Titration Question

[1070752]

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I'm completely stuck on this question, not sure how to go about solving it. Any help would be appreciated.

1. 40 mL of 0.10 M H3PO4 are mixed with 5 mL of 0.20 M NaOH and the mixture brought
up to a total volume of 100 mL with distilled water. (pKas for phosphoric acid are: 2.1,
6.7 and 12.3)

a) What is the pH of the resulting solution?

 

[theonlytycrane]

It's a buffer system:

H3PO4 -> H2PO4- + H+ | pKa = 2.1
H2PO4- -> HPO42- + H+ | pKa = 6.7
HPO42- -> PO43- + H+ | pKa = 12.3

Use Molarity = mol / Volume to find that you have .004 mol of H3PO4. You react this with .001 mol of -OH which results in .003 mol H3PO4 (acid) and .001 mol H2PO4- (conjugate base).

Use the h.h. equation: pH = pKa + log([A-] / [Ha]). Use pKa = 2.1 since we're still in that first dissociation of the buffer system. The log term will be log(.001 / .003) which will be slightly negative, so choose the answer choice given with a pH a little bit less than 2.1.

 

[1070752]

It's a buffer system:

H3PO4 -> H2PO4- + H+ | pKa = 2.1
H2PO4- -> HPO42- + H+ | pKa = 6.7
HPO42- -> PO43- + H+ | pKa = 12.3

Use Molarity = mol / Volume to find that you have .004 mol of H3PO4. You react this with .001 mol of -OH which results in .003 mol H3PO4 (acid) and .001 mol H2PO4- (conjugate base).

Use the h.h. equation: pH = pKa + log([A-] / [Ha]). Use pKa = 2.1 since we're still in that first dissociation of the buffer system. The log term will be log(.001 / .003) which will be slightly negative, so choose the answer choice given with a pH a little bit less than 2.1.

I have two questions:

1. Aren't you supposed to use concentration for the H-H equation, not moles?
2. Why are we only using pKa of 2.1? (shouldn't it be the average pH of all the pKa's? Or are we using 2.1 because the pH of 0.1 M H3PO4 is 1 and that's the closest?)

 

[theonlytycrane]

1. [A-] / [HA] is a ratio so the volume cancels out. Both ways work but you can directly use the mol ratio for a shortcut.

2. We use the pKa of the acid / base pair in the buffer system. The acid is H3PO4 and the conjugate base is H2PO4-. This pair corresponds to the first line with pKa = 2.1.

 

[1070752]

1. [A-] / [HA] is a ratio so the volume cancels out. Both ways work but you can directly use the mol ratio for a shortcut.

2. We use the pKa of the acid / base pair in the buffer system. The acid is H3PO4 and the conjugate base is H2PO4-. This pair corresponds to the first line with pKa = 2.1.

Ah I see, thanks a lot! Would it be done differently if you did the second pKa where H2PO4 was the acid?

 

[theonlytycrane]

In this example we only added .001 mol of -OH. But say we added .005 mol of -OH instead. This would push the acid-base reaction all the way to H2PO4- (.003 mol at the end) and HPO42- (.001 mol at the end). In this case just make sure to keep track of which part of the buffer system we're in (we would use the second pKa in the h.h. equation).

 

[1070752]

In this example we only added .001 mol of -OH. But say we added .005 mol of -OH instead. This would push the acid-base reaction all the way to H2PO4- (.003 mol at the end) and HPO42- (.001 mol at the end). In this case just make sure to keep track of which part of the buffer system we're in (we would use the second pKa in the h.h. equation).

Thank you!