Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq) at 25 °C.
[H+]=
[ClO4-]=
[OH-]=
HClO4 is a strong acid that ionizes 100%; therefore, H3O^+ = HClO4.
For OH, remember (H^+)(OH^-) = Kw = 1E-14
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 gave you the answers two/three pages earlier or you can follow my instructions to get the answers. I'll show you how but I won't give them to you.
To calculate the concentrations of [H3O+], [ClO4-], and [OH-] in the given aqueous solution, we first need to understand the dissociation reaction of HClO4 in water. HClO4 is a strong acid, so it completely dissociates in water.
The dissociation reaction of HClO4 can be represented as follows:
HClO4(aq) → H+(aq) + ClO4-(aq)
From this equation, we can conclude that the concentration of [H+] will be equal to the concentration of [H3O+] because HClO4 dissociates to form H+ ions. Therefore, [H+]=[H3O+].
Given that the solution is 0.130 M in HClO4, the concentration of [H3O+] or [H+] is also 0.130 M.
Now, let's move on to calculate the concentration of [ClO4-]. Since all of the HClO4 dissociates completely in the solution, the concentration of [ClO4-] will be the same as the initial concentration of HClO4, which is 0.130 M.
Finally, let's calculate the concentration of [OH-]. To do this, we need to use the concept of the ionic product of water (Kw). Kw is defined as the product of the concentrations of [H3O+] and [OH-], and at 25 °C, its value is 1.0 x 10^-14 M^2.
So, we have [H3O+]=0.130 M. By rearranging the equation for Kw, we get:
Kw = [H3O+][OH-]
1.0 x 10^-14 M^2 = (0.130 M)([OH-])
To solve for [OH-], divide both sides of the equation by 0.130 M:
[OH-] = (1.0 x 10^-14 M^2) / 0.130 M
[OH-] ≈ 7.69 x 10^-14 M
Therefore, the concentrations are:
[H+] = [H3O+] = 0.130 M
[ClO4-] = 0.130 M
[OH-] ≈ 7.69 x 10^-14 M