Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO-].
I have an initial pH of 2.2
Thanks
Do you know how to solve
pH = -log(H^+) = 2.2 for (H^+). (CH3COO^-) will be the same as (H^+).
To find the initial [H3O+] and [CH3COO-] of an acetic acid solution based on its initial pH of 2.2, we need to consider the acid dissociation constant (Ka) of acetic acid.
The chemical equation for the dissociation of acetic acid (CH3COOH) in water is:
CH3COOH ↔ H3O+ + CH3COO-
The Ka expression for this reaction is:
Ka = [H3O+][CH3COO-] / [CH3COOH]
Since acetic acid is a weak acid, it does not dissociate completely in water. However, we can assume that the initial concentration of acetic acid is nearly equal to its undissociated concentration.
To determine the initial [H3O+] and [CH3COO-], we need to use the general equation relating pH and [H3O+]:
pH = -log[H3O+]
Given that the initial pH is 2.2, we can calculate the initial [H3O+] using the equation:
2.2 = -log[H3O+]
Rearranging the equation gives:
[H3O+] = 10^(-pH)
Substituting the pH value:
[H3O+] = 10^(-2.2)
To find the initial [CH3COO-], we can use the equation for Ka and the [H3O+] value:
Ka = [H3O+][CH3COO-] / [CH3COOH]
Since the initial concentration of acetic acid is nearly equal to the undissociated concentration, we can assume the initial [CH3COO-] can be approximated as equal to the initial [H3O+]:
[CH3COO-] ≈ [H3O+]
Therefore, the initial [CH3COO-] is also approximately 10^(-2.2).
So, the initial [H3O+] is approximately 10^(-2.2) and the initial [CH3COO-] is also approximately 10^(-2.2).