Homework Answers
Accoring to the Henderson-Hasselbulch equation:
pH = pKa + Log([Tyr.O-]/[Tyr.OH])
i.e. 8.5 = 8.8 + Log([Tyr.O-]/[Tyr.OH])
i.e. Log([Tyr.O-]/[Tyr.OH]) = 8.5-8.8 = -0.3
i.e. [Tyr.O-]/[Tyr.OH] = 10-0.3 = 0.5
Now, [Tyr.O-]/([Tyr.O-] + [Tyr.OH]) = 0.334
Therefore, the required percentage = 0.334 * 100 = 33.4%
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1
2
3
4
5
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