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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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fieldhockey32 fieldhockey32
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11 years ago
The pKa for the dissociation of H3PO4 is 2.15. What is the concentration of H2PO4-1 (in M) at pH 3.21 if the original concentration of the phosphate was 2.37 M?
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#1
11 years ago
Using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[AH])
Substituting,
3.21 = 2.15 + log([A-]/[AH])
Solving,
[A-]/[AH] = 11.48, or
[A-] = 11.48[AH].
Since the original concentration was 2.37, then:
[A-] + [AH] = 2.37.
Substituting,
11.48[AH] + [AH] = 2.37 M
Solving,
[AH] = 0.19 M.
Therefore,
[A-] = [H2PO4-] = 2.37 M - 0.19 M = 2.18 M.
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