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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 Preview

    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 ...

    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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