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Electrochemistry of Histidine Titration

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Consider the neutral form of the amino acid Histidine, abbreviated HA. Given a pH of 3.00, how many mL of 0.0500M HClO4 should be added to 25.0 mL of 0.0400M HA?

(hint: H_3A^2+ <--> H_2A^+ <---> HA <---> A^-)

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Solution Summary

The solutions finds the amount acid needed to change the neutral form histidine to the other form, where one of the nitrogen in the ring is protonated.

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Consider the neutral form of the amino acid Histidine, abbreviated HA. Given a pH of 3.00, how many mL of 0.0500M HClO4 should be added to 25.0 mL of 0.0400M HA?
(hint: H_3A^2+ <--> H_2A^+ <---> HA <---> A^-)

Form C is the neutral form, referred to as HA in the equation above. In this form C, we can see that all the NH2 from the chain is in NH3+ form.
Concentration of this neutral form = 0.025 L x 0.04 M = 0.001 moles of HA
At pH =3, what we have is ...

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