A liter of pH 7.20 phosphate buffer is needed for a certain experiment. The Hend

Question

A liter of pH 7.20 phosphate buffer is needed for a certain experiment. The Henderson-Hasselbach equation will be sufficiently accurate for your determination of pH. The pK’s for possibly relevant phosphate species are:

H3PO4 H2PO4- + H+ pK = 2.15

H2PO4- HPO4-2 + H+ pK = 7.20

HPO4-2 PO4-3 + H+ pK = 12.4

0.100 Moles of H3PO4 were dissolved in about 800 mL of water, and the pH was adjusted to 7.20 using a standardized pH meter and a 1 M solution of KOH followed by addition of water to a volume of 1.00 L.

a. (10 pts) Write the equation for the charge balance to show how [K+], [H2PO4-], and [HPO4-2] are related. ([H+] and [OH-], although they are charged species, can be neglected for charge balance when compared to [K+], [H2PO4-], and [HPO4-2])

b. (10 pts) At pH 7.20 what is the ratio of [H2PO4-] to [HPO4-2]?

c. (20 pts) Combine your findings from parts a and b to determine how many mL of 1 M KOH must be added. {An important hint: By mass balance, the total molar concentration of phosphate species at pH 7.2 must equal the total molar concentration of phosphate species that was added from H3PO4.}

PLEASE ANSWER ALL PARTS, I NEED HELP. THANKS IN ADVANCE!

Explanation / Answer

1) the equations will be

H3PO4 --> H+ + H2PO4-

H2PO4- --> HPO4-2 + H+

HPO4-2 --> H+ + PO4-3

The neutralization reaction with KOH will be

H2PO4- + K + + OH- ---> KPO4-2 + H2O

b) according to Hendersen Hassalbalch equation the pH of a buffer is related to pKa as

pH = pKa + log [Salt] / [acid]

Here, [salt] = [HPO4-2 ] ; [acid] = [H2PO4-]

pKa = 7.2 (for this equilibrium)

so pH = 7.2 = 7.2 + log [salt] / [acid]

This means the [salt] = [acid] or [HPO4-2 ] = [H2PO4-]

c) For this

H3PO4 --> H+ + H2PO4-

Ka1 = [H+ ] [ H2PO4-] / [H3PO4 ] = antilog of -pKa = 0.0071

Let x amount of H3PO4 dissociates

[H+] = [H2PO4-] = x

given pH = 7.2

By mass balance

[H3PO4]0 = [H3PO4] + [H2PO4-] + [HPO4-2] +[ PO4-3]

We may ignore concentration of [PO4-3] as Ka <<1

given at pH = 7.2 , [H2PO4-] = HPO4-2]

[H3PO4 ]0 = [H3PO4] + 2 X [[H2PO4-] ]

so if we know the intial concentration of phosphoric acid we may calculate the volume of KOH

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