Question

Part A - pH of a Compound That Contains Two Carboxyls and One Aminium Group A compound has two carboxyls and one aminium group. The pKas of the groups are 2.0, 4.4, and 9.5, respectively. A biochemist has 100.0 mL of a 0.10 M solution of this compound at a pH of 5.3. She adds 40.0 mL of 0.10 M NaOH. To the nearest hundredth of a unit, what will be the pH of the solution after addition of the NaOH (assume full stirring to reach a new equilibrium)?

Answer 1

we have 2 carbonyl and one aminium group

pka carbonyl1 = 2.0

pka carbonyl 2= 4.4

pka aminium gruopu = 9.5

100mL of a 10 M solution pH= 5.3

+ 40. mL NaOH 0.10 M pH= ?

You have 3 pKa's. With luck, only one is relevant at a given time. But it should be clear which one is relevant. In particular. note that the last 2 are very far apart.

So, first, which pK is the relevant one?

Then you can use Henderson-Hasselbalch (HH) to find what is present at the start.

Now add the base. How much? Is it simply going to further titrate the same COO- ? If so, add it in, and use HH again.

If there is excess base, then you will need to move on to the next pKa.

You can only tell by looking at the numbers.

The given pH is closest to the 4.4 pKa. The lower pK is not of interest. The higher one might be, but we are far from it -- so far.

As a rule of thumb, a pKa is relevant if we are within about 1 pH unit of it. Maybe 2, if we want to stretch. The basis of that comes from HH.

So for now, all you need is that one pKa.

we have the HH equation

ph=pka+log(A-/HA)

You also know...

100.0 mL of a 0.10 M solution of this compound then the moles of the Acid are 0.10M*0.1L= 0.01 moles

and of the NaOH 40 mL 0.10 M. the moles of this will be 0.004 moles

Now we calculate the new concentration of each compound with the total volumen of solutio 0.140L

[HA] = 0.07 M and the [NaOH] = 0.028 M = [A-]

Then we calculate the pH using the pKA of 4.4

pH= 4.4 + log (0.028/0.07)

pH = 4.4 + log (0.4)

pH = 4.4 -0.24

it makes no sense that if we start with a pH of 5.3 when we add OH the pH goes down, so we use the equation with the next pKa value

pH= 9.5 - 0.24

pH = 9.26

Then the pH goes from 5.3 to 9.26 when we add 40 mL of NaOH 0.10 M