Question

Phosphoric acid is a triprotic acid with the following pKa values: pKa1=2.148, pKa2=7.198, pKa3=12.375 You wish...

Phosphoric acid is a triprotic acid with the following pKa values: pKa1=2.148, pKa2=7.198, pKa3=12.375 You wish to prepare 1.000 L of a 0.0100 M phosphate buffer at pH 7.720. To do this, you choose to mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture? mass NaH2PO4? mass Na2HPO4? What other combination of phosphoric acid and/or its salts could be mixed to prepare this buffer? (Check all that apply). H3PO4 and NaH2PO4 H3PO4 and Na2HPO4 H3PO4 and Na3PO4 NaH2PO4 and Na3PO4 Na2HPO4 and Na3PO4. Please help me!! Thank you so much :)

Homework Answers

Answer #1

pKa1 = 2.148 = [H+][H2PO4 -] / [H3PO4]
2.148 = x^2 / 0.01 - x
x^2 + 2.148x - 0.02148 = 0
x = [H2PO4-] = 0.00995M

pH = pKa + log[Na2HPO4]/[NaH2PO4]
7.55 = 7.198 + log[Na2HPO4]/[NaH2PO4]
0.352 = log[Na2HPO4]/[NaH2PO4]
2.25 = [Na2HPO4]/[NaH2PO4]
you will need 2.25 x the amount of Na2HPO4 as NaH2PO4
x / y = 2.25
x + y = 0.01

2.25y + y = 0.01
3.35y = 0.01
y = 0.003M = [NaH2PO4]
x = 0.007M = [Na2HPO4]

0.003moles NaH2PO4 x 120g/mole = 0.36g
0.007moles Na2HPO4 x 139g/mole = 0.973g

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