One compound contains an amino group with $\mathrm{p}K_\mathrm{a}$ of $8.8$, and one other ionizable group with a $\mathrm{p}K_\mathrm{a}$ of $5$ to $7$. To $100\ \mathrm{mL}$ of a $0.2\ \mathrm{M}$ solution of this compound at $\mathrm{pH}\ 8.2$ was added $40\ \mathrm{mL}$ of a solution of $0.2\ \mathrm{M}\ \ce{HCl}$ and the $\mathrm{pH}$ changed to $6.2$. What is the $\mathrm{p}K_\mathrm{a}$ of the second ionizable group?

I am a little confused because I am not sure what to do when the $\mathrm{pH}$ does not equal the $\mathrm{p}K_\mathrm{a}$ of the amino group. I'm thinking since the known $\mathrm{p}K_\mathrm{a}$ of $8.8$ is highest at that point there should be no free $\ce{H+}$.

  • 3
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Since you're given that the maximum $\mathrm{pK_a}$ of the second group is $7$, you can assume that the amino group will fully protonate before the second group starts accepting electrons. (Perhaps this is what you meant by "at that point there should be no free $\ce{H+}$"?)

Therefore, from Henderson-Hasselbalch$^\ddagger$, you can calculate the fraction of the amino group that is protonated based on the starting $\mathrm{pH}$ and the amino group $\mathrm{pK_a}$. (Hint: Yes, the amino group should be mostly, but not entirely, protonated.)

It's straightforward to then calculate:

  1. The total number of moles of acid added
  2. The number of moles of not-yet-protonated amino groups before adding the acid

which is equal to

  1. The number of moles of acid consumed to protonate the rest of the amino groups

and, therefore, you can find

  1. The number of moles of acid left over to protonate a portion of the second ionizable group in the molecules.

Having calculated #4, you can then calculate what fraction of the second ionizable group is protonated by the leftover acid. This feeds back into Henderson-Hasselbalch, from which you can calculate the second $\mathrm{pK_a}$ value since you are told the value of the final $\mathrm{pH}$.

$^\ddagger\ \mathrm{pH}=\mathrm{pK_a}+\log{\left({\left[\ce{A-}\right]\over\left[\ce{HA}\right]}\right)}=\mathrm{pK_a}+\log{\left({\left[\ce{A_{\mathrm{total}}}\right]-\left[\ce{HA}\right]\over\left[\ce{HA}\right]}\right)}=\mathrm{pK_a}+\log{\left({1\over f_\mathrm{protonated}}-1\right)}$



1. HH Equation:

1a.) 8.2=8.8+log(B/A)

1b.) solving for ratio of B:A= (B/A)=.251

1c.) %B= (.251/.351)*100%= 20%, thus %A= 80% (.351 is the TOTAL amount of moles for acid and base) it comes from the equation (x/(x+1))

2.Amount Reacted:

KNOWN: .1L, .2M of compound

2a.) .1L*.2M= .02 moles

By using percentages, we have

2b.) .8*.02= .004 moles of base .2*.02= .016 moles of acid

KNOWN: .04L, .2M of HCl

2c.) .04L* .2M= .008 moles HCl

.004 moles of base will get protonated by .004 moles of H+ from HCl

This forms .004 moles of acid

Adding the original acid concentration gives .02 moles of acid

There are still .004 moles of H+ left

This can protonate the acid

This can occur because this is an ionizable group with pKa from 5-7

See top of thread to see original problem

Thus, .004 moles of the H+ protonates .004 of the .02 moles of the original acid

This is your new Base to Acid ratio for the HH eq. (.016/.004)

The .016 comes from .02mol original acid - .004 mol H+

The .004 on the denominator comes from .004 mol formation of the new acid


3a.) 6.2 (given)=pKa + log(.016/.004)

3b.) Solve for pKa

  1. Thus the pKa of the second ionizable group is 5.6

Hopefully this is easy to understand

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pH=pKa + log ([A]/[HA])
pH goes to 6.2
8.2=8.8 + log([A]/[HA])
8.2-8.8 = log ([A]/[HA])
the difference is negative but since it represents just "the difference" we can use positive.
.06= log ([A]/[HA])
so using the new pH,
6.2= pKa + .06
6.2-.06= pKa
pKa= 5.6

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    $\begingroup$ At minimum, this doesn't seem to account properly for the differing volumes of solution used. $\endgroup$ – hBy2Py Oct 4 '15 at 15:08
  • $\begingroup$ If my mental math is correct, this gives a correct numerical result, but the approach is not generally valid. It only works in this case by the happenstance of the particular numerical values (concentrations, volumes, dissociation constants, etc.) used in the problem statement. $\endgroup$ – hBy2Py Oct 4 '15 at 15:55

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