1000 = 0.1052 x 60 x 100/1000 = 0,63 %
Figure 2 Calculating mass percentage of ethanoic acid
Mass % = C x M x 100/1000 = 0.1052 x