Find a pH of a 0.100 M solution of the weak acid HA. The pKa for the acid 5.0
...............HA ==> H^+ + A^-
I...............0.1.........0.......0
C..............-x..........x........x
E...........0.1-x........x.........x
Ka = (H^+)(A^-)/(HA)
Plug in values for Ka and the E line for the others. Solve for (H^+) which is x, then pH = -log(H^+)
Post your work if you get stuck.
To find the pH of a 0.100 M solution of the weak acid HA using the given pKa value, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this equation, [A-] represents the concentration of the conjugate base of the acid, and [HA] represents the concentration of the acid.
First, let's determine the concentration of the conjugate base [A-]. Since we have a 0.100 M solution of the acid HA, we can assume that the weak acid dissociates as follows:
HA ⇌ H+ + A-
In equilibrium, the concentration of [HA] is equal to the initial concentration of 0.100 M, and the concentration of [A-] will also be equal to the initial concentration of [H+]. Therefore, [A-] = [H+] = 0.100 M.
Next, let's substitute the values into the Henderson-Hasselbalch equation:
pH = 5.0 + log(0.100/0.100)
Simplifying,
pH = 5.0 + log(1)
Since log(1) = 0,
pH = 5.0 + 0
pH = 5.0
Therefore, the pH of a 0.100 M solution of the weak acid HA is 5.0.
To find the pH of a 0.100 M solution of the weak acid HA, you can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH is the pH of the solution
pKa is the dissociation constant of the weak acid
[A-] is the concentration of the conjugate base (in this case, the concentration of A- ions)
[HA] is the concentration of the weak acid (in this case, the concentration of HA molecules)
In this case, we are given that the pKa for the acid HA is 5.0.
Now, we need to determine the concentration of the conjugate base ([A-]) and the concentration of the weak acid ([HA]). Since we are given that the solution is 0.100 M, we can assume that the concentration of both the conjugate base and the weak acid are the same.
Therefore, [A-] = [HA] = 0.100 M.
Now, we can substitute these values into the Henderson-Hasselbalch equation:
pH = 5.0 + log(0.100/0.100)
Simplifying this equation, we get:
pH = 5.0 + log(1)
Since the logarithm of 1 is 0, we can simplify further:
pH = 5.0
Therefore, the pH of a 0.100 M solution of the weak acid HA with a pKa of 5.0 is 5.0.