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. 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. At pH 7.20 what is the ratio of [H2PO4-] to [HPO4-2]?

c. 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.}

Explanation / Answer

a)

charge balance:

[H+] + [K+] = [H2PO4-] + 2[HPO4-2] + 3*[PO4-3] +[OH-]

note that in real equilbirium, the values:

[OH-] << [H+] < H3PO4 < PO4-3

then, this can be simplified:

[K+] = [H2PO4-] + 2[HPO4-2]

b)

pH = 7.2, find ratio

pH = pKa + log(HPO4-2 / H2PO4-)

if pH = pKa, then

log(HPO4-2 / H2PO4-) = 1

[HPO4-2] / [H2PO4-] = 1

ratio is 1:1 os 1

c)

find mL of KOH for:

pH = 7.2 mmol of KOH --> half value of H3PO4 added

mol of H3PO4 = MV = 0.1

then we need --> 0.3 mol of KOH for total neutralization

we need only 50% neutralizatino so

0.3/2 = 0.15 mol of KOH required

V = mol/M = 0.15/1 = 0.15 L = 150 mL

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