Question

% dissociation for H₃PO₄

Trial 1:% dissociation for H₃PO₄=23.12%

Trial 2:% dissociation for H₃PO₄=20.6%

%dissociation of H₂PO₄^-

Trial 1: 0.84%

Trial 2: 0.695%

2.) Write the equation for the dissociation reaction where K = Ka1. Consider the % dissociation for H₃PO₄ in the initial sample of cola. Does this value suggest that the amount of [H₂PO₄^-] in solution is negligible prior to addition of NaOH? WHY or WHY NOT and how does that affect the accuracy of using the estimate pH_1/2 = pKa to establish Ka?

3.) Which Ka1 value calculated from your observations is closer to the literature value of Ka1 for H₃PO₄, the one obtained using initial pH and [H₃PO₄] OR the one obtained from V_1/2? Is this as you would expect based on your answer to question 2?

4.) Now consider the % dissociation of H₂PO₄^- at the first equivalence point. Write the equation for the dissociation reaction where K = Ka2. Is HPO₄^2-] negligible as titration of H₂PO₄^- begins? WHY OR WHY NOT?

Apply this information to a discussion of the accuracy of your observed Ka1 and Ka2 values.

5.) Although there are 3 moles of H⁺ in H₃PO₄ the titration only shows 2 equivalence points. Why is that? You may use a qualitative explanation from your text or information provided in lecture, but be explicit in terms of what would be necessary for a 3^rd equivalence point to be observed.

Answer 1

2.) Write the equation for the dissociation reaction where K = Ka1.

Ka = [H+][H2PO4-]/[H3PO4]

Consider the % dissociation for H3PO4 in the initial sample of cola.

In soda -->

Trial 1:% dissociation for H3PO4=23.12%

Trial 2:% dissociation for H3PO4=20.6%

AVG = (23.12+20.6)/2 = 21.86

then

pH = pKa1 + log(H2PO4-/H3PO4)

x = 21.86 --> 0.2186

assume 1 mol of H3PO4, so 0.2186 is H2PO4- and (1-0.2186) = 0.7814 H3PO4

pH = 2.13 + log(0.2186 /0.7814)

pH = 1.58

Does this value suggest that the amount of [H2PO4-] in solution is negligible prior to addition of NaOH?

No... it is not neglegible; it is actually a large % amount of H2PO4- in solution

WHY or WHY NOT and how does that affect the accuracy of using the estimate pH1/2 = pKa to establish Ka?

This affects because H2PO4- must be consodered iniitially

if in equilbiirum, we do not add the "initial" amount of H2PO4-; there will be large error