the phosphoric acid in a 100.00 ml sample of coca cola drink was titrated with 0

Question

the phosphoric acid in a 100.00 ml sample of coca cola drink was titrated with 0.1025 M NaOH . if the first equivalence point occurred after 16.11 mL of base was added, ans the second equivalence point occurred after 32.55 mL of yhe base was added, calculate molar concentrations of H3po4 in the Cola sample based on:

A. the first equivalence point
B. the second equivalence point
C. what is the average concentration?

2. based on the average concentration in question 1c, if you were able to titrate to the third equivalence point of H3po4, what would the total volume of NaOH you would need?

3. based on question 1. how many moles , of NaOH were added when:
a. pH =pKa1
b. pH = pKa2

Explanation / Answer

1. H3PO4 + H2O H3O+ + H2PO4 -

2. H2PO4 - + H2O    H3O+ + HPO4 -2

3. HPO4 -2   + H2O    H3O+   + PO4 -3

1. After the first equivalence point all H3PO4 converts into H2PO4-.

Strength of NaOH = 0.1025 M

Volume of NaOH required = 16.11 mL

No of moles of NaOH present in 16.11 mL = (0.1025 * 16.11/1000) = 0.00165 moles.

According to the reaction (1) one mole NaOH reacts with one mole H3PO4. So from the first equivalent point we see that 0.00165 moles of H3PO4 was present in the sample.

Volume of NaOH required to reach to the second equivalence point = 32.55 mL

Volume of NaOH reacted with H2PO4- = 32.55-16.11 = 16.44 mL

No of moles of NaOH present in 16.44 mL = (0.1025 * 16.44/1000) = 0.00168 moles

According to the reaction (2) one mole NaOH reacts with one mole H2PO4- . So from the second equivalent point we see that 0.00168 moles of H2PO4- was present in the sample.

As all H3PO4 converted into H2PO4- , so no of moles of H2PO4- = no of moles of H3PO4 = 0.00168 moles

Average no of moles of H3PO4 in the 100 mL sample = (0.00165 + 0.00168)/2 = 0.001665 moles

Average concentration of H3PO4 = (0.001665 moles/100 mL)*1000 = 0.01665 M

2. No of moles of H3PO4 present in the original solution = 0.001665 moles. After third equivalence point 0.001665 moles of PO4-3 should be formed in the solution. From the reaction (3), one mole of NaOH reacts with one mole of HPO4-2 .

So 0.001665 moles of NaOH will react with 0.001665 HPO4-2 .

Amount of NaOH solution which contain 0.001665 moles of NaOH = (1000 mL/0.1025 moles) * 0.001665 moles = 16.24 mL

Total volume of NaOH required to reach the 3rd equivalence point = 32.55+16.24 = 48.79 mL

3. a.   From the first reaction:

pH = pKa1 + log [H2PO4-]/[ H3PO4]

pH and pKa1 will be same when [H2PO4-] = [ H3PO4], this happens when the titration reaches at the halfway towards the first equivalence point.

Amount of H3PO4 initially in the solution = 0.001665 moles, at the half equivalence point 0.001665/2 or 8.325*10-4 moles H3PO4 will react with 8.325*10-4 moles of NaOH.

So, 8.325*10-4 moles of NaOH were added when pH = pKa1.

b. Similarly for the second reaction, 8.325*10-4 moles H2PO4- will react with 8.325*10-4 moles of NaOH when pH = pKa2.

So, no of moles of NaOH added when pH = pKa2 is

= no of moles of NaOH required to reach the 1st equivalence point + 8.325*10-4 moles of NaOH

= 0.001665 + 8.325*10-4

= 0.002497 moles

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.