Determine the pH of a .10M solution of KA, the potassium salt of HA. For HA, Ka=1.0x10^-5
To determine the pH of a solution of KA, the potassium salt of HA, we need to consider the dissociation of HA in water.
The given information is the equilibrium constant (Ka) of the acid HA, which is 1.0x10^-5. This indicates that HA is a weak acid, as the Ka value is less than 1.
When HA dissolves in water, it gives rise to an equilibrium as follows:
HA ⇌ H+ + A-
In this equilibrium, HA acts as the acid, and H+ is the hydrogen ion (proton) released by the acid. A- is the conjugate base of HA.
Since KA is the potassium salt of HA, it means that A- is the conjugate base of the acid HA. The presence of a conjugate base in the salt suggests that the solution will be basic rather than acidic.
To find the pH of a basic solution, we need to determine the concentration of hydroxide ions (OH-) and then convert it to pH using the relationship: pH = -log10[OH-].
Since KA is a salt, it completely dissociates in water, into K+ and A-. Therefore, the concentration of A- ions will be equal to the concentration of the KA solution, which is 0.10 M in this case.
However, in a basic solution, the OH- ions come from the dissociation of water, which undergoes auto-ionization.
H2O ⇌ H+ + OH-
Since water is a neutral substance, the concentrations of H+ and OH- ions in pure water are equal, and each is 1.0x10^-7 M. However, in a basic solution, the concentration of OH- increases.
Now, to determine the concentration of the OH- ions in the solution, we need to apply the concept of water auto-ionization and the relationship:
Kw = [H+][OH-]
Given that Kw is 1.0x10^-14 at 25°C, we can rearrange the equation to solve for the OH- concentration as follows:
[OH-] = Kw / [H+]
Substituting Kw = 1.0x10^-14 and [H+] = 1.0x10^-7, we find:
[OH-] = 1.0x10^-14 / 1.0x10^-7 = 1.0x10^-7 M
Since we previously determined that the concentration of A- ions in the KA solution is 0.10 M, we can conclude that the concentration of OH- ions in the solution is 1.0x10^-7 M.
Finally, to calculate the pH of the solution, we use the relationship:
pH = -log10[OH-]
pH = -log10[1.0x10^-7]
pH ≈ 7.00
Therefore, the pH of a 0.10 M solution of KA is approximately 7.00, indicating that it is neutral.