1. The data attached is for the titration of vinegar (density 1.05 g/mL) which c
ID: 558017 • Letter: 1
Question
1. The data attached is for the titration of vinegar (density 1.05 g/mL) which contains acetic acid. HC2H3O2. Here an aliquot of 5.00 mL vinegar is titrated using 0.1029 M NaOH. Calculate the molarity of acetic acid using the correct number of significant figures. Enter only a numerical answer which is in units of mol/L. You must be correct +/- 0.001 M.2. Calculate the % mass acetic acid in your sample assuming a density of 1.05 g/mL for vinegar. Your answer must be within +/- 0.00%.
3. Determine the correct ratio of [C2H3O2-]/[ HC2H3O2] at the point where pH = pKa. The ratio should be to within 0.000.
4. Determine the Ka using the initial pH value. Calculate to within +/- 0.05
5. Determine the Ka value from the pH at one-half equivalence. Hint; See exercise 3.
6. Consider where the vinegar is titrated to pH 5.00 to create a buffer solution. What pH would result from adding 1.00 mL of 0.10 HCl to this solution? Answer must be correct +/- 0.05 units. 1. The data attached is for the titration of vinegar (density 1.05 g/mL) which contains acetic acid. HC2H3O2. Here an aliquot of 5.00 mL vinegar is titrated using 0.1029 M NaOH. Calculate the molarity of acetic acid using the correct number of significant figures. Enter only a numerical answer which is in units of mol/L. You must be correct +/- 0.001 M.
2. Calculate the % mass acetic acid in your sample assuming a density of 1.05 g/mL for vinegar. Your answer must be within +/- 0.00%.
3. Determine the correct ratio of [C2H3O2-]/[ HC2H3O2] at the point where pH = pKa. The ratio should be to within 0.000.
4. Determine the Ka using the initial pH value. Calculate to within +/- 0.05
5. Determine the Ka value from the pH at one-half equivalence. Hint; See exercise 3.
6. Consider where the vinegar is titrated to pH 5.00 to create a buffer solution. What pH would result from adding 1.00 mL of 0.10 HCl to this solution? Answer must be correct +/- 0.05 units. 14.00 T 12.00 8.00 6.00 4.00 2.00 0.00 0.0 5.0-10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0
Explanation / Answer
Titration of vinegar (acetic acid) with NaOH
1. From the plot,
volume of NaOH used to reach equivalence point = 36 ml
moles of NaOH used = 0.1029 M x 36 ml = 3.7044 mmol
moles of acetic acid present in vinegar = 3.7044 mmol
molarity of acetic acid = 3.7044 mmol/5 ml = 0.741 M
2. mass of vinegar taken = 5 ml x 1.05 g/ml = 5.25 g
mass% acetic acid = 0.222 g x 100/5.25 g = 4.23%
3. pH = pKa at half equivalence point
From the graph,
pH = pKa = 4.75
(C2H3O2-/HC2H3O2) = 1
the amount of [C2H3O2-] formed in solution = amount of [HC2H3O2] left in solution
4. Initial pH = 2.7
pH = -log[H+]
[H+] = 2 x 10^-3 M
[H+] = [C2H3O2-] = 2 x 10^-3 M
Ka = [H+][C2H3O2-]/[HC2H3O2] = (2 x 10^-3)^2/0.741
x = [H+] = 5.398 x 10^-6 M
pH = -log[H+] = 5.268
5. at half equivalence point,
pH = pKa
From graph, pH = 1/2 eq. point = 4.75
So, pKa = 4.75
Ka = -inv.log(-pKa) = 1.77 x 10^-5
6. Buffer pH = 5
Volume of NaOH added = 23 ml
added mmol of NaOH = 0.1029 M x 23.0 ml = 2.37 mmol
initial mmol of acetic acid = 0.741 M x 5 ml = 3.705 mmol
remaining mmol acetic acid = 1.335 mmol
after 0.1 M x 1 ml HCl added = 0.1 mmol
pH = pKa + log(C2H3O2-/HC2H3O2)
= 4.75 + log[(2.37 - 0.1)/(1.335 + 0.1)]
= 4.95
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