pKa: 2.148 pKa2 = 7.198 pKa3 = 12.375 You wish to prepare 1.000 L of a 0.0500 M
ID: 838762 • Letter: P
Question
pKa: 2.148 pKa2 = 7.198 pKa3 = 12.375
You wish to prepare 1.000 L of a 0.0500 M phosphate buffer at pH 7.600. 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 NaHPO4
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
Explanation / Answer
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
check this, i'm not 100% sure of thsi
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