Question: Compound X has two ionizable groups, with one pK_a of 3.8 and a second

Question

Question: Compound X has two ionizable groups, with one pK_a of 3.8 and a second pK_a between 7 and 10. A biochemist has 10 ml of a 1.0 M solution of compound X, at a pH of 7.9. She adds 10 ml of 1.0 M HCl, which changes the pH to 3.2. What is the pK_a of the second ionizable group?

Second pK_a=?

Answer: 8.5

8 ml of the HCl was used to titrate the pK_a 3.8 group; so 2 ml was enough to titrate the rest of the high pK_a group. The upper pK_a group must have begun the experiment as 80% acid form, 20% base form.

Would somebody please walk me through this?

Explanation / Answer

First let us calculate the amount of acid in your sample:

100mL * L/1000mL * .1mol/L = .01 mol acid

Okay, you are given that you are to assume that the acid is diprotic (two H+ ions per molecule). That means it has a first ionizable H+ ion with pKa1 and a second with pKa2.

Therefore, you have .02 mol potential H+ ions.

Now, there are two basic principles you need to know to do this problem:

First, pH = pKa when [H+] = [AH]. This means, when half of the H+ ions have been removed the pH = pKa.

Also, if you have a mixture of acids with very different K values and similar concentrations, you only pay attention to the strongest until it is gone. A diprotic acid can be considered two acids, one much stronger than the other.

Using these two pieces of information, you know that half of your first protons have been removed (pH = pKa1) BEFORE YOU STARTED THE TITRATION! That was the really confusing part.

Therefore, you lose .005 mol potential H+ ions (half of the first .01 mol H+ ions). So you now have .015 mol H+ ions left in solution or bonded to your acid molecules.

Now you are ready to add the NaOH. When you do, you find that it removes an additional .0075 mol H+ (.075L + .1M NaOH = .0075 mol OH-), leaving you with .0075 mol H+ still available to the solution. All of this comes from your second proton, because the .01 mol H+ from the first is now gone.

******

The next step is to calculate the amount of (HA)- and (A)2- so that you can do a RICE. 0.0125 mol of H+ ions have been removed from the initial .01 mol H2A, so all of the first H+s have been removed and .0025 mol of the second H+ have been removed. That means that under "initial" condidtions (for RICE), there are 0.0025 mol (A)2-. This leaves 0.0075 mol (HA)-.

.075 L of NaOH was added to our .1 L of acid solution, and we assume that volumes are additive. Therefore we have .175 L of final solution.

Since final concentrations are needed for RICE, we divide amount of (HA)- and (A)2- by the volume (M = mol/L):

[(HA)-] = .0075 mol / .175 L = .0428 M
[(A)2-] = .0025 mol / .175 L = .0143 M

*****

Now it is a simple (or at least simpler) matter to find the Ka2.

R (HA)- <--> (H)+ + (A)2-
I .043 M 0 M .014
C -x +x +x
E .043-x x .014+x

(That's a RICE, but the formatting makes it look weird.)

pH = -log[H+] = -log(x), so x = 10^-pH = 10^-(6.72) = 1.9 E-7

Therefore x is [H+] = 1.9 E-7

So Ka2 = x(.014+x)/(.043-x) = (1.9E-7)(.014)/(.043) = 6.18 E-8

pKa2 = -log(Ka2) = -log (6.18 E-8) = 7.21

This fits in the range and is a good value for a pKa2.


Note:
Be sure to always pay attention to units when doing chem or physics. Liters times Molarity cannot be Molarity. Also remember that Molarity is mol/L, not mol*L.

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