Question

 

Table 1: Hot Sauce Titration mass of hotsauce (g) 0.8

Concentration of NaOH solution used 0.1M

Volume needed of NaOH to neutralize the sauce_______?

pH of NaOH 12.5

Table 2: Hot Sauce pH Titration Data Trial 1

Increments of NaOH Added (mL)	Total NaOH Added (mL)	Hot Sauce pH
0	0	3.7
1.0	1.0	4.1
1.0	2.0	4.4
1.0	3.0	4.8
1.0	4.0	5.6
1.0	5.0	9.7
1.0	6.0	11.2
1.0	7.0	11.5
1.0	8.0	11.7
1.0	9.0	11.9
1.0	10.0	11.9
1.0	11.0	12.0
1.0	12.0	12.0
1.0	13.0	12.1
1.0	14.0	12.1
1.0	15.0	12.2
1.0	16.0	12.2
1.0	17.0	12.2
1.0	18.0	12.3
1.0	19.0	12.5

Table 3: Hot Sauce pH Titration Data Trial 2

Increments of NaOH Added (mL)	Total NaOH Added (mL)	Hot Sauce pH
0	0	3.7
1.0	1.0	4.0
1.0	2.0	4.1
1.0	3.0	5.9
1.0	4.0	7.5
1.0	5.0	9.3
1.0	6.0	10.8
1.0	7.0	11.9
1.0	8.0	11.9
1.0	9.0	12.0
1.0	10.0	12.0
1.0	11.0	12.0
1.0	12.0	12.1
1.0	13.0	12.1
1.0	14.0	12.2
1.0	15.0	12.2
1.0	16.0	12.3
1.0	17.0	12.3
1.0	18.0	12.4
1.0	19.0	12.5

Post-Lab Questions

Construct a graph from Tables 2 and 3 on a computer program such as Microsoft Excel®. Send your graph with your Post-Lab Questions to your lab instructor.

What are the equivalence points for both trials? Use your graphs to determine this. Calculate their average to answer Question 3.

Using the determined equivalence point from question 2 and the balanced reaction of acetic acid and sodium hydroxide, calculate the molarity of the acetic acid in your hot sauce packet. Assume the density of the hot sauce is 1.00 g/mL and use the mass of taco sauce from Table 1 for your calculation.

Answer 1

Ans.1). Hot sauce is actually consisted of acetic acid which is titrated with NaOH.

The graph is plotted between pH of Hot sauce versus the volume of titrant (mL) added. pH is taken along the y-axis and volume of NaOH added along the x-axis. As the volume is same for both trials, observations for pH of Hotsauce for both trials are plotted in one graph as shown below :

Ans.2). The region where very small addition of titrant cause a very rapid rise in the pH ( steep slope), is determined and then the midpoint of that region is called as equivalence point.

In graph, Series 1 = Trial 1 (blue line) and Series 2 = Trial 2 (Red line) observations.

Equivalence Point For Trial 1. = midpoint of region 4.8 to 11.2 = 8.0 mL

Equivalence Point For Trial 2 = midpoint of region 4.1 to 11.9 = 8.0 mL

Average Equivalence Point = 8.0 mL

Volume needed of NaOH to neutralize the sauce = 8.0 mL

Ans 3). the balanced reaction :

CH₃COOH + NaOH ----> CH₃COO^-Na⁺ + H₂O

Molarity of NaOH, M₁ = 0.1 M

Volume of NaOH, V₁ = 8.0 mL

Volume of CH₃COOH (Acetic acid), V₁ =given mass / density = 0.8 / 1.00 = 0.8 mL

Hence , Applying Molarity equation,

M₁V₁ = M₂V₂

Molarity of CH₃COOH, M₂ = M₁V₁ / V₂

M₂ = (0.1 x 8) / 0.8 = 1 M

12 10 12 16 18 20 Volume of 0.1M NaOH