Table 1: Hot Sauce and Ketchup Titration Data Hot Sauce (Trial 1) Hot Sauce (Tri
ID: 572399 • Letter: T
Question
Table 1: Hot Sauce and Ketchup Titration Data
Hot Sauce
(Trial 1)
Hot Sauce
(Trial 2)
Ketchup
(Trial 1)
Ketchup
(Trial 2)
Mass of sauce (g)
1.5g
1.5g
1.5g
1.5g
Concentration of NaOH used (M)
0.1M
0.1M
0.1M
0.1M
(mL) NaOH needed to reach equivalence point
6.5 mL
8mL
3 mL
2.75
(mol) NaOH needed to reach equivalence point [show work below]
0.163 mol
0.2 mol
0.075mol
0.069mol
Concentration of C2H4O2 (mol/g of sauce) [show work below]
0.433mol
0.533mol
0.2mol
0.183mol
Average Concentration of C2H4O2 (mol/g of sauce)
0.483M
0.383M
pH of solution at equivalence point
6.89
6.83
6.36
6.12
NaOH needed to reach half-equivalence point (mL)
3.25mL
4.0mL
1.5mL
1.375mL
pH of solution at half- equivalence point
3.45
3.42
3.18
3.06
Concentration Calculations:
mol NaOH needed (Hot Sauce Trial 1):
6.5mL NaOH * (1g/1mL)*(1 mol/40g NaOH)=0.163 mol
Concentration of C2H4O2 (Hot Sauce Trial 1): NaOH+C2H4O2=H2O+NaC2H3O2
1:1 ratio NaOH:C2H4O2
6.5*10^-4mol/0.0015L=0.433M
mol NaOH needed (Hot Sauce Trial 2):
8mL NaOH * (1g/1mL)*(1 mol/40g NaOH)=0.2 mol
Concentration of C2H4O2 (Hot Sauce Trial 2):
8*10^-4mol/0.0015L=0.533M
mol NaOH needed (Ketchup Trial 1):
3.0mL NaOH * (1g/mL)*(1mol/40g NaoH)=0.075
Concentration of C2H4O2 (Ketchup Trial 1):
3*10^-4mol/0.0015L=0.2M
mol NaOH needed (Ketchup Trial 2):
2.75mL NaOH * (1g/mL)*(1mol/40g NaoH)=0.069
Concentration of C2H4O2 (Ketchup Trial 2):
8*10^-4mol/0.0015L=0.533M
1. Using your own half-equivalence point data from your experiment, what is the experimental pKa for acetic acid in this reaction? Using this experimental pKa value, what is your experimental Ka for acetic acid?
2. Look up the accepted “actual” Ka value for acetic acid. How does your value compare? Calculate the percent error for your experimental value. (You must show all work to receive credit.)
Table 1: Hot Sauce and Ketchup Titration Data
Hot Sauce
(Trial 1)
Hot Sauce
(Trial 2)
Ketchup
(Trial 1)
Ketchup
(Trial 2)
Mass of sauce (g)
1.5g
1.5g
1.5g
1.5g
Concentration of NaOH used (M)
0.1M
0.1M
0.1M
0.1M
(mL) NaOH needed to reach equivalence point
6.5 mL
8mL
3 mL
2.75
(mol) NaOH needed to reach equivalence point [show work below]
0.163 mol
0.2 mol
0.075mol
0.069mol
Concentration of C2H4O2 (mol/g of sauce) [show work below]
0.433mol
0.533mol
0.2mol
0.183mol
Average Concentration of C2H4O2 (mol/g of sauce)
0.483M
0.383M
pH of solution at equivalence point
6.89
6.83
6.36
6.12
NaOH needed to reach half-equivalence point (mL)
3.25mL
4.0mL
1.5mL
1.375mL
pH of solution at half- equivalence point
3.45
3.42
3.18
3.06
Concentration Calculations:
mol NaOH needed (Hot Sauce Trial 1):
6.5mL NaOH * (1g/1mL)*(1 mol/40g NaOH)=0.163 mol
Concentration of C2H4O2 (Hot Sauce Trial 1): NaOH+C2H4O2=H2O+NaC2H3O2
1:1 ratio NaOH:C2H4O2
6.5*10^-4mol/0.0015L=0.433M
mol NaOH needed (Hot Sauce Trial 2):
8mL NaOH * (1g/1mL)*(1 mol/40g NaOH)=0.2 mol
Concentration of C2H4O2 (Hot Sauce Trial 2):
8*10^-4mol/0.0015L=0.533M
mol NaOH needed (Ketchup Trial 1):
3.0mL NaOH * (1g/mL)*(1mol/40g NaoH)=0.075
Concentration of C2H4O2 (Ketchup Trial 1):
3*10^-4mol/0.0015L=0.2M
mol NaOH needed (Ketchup Trial 2):
2.75mL NaOH * (1g/mL)*(1mol/40g NaoH)=0.069
Concentration of C2H4O2 (Ketchup Trial 2):
8*10^-4mol/0.0015L=0.533M
Explanation / Answer
1. In case of hot sauce, pH at half-equivalence point = (3.45+3.42)/2 = 3.435
According to Henderson-Hasselbulch equation: pH = pKa + Log([NaC2H3O2]/[C2H4O2])
At half-equivalence point, [NaC2H3O2] = [C2H4O2]
i.e. [NaC2H3O2]/[C2H4O2] = 1
Therefore, 3.435 = pKa + Log(1)
i.e. pKa = 3.435 + 0 = 3.435
pKa = -Log(Ka)
i.e. Ka = 10-pKa = 10-3.435 = 3.673*10-4
In case of ketchup, pH at half-equivalence point = (3.18+3.06)/2 = 3.12
At half-equivalence point, pH = pKa
i.e. pKa = 3.12
i.e. Ka = 10-3.12 = 7.586*10-4
2. The actual Ka value for acetic acid = 1.8*10-5
In case of hot sauce, The experimetnal Ka value for acetic acid = 3.673*10-4 > actual Ka value for acetic acid
Note: The experimental value is much more greater than the actual value.
Now, there will be a large deviation if you calculate the percent errors, as shown below.
%error = {(experimental value-actual value)/actual value}*100
The percent error for your experimental value = {(3.673*10-4-1.8*10-5)/1.8*10-5)}*100 = 1940.6%
In case of ketchup, the percent error for your experimental value = {(7.586*10-4-1.8*10-5)/1.8*10-5)}*100 = 4114.4%
Note: Please be sure to cross check the pH values at half-equivalence points for both hot sauce and ketchup, then repeat the calculations as shown above.
All the very best!!!!
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.