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I am struggling with a task. I have the amino acid Glycin with pKa1=2,3 and pKa2=9,6.

I have 1.0 liter 0.20 mol/L solution of glycin. I add 0.1 liter 1 mol/L HCl solution. What is the pH in this buffer.

I have calculated that the isoelectric point is 5.95 for this amino acid. Using the equation pH=pKa + log [base]/[acid], where do I go from here? Should the pKa used be 5.95?

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    $\begingroup$ 5.95 is pI, not pKa. As pKa2 for R-NH3+ is so high, you can formally consider it just as a salt R-COO- with pKa 2.3, resp. pKb 11.7, with the reaction R-COO- + H+ <=> R-COOH. $\endgroup$
    – Poutnik
    Commented Mar 18 at 4:06
  • $\begingroup$ Are you asking about glycin or glycine? You write the former, but provide $\mathrm{p}K_\mathrm{a}$ values for the latter. $\endgroup$
    – andselisk
    Commented Mar 20 at 17:28

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Glycine in aqueous neutral solution (i.e. add water to glycine) occurs as zwitter ion: enter image description here

The $\mathrm{p}K_\mathrm{a}$ of the ammonium group is 9.6. The $\mathrm{p}K_\mathrm{a}$ of the carboxylic acid is 2.3 (quite different from acetic acid, which is around 4.7). Given that the carboxylic acid group is a stronger acid than the ammonium group, the zwitter ion forms, as shown above.

Adding HCl to the zwitter ion will protonate the carboxylate, so the $\mathrm{p}K_\mathrm{a}$ of 2.3 is relevant (the ammonium is already protonated, and remains protonated as you add strong acid). In the problem, you are adding just enough HCl to protonate half of the carboxylate. The result is a 1:1 buffer, a mixture of the zwitter ion (net neutral) and the twice protonated species (net charge of +1).

With this information, you should be able to plug in concentrations and $\mathrm{p}K_\mathrm{a}$ into the buffer equation (the Henderson-Hasselbalch equation).

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