Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.130 M in HClO4(aq) at 25 °C.
ClO4= 1.30M
H30=1.30
OH=1E-14/1.30
[H+]=
[ClO4-]=
[OH-]=
To calculate the concentration of ions in the solution, we need to consider the dissociation of HClO4 in water. HClO4 undergoes complete ionization, meaning it dissociates completely into its ions.
The balanced equation for the dissociation of HClO4 is:
HClO4(aq) -> H+(aq) + ClO4-(aq)
Since HClO4 is a strong acid, it completely dissociates into H+ and ClO4- ions.
Therefore, in a 0.130 M solution of HClO4, the concentration of [H3O+] (which represents the same concentration as H+) is also 0.130 M.
Since the concentration of H+ and OH- in water are related by the autoionization of water:
H2O(l) ⇌ H+(aq) + OH-(aq)
where the equilibrium constant, Kw = [H+][OH-] = 1.0 x 10^-14 at 25 °C.
Using the above equation and the fact that the solution is neutral, we can find the concentration of OH- ions.
[H+][OH-] = Kw
0.130 M * [OH-] = 1.0 x 10^-14
Rearranging the equation, we find:
[OH-] = (1.0 x 10^-14) / 0.130 M
Calculating this, we get: [OH-] = 7.69 x 10^-14 M
Finally, to find the concentration of ClO4- ions, we use the fact that every ClO4- ion is produced for each HClO4 molecule that dissociates. Therefore, the concentration of [ClO4-] is also 0.130 M.
In summary:
[H3O+] = 0.130 M
[ClO4-] = 0.130 M
[OH-] = 7.69 x 10^-14 M
To calculate the concentrations of [H3O+], [ClO4-], and [OH-] in the aqueous solution of HClO4, we need to consider the dissociation reaction of HClO4 in water.
The dissociation reaction of HClO4 in water can be represented as follows:
HClO4(aq) ⇌ H+(aq) + ClO4-(aq)
According to this equation, HClO4 dissociates to produce H+ ions and ClO4- ions in water. Since HClO4 is a strong acid, it will dissociate completely, meaning that there will be no undissociated HClO4 left in the solution.
Given that the initial concentration of HClO4 is 0.130 M, it means that [HClO4] = 0.130 M.
Since HClO4 dissociates completely, the concentration of H+ ions ([H3O+]) in the solution will be equal to the initial concentration of HClO4.
Therefore, [H3O+] = [H+] = 0.130 M.
Now let's consider the concentration of ClO4- ions. Since HClO4 dissociates into one H+ ion and one ClO4- ion, the concentration of ClO4- ions will also be 0.130 M.
To calculate the concentration of OH- ions ([OH-]), we need to use the concept of Kw, which is the ion product of water.
At 25 °C, Kw = 1.0 x 10^-14 M^2, which represents the equilibrium constant for the autoionization of water:
2H2O(l) ⇌ H3O+(aq) + OH-(aq)
Since Kw = [H3O+][OH-], and we know that [H3O+] = 0.130 M, we can rearrange the equation to solve for [OH-]:
[OH-] = Kw / [H3O+]
Plugging in the values, we have:
[OH-] = (1.0 x 10^-14 M^2) / (0.130 M)
Calculating this value, we find:
[OH-] ≈ 7.7 x 10^-14 M
So, the concentrations in the aqueous solution are approximately:
[H3O+] = 0.130 M
[ClO4-] = 0.130 M
[OH-] ≈ 7.7 x 10^-14 M