You discovered a new diprotic acid, which you have named edmontonic acid. You want to characterize the pKa’s of edmontonic acid, so you perform a titration experiment. Your data are summarized in the table below. Your titration sample contains 150 mmol of edmontonic acid.
amount NaOH added (mmol) 180.0 and pH 4.168
amount of NaOH added 405.0
and pH 6.758
Determine the pKa’s of edmontonic acid. Show your work. (5)
To determine the pKa values of edmontonic acid, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where:
pH: The measured pH
pKa: The pKa value of the acid
[A-]: The concentration of the conjugate base
[HA]: The concentration of the acid
Let's start by calculating the concentration of the acid ([HA]) and the conjugate base ([A-]) for each data point.
For the first data point:
amount NaOH added = 180.0 mmol
Since edmontonic acid is diprotic, we need to calculate the concentration of each species at half-equivalence point. At half-equivalence, half of the acid has been neutralized and is converted to its conjugate base.
[HA] = 150 mmol - (180.0 mmol / 2) = 150 mmol - 90.0 mmol = 60.0 mmol
[A-] = 180.0 mmol / 2 = 90.0 mmol
pH = 4.168
Let's substitute the values into the Henderson-Hasselbalch equation:
4.168 = pKa + log (90.0 mmol / 60.0 mmol)
Now, we can solve for pKa:
4.168 = pKa + log (1.5)
log (1.5) ≈ 0.176091
4.168 ≈ pKa + 0.176091
pKa ≈ 4.168 - 0.176091
pKa ≈ 3.992
For the second data point:
amount NaOH added = 405.0 mmol
Using the same reasoning as before, at half-equivalence:
[HA] = 150 mmol - (405.0 mmol / 2) = 150 mmol - 202.5 mmol = -52.5 mmol (negative because excess base)
[A-] = 405.0 mmol / 2 = 202.5 mmol
pH = 6.758
Using the Henderson-Hasselbalch equation:
6.758 = pKa + log (202.5 mmol / -52.5 mmol)
log (2.5) ≈ 0.39794
6.758 ≈ pKa + 0.39794
pKa ≈ 6.758 - 0.39794
pKa ≈ 6.360
Therefore, the pKa values of edmontonic acid are approximately 3.992 and 6.360.
To determine the pKa values of edmontonic acid, we can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a solution to the pKa of the acid and the concentrations of the acid and its conjugate base. It is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the measured pH of the solution
- pKa is the dissociation constant of edmontonic acid
- [A-] is the concentration of the conjugate base of edmontonic acid
- [HA] is the concentration of edmontonic acid
In this case, we have two pKa values to determine, since edmontonic acid is diprotic. Let's calculate each pKa value individually:
1. First, we need to find the concentration of edmontonic acid at each point of the titration. We know that initially, we have 150 mmol of edmontonic acid. As we add NaOH, it reacts with the acid, forming its conjugate base. Therefore, at 180.0 mmol of NaOH added, we have consumed 180.0 mmol of edmontonic acid, leaving us with 150.0 mmol - 180.0 mmol = -30.0 mmol of edmontonic acid. At 405.0 mmol of NaOH added, we have consumed a total of 405.0 mmol of edmontonic acid, leaving us with 150.0 mmol - 405.0 mmol = -255.0 mmol of edmontonic acid.
2. Next, we can calculate the concentrations of the conjugate base [A-]. At 180.0 mmol of NaOH added, we can assume that all of the 180.0 mmol of NaOH has reacted and formed the conjugate base. Therefore, [A-] = 180.0 mmol. At 405.0 mmol of NaOH added, we can similarly assume that all of the 405.0 mmol of NaOH has reacted, giving us [A-] = 405.0 mmol.
3. Now, we can calculate the pH at each point of the titration. We are given pH values of 4.168 and 6.758 at the respective points. Let's denote these pH values as pH1 and pH2.
4. Using the Henderson-Hasselbalch equation, we can rearrange it to calculate the pKa values:
pKa = pH - log([A-]/[HA])
For the first pKa value, substituting in the values we have:
pKa1 = pH1 - log([A-]/[HA])
Similarly, for the second pKa value:
pKa2 = pH2 - log([A-]/[HA])
5. Now, let's substitute the known values and calculate the pKa values:
For pKa1:
pKa1 = 4.168 - log(180.0/-30.0)
pKa1 = 4.168 + 1.774
pKa1 ≈ 5.94
For pKa2:
pKa2 = 6.758 - log(405.0/-255.0)
pKa2 = 6.758 + 0.579
pKa2 ≈ 7.34
Thus, the pKa values of edmontonic acid are approximately 5.94 and 7.34.