A particular weak acid (HA) with a Ka of 2.0e-5 is 75% titrated with a strong base to produce a buffer solution. What is the pH of this final solution? Show all work.
Use the HH equation.
I would start with any convenient number for the initial concentration of HA. Any number will work but I would choose 1 to make it easy.
So if we start with 1M HA it will be 0.25 at 75% titrated and the salt (the base) will be what? Plug those values into the HH equation and solve for pH.
To find the pH of the final solution, we need to understand that a buffer solution is one where the concentration of the acid and its conjugate base are approximately equal. In this case, since 75% of the weak acid (HA) is titrated with a strong base, we have converted 75% of HA into its conjugate base (A-).
Let's denote the initial concentration of the weak acid as [HA]0, and the concentration of the conjugate base as [A-]0. Since 75% of HA is converted, the concentration of HA remaining is 25% of the initial concentration, which can be written as 0.25 * [HA]0. The concentration of A- can be found using the stoichiometry of the reaction, so it would be 0.75 * [HA]0.
The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentrations of the acid and its conjugate base, and it is given by:
pH = pKa + log([A-]/[HA])
Where pKa is the negative logarithm of the acid dissociation constant (Ka).
In this case, the Ka is given as 2.0e-5, so the pKa would be:
pKa = -log(2.0e-5)
Now, let's substitute the values we found into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
= -log(2.0e-5) + log((0.75 * [HA]0) / (0.25 * [HA]0))
Simplifying this expression, we get:
pH = -log(2.0e-5) + log(3)
Using the properties of logarithms, we can simplify it further:
pH = -log(2.0e-5) + log(3)
≈ -(-4.70) + 0.48
≈ 4.70 + 0.48
≈ 5.18
Therefore, the pH of the final solution is approximately 5.18.