1. The phosphorie acid in a 100.00-mL sample of Cola drink was titrated with 0 1
ID: 556676 • Letter: 1
Question
1. The phosphorie acid in a 100.00-mL sample of Cola drink was titrated with 0 1025 M NaOH. If the first equivalence point occurred after 16.11 mL of base was added, and the second equivalence poim occurred after 32.55 mL of base was added, calculate the molar concentrations of H,PO, in the Cola sample based on: a. The first equivalence point b The second equivalence point 1 c. What is the average concestrution? 2. Based on the average concentration in question Ic, if you were able to titrate to the third equivalence point of H,PO4. What woulkd the tot of NaOll yoa would need? 48. 58 ml.Explanation / Answer
Titration of cola sample
1. H3PO4 in cola sample is titrated with NaOH
a. First equivalence point
volume of NaOH consumed = 16.11 ml
moles of NaOH = 0.1025 M x 0.01611 L = 0.00165 mol
moles H3PO4 present = 0.00165 mol
molar concentration of H3PO4 in cola sample = 0.00165 mol/0.1 L = 0.0165 M
b. First equivalence point
volume of NaOH consumed = 32.55 ml
moles of NaOH = 0.1025 M x 0.03255 L = 0.00334 mol
moles H3PO4 present = 0.00334 mol/2 = 0.00167 mol
molar concentration of H3PO4 in cola sample = 0.00167 mol/0.1 L = 0.0167 M
c. average concentration of H3PO4 in cols sample = 0.00166 M
2. Volume of NaOH needed to reach the third equivalence point = 3 x moles of H3PO4
moles H3PO4 = 0.0166 M x 0.1 L = 0.00166 mol
Volume NaOH needed for third equivalence point = 3 x 0.00166 mol/0.1025 M = 48.6 ml
3. Based on the above calculations,
a. pH = pKa1 = first 1/2 equivalence point
Volume of NaOH added = 48.6/3 x 2 = 8.1 ml
moles NaOH consumed = 0.1025 M x 0.0081 L = 0.00083 mol
a. pH = pKa2 = second 1/2 equivalence point
Volume of NaOH added = [(48.6/3 x 2) + 8.1] = 24.3 ml
moles NaOH consumed = 0.1025 M x 0.0243 L = 0.0025 mol
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