You need to determine the pKa of a weak, monoprotic acid. You add 0.500 moles of the acid to 1.00 L of water. The resulting pH was 1.87. What is the pKa of this weak acid? You may ignore the autoionization of water.
2.74
3.43
5.38
2.01
1.87
......HA ==> H^+ + A^-
I.....0.5....0......0
C.....-x.....x......x
E...0.5-x....x......x
Enter the E line into Ka expression and solve for Ka. The problem tells you the pH which will give you the (H^+) and (A^-). Then convert to pKa.
To determine the pKa of a weak acid, we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of an acid. The Henderson-Hasselbalch equation is:
pH = pKa + log([A-]/[HA])
Where pH is the measured pH of the solution, pKa is the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.
In this case, you are given the pH of the solution, which is 1.87. We can assume that the concentration of the conjugate base ([A-]) is negligible compared to the concentration of the acid ([HA]) because the acid is not completely dissociated. Therefore, we can rewrite the Henderson-Hasselbalch equation as:
pH = pKa + log(1/[HA])
To find the pKa, we need to solve for it. Rearrange the equation to isolate pKa:
pKa = pH - log(1/[HA])
Substitute the given values into the equation:
pKa = 1.87 - log(1/[HA])
Since you are given that 0.500 moles of the acid are added to 1.00 L of water, the concentration of the acid ([HA]) can be calculated as:
[HA] = 0.500 moles / 1.00 L = 0.500 M
Substitute [HA] into the equation:
pKa = 1.87 - log(1/0.500)
Calculate the log:
pKa = 1.87 - log(2)
Using a scientific calculator or logarithm table, find the logarithm base 10 of 2. It is approximately 0.301.
pKa = 1.87 - 0.301
pKa = 1.57
Therefore, the pKa of this weak, monoprotic acid is approximately 1.57.