Question

Acid Base titration Lab: Add NaOH to a hot sauce mixture, then to a ketchup mixture. ****The hot sauce mixture consisted of 20 mL DH20 AND 1.5g Hot sauce. Ketchup = 20 mL DH20 and 1.5 g ketchup. the below columns are the mLs added of NaOH, and the pH value at each point. 3rd column is derivative

A.) For each trial calculate mol NaOH needed to reach equiv. point, and the average concentration of C2H4O2 in each trial.

B.) What is the concentration of NaOH used (M) in each trial?

HOT SAUCE TRIAL-

mL pH derivative

0.00 3.50 0.49

0.5 3.76 0.44

1.0 3.98 0.37

1.5 4.13 0.31

2.0 4.27 0.26

2.5 4.39 0.24

3.0 4.50 0.23

3.5 4.62 0.23

3.75 4.68 0.22

4.0 4.73 0.21

4.25 4.78 0.21

4.5 4.83 0.22

4.75 4.89 0.23

5.0 4.95 0.23

5.25 5.01 0.24

5.5 5.07 0.26

5.75 5.13 0.28

6.0 5.21 0.31

6.25 5.29 0.33

6.5 5.37 0.37

6.75 5.47 0.44

7.0 5.58 0.54

7.25 5.72 0.73

7.5 5.93 1.13

7.75 6.20 2.00

8.0 6.70 3.59

8.25 8.05 4.84

8.5 9.60 4.25

8.75 10.38 2.75

9.0 10.83 1.59

9.25 11.09 0.90

9.5 11.27 0.55

10.0 11.49 0.39

10.5 11.64 0.30

11.0 11.75 0.24

KETCHUP TRIAL:

mL pH Derivative

0.0 3.95 0.50

0.5 4.21 0.49

1.0 4.45 0.48

1.5 4.68 0.47

2.0 4.91 0.49

2.25 5.04 0.56

2.5 5.19 0.66

2.75 5.36 0.85

3.0 5.59 1.26

3.25 5.89 2.12

3.5 6.45 3.57

3.75 7.85 4.28

4.0 9.01 3.42

4.25 9.61 2.22

4.5 9.98 1.42

4.75 10.26 0.96

5.0 10.47 0.67

5.5 10.75 0.52

6.0 10.94 0.44

Answer 1

A) C₂H₄O₂ refers to acetic acid and reacts with NaOH on a 1:1 molar basis as below.

C₂H₄O₂ (aq) + NaOH (aq) ------> NaC₂H₃O₂ (aq) + H₂O (l)

Molar mass of C₂H₄O₂ = (2*12.01 + 4*1.008 + 2*15.9994) g/mol = 60.0508 g/mol.

Mole(s) of C₂H₄O₂ corresponding to 1.5 g acid = (1.5 g)/(60.0508 g/mol) = 0.02498 mole.

Mole(s) of NaOH required for titration = mole(s) of C₂H₄O₂ taken = 0.02498 mole ≈ 0.025 mole (ans).

B) Plot the derivative with respect to volume as below for hot sauce.

Plot of derivative of pH with volume vs volume of NaOH added (mL) for hot sauce

The maxima of the plot gives the volume of NaOH required for the titration. The maxima is at (4.84, 8.25); therefore, the volume of NaOH required to reach the equivalence point is 8.25 mL.

Molar concentration of NaOH = (moles of NaOH)/(volume of NaOH in mL) = (0.025 mole)/[(8.25 mL)*(1 L/1000 mL)] = 3.030 mol/L ≈ 3.03 M (ans).

Next, plot the derivative with respect to volume as below for ketchup.

Plot of derivative of pH with volume vs volume of NaOH added (mL) for ketchup.

The maxima of the plot gives the volume of NaOH required for the titration. The maxima is at (4.28, 3.75); therefore, the volume of NaOH required to reach the equivalence point is 3.75 mL.

Molar concentration of NaOH = (moles of NaOH)/(volume of NaOH in mL) = (0.025 mole)/[(3.75 mL)*(1 L/1000 mL)] = 6.667 mol/L ≈ 6.67 M (ans).