# Chem. Lab. Acid-Base Titration

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 60 x 100/1000

= 0,63 %

Version 2*

Figure 3 Calculating the molar mass of CH3COOH

M(CH3COOH)= (2 x 12) + (2 x 16) + 4 = 60 g/mol

Figure 3 Calculating the molar mass of CH3COOH

M(CH3COOH)= (2 x 12) + (2 x 16) + 4 = 60 g/mol

Figure 5 Calculating the molar quantity of CH3COOH:

m (CH3COOH) = 2.5 x 10-3 mol x 60 g/mol

= 0,15 g

Figure 5 Calculating the molar quantity of CH3COOH:

m (CH3COOH) = 2.5 x 10-3 mol x 60 g/mol

= 0,15 g

Figure 4 Calculating the molar quantity of CH3COOH:

n (CH3COOH) = 0.100 M x (0.0250 L) = 2.50 x 10-3 mol

Figure 4 Calculating the molar quantity of CH3COOH:

n (CH3COOH) = 0.100 M x (0.0250 L) = 2.50 x 10-3 mol

Figure 6 Calculating the molar quantity of CH3COOH:

Mass % = m (CH3COOH) / m (solution) x 100

= 0,15 / 25 x 100

= 0,60 %

Figure 6 Calculating the molar quantity of CH3COOH:

Mass % = m (CH3COOH) / m (solution) x 100

= 0,15 / 25 x 100

= 0,60 %

*Two different methods for the calculation of the mass percentage were used to be certain of the result’s efficiency

**The version 1 is the one I will take into consideration, due to it being more accurate.

Graphing data

pH

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _Equivalency point_ _ (7,26.3)

Volume of alkali added (cm3) /±0.01/

Conclusion and Evaluation (CE)

Conclusion

In the experiment we calculated the concentration of ethanoic acid in the diluted vinegar, as well as calculated the mass percentage of ethanoic acid. For the experiment the diluted ethanoic acid needs to be neutralized by adding an alkali (in our case sodium hydroxide - NaOH). We can observe when it reaches the point of equivalency when the phenolphthalein added in the diluted vinegar turns purple. In conclusion, for the first task, I can...

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