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.
...................HA ==> H^+ + A^-
I.............0.001 M.......0.........0
C................-x.............x..........x
E............0.001-x.........x..........x
Plug the E line into the Ka expression and solve for x = (H^+). Convert H^+ to pH the usual way. You did that in the question just below.
degree of ionization = (H^+)/(HA) in the E line above.
pKa = -log Ka and you have Ka.
You get pKb from pKa + pKb = pKw = 14. You know pKa and pKw, solve for pKb. Post your work if you get stuck.
how did we get the molarity (0.001)
I did not know how to solve the question because there is no molarity can you explain please.
To know more knowledge and understand chemistry
To solve these questions, we need to understand the concepts of Ka, pH, degree of ionization, pKa, and pKb.
(a) pH calculation:
The pH of a solution can be determined using the concentration of hydrogen ions ([H+]). In the case of a weak acid, we can use the degree of ionization to calculate the concentration of [H+] ions.
The equation for the ionization of a weak acid HA is:
HA ⇌ H+ + A-
The equilibrium constant (Ka) is given as 1.6 x 10^-6, which is the ratio of the concentrations of the products ([H+][A-]) to the concentration of the reactant ([HA]).
Ka = [H+][A-] / [HA]
Since the concentration of A- can be assumed negligible compared to HA, we can simplify the equation to:
Ka = [H+] / [HA]
Given that the initial concentration of HA is 10^-3 M, we can denote the degree of ionization as 'x'. This means [HA] will decrease by 'x' and [H+] will increase by 'x'.
So, after ionization:
[H+] = 'x'
[HA] = 10^-3 - 'x'
Substituting these values into the equation for Ka:
1.6 x 10^-6 = 'x' / (10^-3 - 'x')
To solve this equation, we can make the approximation that 'x' is much smaller than 10^-3:
1.6 x 10^-6 ≈ 'x' / (10^-3)
1.6 x 10^-3 ≈ 'x'
x ≈ 1.6 x 10^-3
So, the concentration of [H+] after ionization is approximately 1.6 x 10^-3 M.
The pH of a solution is given by the equation:
pH = -log[H+]
Using the value of [H+], we can calculate the pH:
pH = -log(1.6 x 10^-3)
pH ≈ 2.8
Therefore, the pH of the 10^-3 M solution of the weak acid HA is approximately 2.8.
(b) Degree of ionization (α) calculation:
The degree of ionization represents the extent to which a weak acid dissociates into ions in a solution. It is calculated by dividing the concentration of [H+] ions after ionization by the initial concentration of the weak acid [HA].
α = [H+] / [HA]
Using the values we obtained previously:
α = (1.6 x 10^-3) / 10^-3
α = 1.6
The degree of ionization (α) cannot exceed 1, so we can say that the degree of ionization of the 10^-3 M solution of the weak acid HA is approximately 1.
(c) pKa and pKb calculation:
pKa is the negative logarithm of the acid dissociation constant Ka, while pKb is the negative logarithm of the base dissociation constant Kb. In the case of a weak acid like HA, the pKa value is related to its strength.
pKa = -log(Ka)
Using the given value of Ka:
pKa = -log(1.6 x 10^-6)
pKa ≈ 5.8
Since HA is an acid, it does not have a Kb value and, therefore, there is no pKb to calculate.
Therefore, the pKa of the weak acid HA is approximately 5.8.