The Ka for a weak acid, HA, is 1.6x10-6. What are the (a) pH and (b) degree of ionization of the acid in a 10-3 M solution? (c) Calculate pKa and pKb.
To answer part (a), we need to find the pH of a 10^-3 M solution of the weak acid HA. The concentration of the acid is given, and we can assume that the degree of ionization is small, so we can neglect the change in concentration of HA.
To find the pH, we need to calculate the concentration of H+ ions in the solution. Since HA is a weak acid, it partially dissociates in water to form H+ ions and A- ions. The equilibrium expression for the dissociation of HA is as follows:
HA ⇌ H+ + A-
The Ka expression for HA is given by:
Ka = [H+][A-] / [HA]
Since the initial concentration of HA is 10^-3 M, we can assume that the concentration of H+ ions is much smaller than the initial concentration of HA. Therefore, we can approximate the equilibrium concentration of H+ ions to be approximately equal to [H+].
Let's denote the concentration of H+ ions as [H+]. We can express the equilibrium concentrations as follows:
[H+] = [A-]
[HA] - [H+] ≈ [HA]
Since the equilibrium concentration of [A-] is equal to [H+], we can substitute these values into the Ka expression:
Ka = [H+][H+] / [HA]
Since [H+] = [A-], we can simplify this expression:
Ka = [H+]^2 / [HA]
Given that the Ka for HA is 1.6x10^-6, we can set up the following equation:
1.6x10^-6 = [H+]^2 / 10^-3
Now we can solve for [H+]:
[H+]^2 = 1.6x10^-9
[H+] = √(1.6x10^-9) ≈ 1.3x10^-5 M
The pH of a solution is defined as the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log[H+]
pH = -log(1.3x10^-5)
pH ≈ 4.9
So, the (a) pH of the 10^-3 M solution of HA is approximately 4.9.
To answer part (b), we need to calculate the degree of ionization of the acid in the 10^-3 M solution. The degree of ionization is defined as the ratio of the concentration of ionized acid (H+ ions) to the initial concentration of the acid (HA).
Degree of ionization = [H+] / [HA]
Degree of ionization = (1.3x10^-5) / (10^-3)
Degree of ionization ≈ 0.013 or 1.3%
So, the (b) degree of ionization of the acid in the 10^-3 M solution is approximately 1.3%.
To calculate the pKa, we can use the formula:
pKa = -log(Ka)
pKa = -log(1.6x10^-6)
pKa ≈ 5.8
The pKa value represents the negative logarithm (base 10) of the acid dissociation constant (Ka). The smaller the pKa value, the stronger the acid.
To calculate the pKb, we can use the relationship between the pKa and pKb values:
pKa + pKb = pKw
In water, pKw is equal to 14. So, we can rearrange the equation to solve for pKb:
pKb = pKw - pKa
pKb = 14 - 5.8
pKb ≈ 8.2
So, the (c) pKa is approximately 5.8 and the pKb is approximately 8.2.