Acid_Base Titration Hello. I need some help in completing this table, I did most
ID: 572169 • Letter: A
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Acid_Base Titration Hello. I need some help in completing this table, I did most of it, but I'm still confused, can you please explain. Thank you!
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]
0.0249 mol
0.0249 mol
0.0249mol
0.0249mol
Average Concentration of C2H4O2 (mol/g of sauce)
pH of solution at equivalence point
6.89
6.83
6.36
6.12
NaOH needed to reach half-equivalence point (mL)
pH of solution at half- equivalence point
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):
1.5/60.05=0.0249 mol
mol NaOH needed (Hot Sauce Trial 2):
6.5mL NaOH * (1g/1mL)*(1 mol/40g NaOH)=0.163 mol
Concentration of C2H4O2 (Hot Sauce Trial 2):
1.5/60.05=0.0249 mol
mol NaOH needed (Ketchup Trial 1):
3.0mL NaOH * (1g/mL)*(1mol/40g NaoH)=0.075
Concentration of C2H4O2 (Ketchup Trial 1):
1.5/60.05=0.0249 mol
mol NaOH needed (Ketchup Trial 2):
3.0mL NaOH * (1g/mL)*(1mol/40g NaoH)=0.069
Concentration of C2H4O2 (Ketchup Trial 2):
1.5/60.05=0.0249 mol
HOT SAUCE TRIAL 1
0.1 M NaOH pH Derivate
0 p D
mL
0.00 3.57 0.56
0.50 3.88 0.49
1.0 4.09 0.40
1.5 4.27 0.34
2.0 4.43 0.30
2.5 4.56 0.28
3.0 4.70 0.27
3.5 4.83 0.27
3.75 4.90 0.27
4.0 4.96 0.29
4.25 5.04 0.31
4.5 5.12 0.33
4.75 5.20 0.37
5.0 5.30 0.43
5.25 5.41 0.49
5.5 5.54 0.59
5.75 5.70 0.79
6.0 5.90 1.31
6.25 6.19 2.49
6.5 6.89 4.26
6.75 8.71 4.08
7.25 10.70 2.50
7.75 11.24 1.17
8.25 11.49 0.54
8.75 11.63 0.29
9.25 11.74 0.20
9.75 11.82 0.16
10.25 11.88 0.14
HOT SAUCE TRIAL 2
0.1M NaOH pH Derivate
0 p D
mL
0.00 3.51 0.53
0.50 3.80 0.47
1.0 4.01 0.39
1.5 4.18 0.32
2.0 4.32 0.27
2.5 4.45 0.23
3.0 4.56 0.19
4.0 4.70 0.18
4.25 4.76 0.21
4.50 4.83 0.18
5 4.88 0.18
5.25 4.95 0.24
5.50 5.01 0.27
5.75 5.08 0.29
6 5.16 0.33
6.25 5.25 0.36
6.5 5.34 0.39
6.75 5.44 0.45
7 5.55 0.57
7.25 5.71 0.79
7.5 5.92 1.37
7.75 6.23 2.48
8 6.83 3.98
8.5 9.89 3.49
9 10.97 2.21
9.5 11.33 1.34
KETCHUP TRIAL 1
0.1 NaOH pH Derivate
0 p D
mL
0.00 3.87 0.68
0.50 4.23 0.63
1.0 4.52 0.59
1.5 4.80 0.59
2.0 5.09 0.64
2.25 5.26 0.81
2.50 5.48 1.24
2.75 5.78 2.19
3.0 6.36 3.65
3.25 7.88 3.64
3.75 9.71 2.43
4.25 10.39 1.36
4.75 10.74 0.79
5.25 10.93 0.50
KETCHUP TRIAL 2
0.1M NaOH pH Derivate
0 p D
mL
0.00 3.87 0.71
0.50 4.25 0.66
1.0 4.55 0.63
1.5 4.84 0.64
2.0 5.16 0.74
2.25 5.38 1.19
2.5 5.65 2.43
2.75 6.12 5.19
3.0 8.89 5.47
3.25 9.63 2.78
3.5 10.05 1.41
4.0 10.56 0.80
4.5 10.83 0.51
5.0 11.00 0.35
5.5 11.11 0.27
Acid_Base Titration Hello. I need some help in completing this table, I did most of it, but I'm still confused, can you please explain. Thank you!
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]
0.0249 mol
0.0249 mol
0.0249mol
0.0249mol
Average Concentration of C2H4O2 (mol/g of sauce)
pH of solution at equivalence point
6.89
6.83
6.36
6.12
NaOH needed to reach half-equivalence point (mL)
pH of solution at half- equivalence point
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):
1.5/60.05=0.0249 mol
mol NaOH needed (Hot Sauce Trial 2):
6.5mL NaOH * (1g/1mL)*(1 mol/40g NaOH)=0.163 mol
Concentration of C2H4O2 (Hot Sauce Trial 2):
1.5/60.05=0.0249 mol
mol NaOH needed (Ketchup Trial 1):
3.0mL NaOH * (1g/mL)*(1mol/40g NaoH)=0.075
Concentration of C2H4O2 (Ketchup Trial 1):
1.5/60.05=0.0249 mol
mol NaOH needed (Ketchup Trial 2):
3.0mL NaOH * (1g/mL)*(1mol/40g NaoH)=0.069
Concentration of C2H4O2 (Ketchup Trial 2):
1.5/60.05=0.0249 mol
Explanation / Answer
Calculation for concentration of C2H4O2 in sample
Hot sauce (Trial 1)
volume of 0.1 M NaOH used to reach equivalence point = 6.5 ml
moles of NaOH used = molarity x volume = 0.1 M x 0.0065 L = 6.5 x 10^-4 mol
moles of C2H4O2 present = moles of NaOH consumed = 6.5 x 10^-4 mol
Volume of hot sauce taken = mass x density = 1.5 g x 1 g/ml = 1.5 ml = 0.0015 L
Concentration of C2H4O2 in hot sauce = moles/volume = 6.5 x 10^-4 mol/0.0015 L = 0.433 M
Similarly,
Hot sauce (Trial 2)
volume of 0.1 M NaOH used to reach equivalence point = 8 ml
moles of NaOH used = molarity x volume = 0.1 M x 0.008 L = 8 x 10^-4 mol
moles of C2H4O2 present = moles of NaOH consumed = 8 x 10^-4 mol
Volume of hot sauce taken = mass x density = 1.5 g x 1 g/ml = 1.5 ml = 0.0015 L
Concentration of C2H4O2 in hot sauce = moles/volume = 8 x 10^-4 mol/0.0015 L = 0.533 M
average concentration of C2H4O2 in hot sauce = (0.433 + 0.533)/2 = 0.483 M
----
Volume of NaOH needed to reach half-equivalence point = half of volume needed to reach equivalence point
= 1/2(6.5) = 3.25 ml [for Hot sauce Trial 1]
Similarly,
Volume of NaOH needed to reach half-equivalence point = half of volume needed to reach equivalence point
= 1/2(8) = 4 ml [for Hot sauce Trial 2]
---
pH at half-equivalence point = 4.76 [pH at volume of NaOH added 3.25 ml for Hot sauce Trial 1]
and,
pH at half-equivalence point = 4.70 [pH at volume of NaOH added 4 ml for Hot sauce Trial 1]
In a similar fashion the calculation for the Ketchup trial runs can be performed.
[formula used,
molarity = moles/L of solution]
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