Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq) at 25 °C.
[H+]=
[ClO4-]=
[OH-]=
Bob Pursley did this for you two/three pages earlier and I told you how to do it about one page earlier. The only difference is I told you how to do it; Bob Pursley GAVE you the answers.
To calculate the concentrations of [H3O+], [ClO4-], and [OH-] in the given aqueous solution, we need to consider the dissociation of HClO4 in water.
The chemical equation for the dissociation of HClO4 in water is:
HClO4(aq) + H2O(l) -> H3O+(aq) + ClO4-(aq)
Since HClO4 is a strong acid, it fully dissociates in water. This means that for every 1 mol of HClO4 that dissolves, 1 mol of H3O+ and 1 mol of ClO4- is produced.
Given that the concentration of HClO4 in the solution is 0.130 M, we can immediately conclude that the concentration of H3O+ is also 0.130 M and the concentration of ClO4- is also 0.130 M.
Now, let's calculate the concentration of OH-.
In water, there is always a small concentration of H2O molecules that undergo autoionization, producing H3O+ and OH-. The autoionization of water can be represented as:
H2O(l) ⇌ H3O+(aq) + OH-(aq)
In pure water at 25 °C, the concentration of H3O+ and OH- is equal, which is given by 1.0 x 10^-7 M.
However, in the presence of HClO4, the concentration of H3O+ increases, and the concentration of OH- decreases due to the reaction between the added H3O+ and OH- ions.
To calculate the concentration of OH- in the given solution, we need to know the concentration of H3O+.
Since HClO4 is a strong acid and fully dissociates, we know that the concentration of H3O+ is 0.130 M.
As a result, the concentration of OH- can be calculated by using the following equation:
[H3O+][OH-] = Kw
Where Kw is the ion product of water, equal to 1.0 x 10^-14 at 25 °C.
Substituting the values, we can rearrange the equation to solve for [OH-]:
(0.130 M)[OH-] = 1.0 x 10^-14
[OH-] = (1.0 x 10^-14) / (0.130 M)
[OH-] ≈ 7.69 x 10^-14 M
Therefore, the concentrations in the given solution can be summarized as:
[H+] = 0.130 M
[ClO4-] = 0.130 M
[OH-] ≈ 7.69 x 10^-14 M