Introduction:
Acids and bases are often described using the Bronsted-Lowry theory which states that acids are proton donors, while bases are proton acceptors (Thorne, 78). This means that an acid has additional H+ ion(s) that can be given, whereas a base lacks these ions and will accept them. Acids and bases can further classified as strong or weak. A weak acid, such as HF, will not ionize completely, meaning that only some of the H+ ions will be lost. A strong acid will however, will release more of their H+ ions (Tro, 664). This concept is similar for bases; when strong, a base will dissociate more fully, whereas a weak base will not.
Titration is a process used to identify an unknown amount of an acid or base in a solution. This is done …show more content…
In this experiment, the reaction, NaOH + CH3COOH → NaCH3COO + H2O takes place. Using the molarity and volume of the NaOH solution, the moles of NaOH can be calculated (mol = M*V). Because the reaction is a 1:1 ratio, the moles of CH3COOH is also identified (mol CH3COOH = mol NaOH). Using the molar mass of CH3COOH, the mass of acetic acid can be calculated, and compared to the mass of the overall vinegar solution (Thorne 86-87).
These values can be compared to the theoretical values which state that the equilibrium constant of acetic acid is 1.8 x 10-5, and the percent mass of acetic acid in vinegar is 5%.
Calculations/Discussion:
Recorded data:
Concentration of standardized NaOH Solution: 0.10 M
Mass of Vinegar Used: 4.03 grams
Volume of NaOH (mL) pH Volume of NaOH (mL) pH Volume of NaOH (mL) pH
0.0 3.0 26.0 5.1 35.8 7.3
2.9 3.6 28.9 5.4 35.9 - Eq. Point 8.0
5.9 3.9 30.9 5.6 36.0 8.8
8.9 4.2 32.9 5.7 37.9 11.1
11.9 4.4 34.9 6.2 40.0 11.4
14.0 4.4 35.1 6.3 41.9 11.6
16.9 4.6 35.4 6.6 43.9 …show more content…
This change in color indicates a significant change in pH, which is supported by the readings of the pH pen (shown in the graph above). The equivalence point for this experiment took place at 35.9 mL of NaOH titrant, and a pH of 8.0.
Calculation of pKa:
Ka = 10-(pH at half volume) pKa = pH at half equivalence point volume (17.9) = 4.7
Ka = 10-4.7 = 1.99 x 10-5 Percent Error = [(Experimental - Theoretical) / Theoretical] x 100%
Percent Error = [(1.99 x 10-5 - 1.8 x 10-5) / 1.8 x 10-5] x 100% = 10.5%
Calculation of moles of NaOH and CH3COOH:
NaOH + CH3COOH → NaCH3COO +
Acids and bases are often described using the Bronsted-Lowry theory which states that acids are proton donors, while bases are proton acceptors (Thorne, 78). This means that an acid has additional H+ ion(s) that can be given, whereas a base lacks these ions and will accept them. Acids and bases can further classified as strong or weak. A weak acid, such as HF, will not ionize completely, meaning that only some of the H+ ions will be lost. A strong acid will however, will release more of their H+ ions (Tro, 664). This concept is similar for bases; when strong, a base will dissociate more fully, whereas a weak base will not.
Titration is a process used to identify an unknown amount of an acid or base in a solution. This is done …show more content…
In this experiment, the reaction, NaOH + CH3COOH → NaCH3COO + H2O takes place. Using the molarity and volume of the NaOH solution, the moles of NaOH can be calculated (mol = M*V). Because the reaction is a 1:1 ratio, the moles of CH3COOH is also identified (mol CH3COOH = mol NaOH). Using the molar mass of CH3COOH, the mass of acetic acid can be calculated, and compared to the mass of the overall vinegar solution (Thorne 86-87).
These values can be compared to the theoretical values which state that the equilibrium constant of acetic acid is 1.8 x 10-5, and the percent mass of acetic acid in vinegar is 5%.
Calculations/Discussion:
Recorded data:
Concentration of standardized NaOH Solution: 0.10 M
Mass of Vinegar Used: 4.03 grams
Volume of NaOH (mL) pH Volume of NaOH (mL) pH Volume of NaOH (mL) pH
0.0 3.0 26.0 5.1 35.8 7.3
2.9 3.6 28.9 5.4 35.9 - Eq. Point 8.0
5.9 3.9 30.9 5.6 36.0 8.8
8.9 4.2 32.9 5.7 37.9 11.1
11.9 4.4 34.9 6.2 40.0 11.4
14.0 4.4 35.1 6.3 41.9 11.6
16.9 4.6 35.4 6.6 43.9 …show more content…
This change in color indicates a significant change in pH, which is supported by the readings of the pH pen (shown in the graph above). The equivalence point for this experiment took place at 35.9 mL of NaOH titrant, and a pH of 8.0.
Calculation of pKa:
Ka = 10-(pH at half volume) pKa = pH at half equivalence point volume (17.9) = 4.7
Ka = 10-4.7 = 1.99 x 10-5 Percent Error = [(Experimental - Theoretical) / Theoretical] x 100%
Percent Error = [(1.99 x 10-5 - 1.8 x 10-5) / 1.8 x 10-5] x 100% = 10.5%
Calculation of moles of NaOH and CH3COOH:
NaOH + CH3COOH → NaCH3COO +