Question

5. Determine the average value for the molarity of the acetic acid in vinegar. Show your calculation 6. Determine the deviation between each individual molarity for acetic acid and the average value from question 4, above. Notice that the formula below uses absolute values: you only care about how big the difference is not whether cach trial value was higher or lower than the average. Deviation = (individual trial value)-(average value from #5) Trial Ideviation Trial 2 deviation Trial 3 deviation 7. Determine your percent deviation, using the formula below. (total of 3 deviations from #6) % deviation x 100% (average value of molarity from #5) Percent deviation 8. Examine the three versions of the equation for the reaction between KHP and sodium hydroxide, shown on Page 1 of this lab. Answer the following questions regarding just the nel lonic equation (the 3rd one). a) Which species in the equation is acting as a proton donor for the forward reaction? b) Identify the two conjugate acid-base pairs for this equation. 9. Below is the net ionic equation for the reaction between hydroxide ion and acetic acid. This was the reaction that you investigated for Part 2 of the experiment. Identify the two conjugate acid base pairs in this reaction. OH (aq) + HC3H6Oaq) + C3H6O4 (aq) + HBO) 10. The pH of vinegar is about 2.4, which is higher (less acidie) than you might predict based on the average molarity you calculated in Question 5, above. Suggest an explanation for this, based on what you know about acetic acid.

Answer 1

Data table:

Question 1:

formula Trail 1 Trail 2 Trail 1 Initial buret reading, mL given 0.00 0.001 0.001 Final buret reading, mL given 15.85 15.76 15.92 Volume of NaOH, mL = Final - initial buret reading 15.85 15.76 15.92 Volume in L = mL/1000 0.01585 0.01576 0.01592 Molarity of NaOH, M given 0.5237 0.5237 0.5237 Moles of NaOH = Molarity * volume in L 0.00830 0.00825 0.00834 Moles of acetic acid in 10ml NaOH Moles = acetic acid moles 0.00830 0.00825 0.00834 Molarity of Acetic acid, M = Moles / volume in L 0.830 0.825 0.834

Volume of NaOH used (mL) = Final buret reading -Initial buret reading

= 15.85 - 0.00 = 15.85mL

Volume of NaOH in L = 15.58mL/1000

= 0.01585 L

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Question 2.

Moles of NaOH = n = Molarity x volume in L

For trial 1,

n = 0.5237 x 0.01585 = 0.0083 mol

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Question 3.

In reaction of NaOH and acetic acid, number of moles of NaOH = number of moles of Acetic acid.

For trial 1, moles of acetic acid = 0.0083mol

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Question 4.

Molarity of Acetic acid in Vinegar = number of moles / Volume in L.

For trail 1,

Molarity of acetic acid = 0.00830/0.01 = 0.830M

Trail 2,

Molarity of acetic acid = 0.00825/0.01 = 0.825M

For trail 3,

Molarity of acetic acid = 0.00834/0.10 = 0.834M

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Question 5:

Average Molarity of acetic acid in Vinegar = sum of Molarity of each trail /3

= (0.830 + 0.825 + 0.834) / 3

= 0.830 M

Average = 0.830M

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Question 6.

Deviation for,

Trail 1 = 0.830 - 0.830 = 0.000 M

Trail 2 = 0.825 - 0.830 = -0.005M

Trail 3 = 0.834 - 0.830 = 0.004M

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Question 7.

Percent deviation = [(0.000+0.004-0.005)/0.830]*100 = 0.001/0.830

= 0.12 %

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Question 8.

For titration between KHP and NaOH,  

Proton doner = KHP.

Two acid-base pairs.

1. C2H3O2 acid & water as a base

2. C2H2O2- acid and hydronium ion base

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Question 9:

Pair 1:Acetic acid acts as the proton donor and, therefore, is a Brønsted-Lowry acid. The water acts as a proton acceptor and is a Brønsted-Lowry base.

Pair 2: acetate anion acts as a proton acceptor and, therefore, is a Brønsted-Lowry base while hydronium ion acts as a proton donor so it's base.

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Question 10.

The pH levels are different because difference in acidity vs H3O+ions.

a. The molar molar concentrations of acid ions vary at the same acidity

b. Acid dissociation constants vary so different amounts of [H30+] come out in equilibrium.