Question

A compound has two carboxyls and one aminium group. The pK_as 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

As the pH of the innitial solution is 5.3, with this we can know the original concentrations. I'll call it HA and A^- so:

With the HH equation we can solve this:

pH = pKa + log [A^-] / [HA]

5.3 = 4.4 + log [A] / [HA]

[A^-]/[HA] = 10^(5.3-4.4)

[A^-]/[HA] = 7.94

[A^-] = 7.94[HA]..........(1)

We know that the original concentration of this buffer is 0.1M:

0.1 = [A^-] + [HA] so replacing (1):

0.1 = 7.94[HA] + [HA]

0.1 = 8.94[HA]

[HA] = 0.0112 M

[A^-] = 7.94 * 0.0112 = 0.0889 M

Now that we knoe the original concentrations in order to get the pH of 5.3, let's see what happens when the NaOH is added:

HA + OH ------> A^- + H₂O

The OH^- is a base, and will going to further titrate the same A^-, so HH equation is modified, now to work with this better, let's see this in number of moles:

moles HA = 0.0112 M * 0.1 = 0.00112 moles

moles A^- = 0.0889 M * 0.1 = 0.00889 moles

moles OH^- = 0.1 * 0.04 = 0.004 moles

total volume = 140 mL or 0.140 L so:

As we are adding base, is logic to think in the fact that the pH will rise from 5.3, so we can use now the pKa3. And for the reaction we know that we have a buffer of HA + A so HH equation will be like this:

pH = pKA + log [OH] / [B]

so, for the buffer we have:

B = 0.00112 + 0.00889 = 0.01 moles ---> [B] = 0.01 / 0.140 = 0.0714 M

[OH] = 0.004 - 0.00112 = 2.88x10^-3 moles --> [OH] = 2.88x10^-3 / 0.140 = 0.0206 M

Finally the concentration of buffer will be: 0.0714 - 0.0206 = 0.0508 M

pH = 9.5 + log (0.0206 / 0.0508)

pH = 9.11

Hope this helps. In a comment tell me if there's something you want to be fixed or explained better.