An unknown compound, X, is though to have a carboxyl group with a pKa of 2.0 and another ionizable group with a pKa between 5 and 8. When 75 mL of 0.1M NaOH is added to 100 mL of a 0.1M solution of X at pH 2.0, the pH increases to 6.72. Calculate the pKa of the second ionizable group of X.

This is what i got so far 0.0075 mol Naoh is added
there is 0.01 moles of solution X , but because of the relationship pH=pKa there is 0.005 moles of carboxylic acid left? That's all i have so far. please help.

You may want a second opinion on this.

If you had 5 millimoles of the carboxylic acid then you must have titrated the rest of the COOH group with the 7.5 millimoles NaOH which leaves 2.5 mmoles NaOH to titrate the other acid group whatever that may be. You must have 10 mmoles of that still un-titrated (you had 10 millimoles of the original and that is gone but you still have 10 mmols of the other acid remaining. So you form 2.5 mmols of the salt, you have 10-2.5 = 7.5 mmoles of the other acid at pH 6.72.
6.72 = pK2 + log (B/A)
I would substitute 2.5 for base and 7.5 for acid (but check those numbers) and solve for pK2. I see that gives an answer between 5 and 8 which satisfies that part of the problem. Check my thinking.

Hi DrBob222,

Could you please explain want you typed:
*You must have 10 mmoles of that still un-titrated (you had 10 millimoles of the original and that is gone but you still have 10 mmols of the other acid remaining.

Are you saying I have 10 mmoles of each acid, if so how did you obtain 10 mmoles

Suppose you have 100 mL of 0.1M H2SO4. You titrate it with 0.1M NaOH.

It takes 10 mmols NaOH to neutralize the first H ion AND it takes 10 mmols of NaOH to neutralize the second H. So after you neutralize the first H you STILL have 10 mmols of H to neutralize. That's what I'm saying with the carboxylic acid + the other acid. After you've neutralized the COOH group you still have 10 mmoles of the "other" acid to neutralize. The only difference is that both H ions are listed together in H2SO4 but in your problem the COOH is one acid and there is another acid group separate from that. But it doesn't change the logic.

To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of an ionizable group and the concentrations of the acid and its conjugate base.

The Henderson-Hasselbalch equation is given by:

pH = pKa + log([A-]/[HA])

Where:
pH = the pH of the solution
pKa = the pKa value of the ionizable group
[A-] = the concentration of the conjugate base of the acid
[HA] = the concentration of the acid

In this case, the carboxyl group has a pKa of 2.0. We know that the initial pH of the solution is 2.0. When 75 mL of 0.1M NaOH is added, it reacts with the carboxyl group and forms its conjugate base, reducing the concentration of the acid [HA].

Let's calculate the initial concentration of the acid [HA] before NaOH is added:
[HA] = 0.1 M x 0.1 L = 0.01 mol

Since the pH of the solution increases to 6.72 after NaOH is added, we can use the Henderson-Hasselbalch equation to find the concentration of the conjugate base [A-]:

6.72 = 2.0 + log([A-]/0.01)
log([A-]/0.01) = 6.72 - 2.0
log([A-]/0.01) = 4.72

Now, let's calculate [A-]:

[A-]/0.01 = 10^4.72
[A-] = 10^4.72 * 0.01
[A-] ≈ 50.118 mol

At the same time, the NaOH also reacts with the second ionizable group of compound X, reducing its concentration. Let's represent the second ionizable group as [HB]. Since we don't know the exact pKa of this group yet, let's denote it as pKa2.

Now, let's calculate the concentration of [HB] after NaOH is added:
[HB] = 0.01 mol - 0.0075 mol (amount of NaOH reacted with the carboxyl group)
[HB] = 0.0025 mol

Using the Henderson-Hasselbalch equation with the new pKa (pKa2) and the concentrations of [HB] and [B-], we can solve for pKa2:

6.72 = pKa2 + log([B-]/[HB])
6.72 = pKa2 + log(50.118/0.0025)
6.72 = pKa2 + log(20047.2)

Rearranging the equation:

pKa2 = 6.72 - log(20047.2)

Calculating pKa2:

pKa2 = 6.72 - 4.301
pKa2 ≈ 2.42

Hence, the pKa of the second ionizable group of compound X is approximately 2.42.