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smithemm smithemm
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12 years ago
Use pKa = 2.35 and pH = 7.00

use the henderson-hasselbalch equation.
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#1
12 years ago
pH = pKa + log ([A-]/[HA])
7.00 = 2.35 + log ([A-]/[HA])
4.65 = log ([A-]/[HA])
10^4.65 = ([A-]/[HA]) = 45,700/1

This is not a good choice to make a buffer of pH = 7.00 because there is such a small amount of HA present. It wouldn't take much base to exceed the capacity of the buffer. Ideally, try to choose a conjugate pair that has a pKa close to the desired pH so that the ratio of base/acid is close to 1/1.
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