Achieve 2 #9/10

[Dalia Torres 3I]

Can someone please explain how to figure these problems out or at least where to start?? I’ve red other explanations but I’m having a hard time trying to grasp these problems. Thank you!!

[Sam Leistiko 2B]

You can think about this question in two ways, either quantitatively using equations, or conceptually using Le Chatelier's principle.

To think about this quantitatively, we must use the Henderson-Hasslebach equation. The HH equation is as follows: pH = pKa + log ([A-] / [HA]).
For bases, it is: pOH = pKb + log ([BH+]/[B]).

For problem 9, we are given that the pH of the solution is 5.54, and that the pKa of the acid is 4.31. If we plug this into the HH equation, we get:

5.54 = 4.31 + log([A-]/[HA]). A simple algebraic manipulation yields: 1.23 = log([A-]/[HA]). Based on how the log function works, we know that for log(x) to be a positive number, x must be greater than 1. In this case, log(x) = log([A-]/[HA]), and x = [A-]/[HA]. So, for log([A-]/[HA]) to be greater than 1, [A-]/[HA] must be greater than 1, which means that [A-] must be greater than [HA]. Therefore, in this case, the charged species is more prevalent.

Another way to think about this is using le chatelier's principle. I want to warn that this explanation does not perfectly capture the underlying concept, but can be useful in trying to solve this type of problem. Because the pH is higher than the pKa, you can essentially think of this situation as though the acid is in a solution, but the solution is less acidic than it would be if the acid had been placed in pure water. This means that there is a lower concentration of H3O+ ions in this solution than the acid wants there to be. So, because [H3O+] is lower, and [H3O+] is the product of acid dissociation in water, this means that Q < K and equilibrium will shift right, favoring the dissociation of the acid into charged [A-] and [H3O+].