In a 1.0x10^-4 M solution of HClO(aq), identify the relative molar amounts of these species:HClO, OH-, H3O+, OCl-, H2O

most -> least: H20, HClO, H3O+, OCl-, OH-

........HClO ==> H^+ + OCl^-

I.....1.0E-4.....0........0
C........-x.......x.......x.
E......1E-4-x.....x.......x

Ka = (H^+)(OCl^-)/(HClO)
Substitute and solve for x = (H^+) = (OCl^-). (HClO) = 1.00E-4-x
You can get OH from
(H^+)(OH^-) = Kw

Thank You!!!

Well, in a solution of HClO(aq), you can definitely find some lively characters! Let's break it down:

1. HClO: The main star of the show here is hypochlorous acid. In this case, its relative molar amount will depend on the concentration you mentioned, which is 1.0x10^-4 M.

2. OH-: This is the hydroxide ion that loves to bring the "basic" to the party. In this solution, the relative molar amount of OH- will be a bit limited since the solution is acidic.

3. H3O+: Ah, the hydronium ion, also known as the "acidic" gang leader. It appears when HClO donates a proton, so its relative molar amount will be present but also a bit limited compared to HClO.

4. OCl-: The hypochlorite ion is formed when HClO loses a proton. Its relative molar amount will be linked to H3O+ and dependent on the acid dissociation constant of HClO.

5. H2O: The quiet and ever-present water molecule. Its relative molar amount in the solution will always be quite high, as it acts both as a solvent and a spectator.

So, to summarize the relative molar amounts in this 1.0x10^-4 M solution:
- HClO: the highest concentration
- H3O+: present but lower than HClO
- OH-: present but lower than HClO and H3O+
- OCl-: dependent on the equilibrium of HClO and H3O+
- H2O: the most abundant, just minding its own business.

To identify the relative molar amounts of the species HClO, OH-, H3O+, OCl-, and H2O in a 1.0x10^-4 M solution of HClO(aq), we need to consider the dissociation reaction of HClO in water and use the dissociation constant (Ka) to determine the concentrations of the species.

HClO(aq) ⇌ H+(aq) + ClO-(aq)

The Ka expression for this reaction is:

Ka = [H+][ClO-] / [HClO]

We know the initial concentration of HClO is 1.0x10^-4 M. Let's assume x represents the concentration of H+ and ClO- ions formed in the solution. Initially, the concentration of these ions is zero.
Thus, the equilibrium concentrations will be:

[HClO] = 1.0x10^-4 - x
[H+] = x
[ClO-] = x

We need to determine the value of x, so we can calculate the equilibrium concentrations of the species. This can be achieved by using the Ka expression:

Ka = [H+][ClO-] / [HClO]
Ka = (x)(x) / (1.0x10^-4 - x)

Since Ka is a constant value for a given temperature, we can solve this equation to find the value of x, which will represent the concentration of H+ and ClO- ions.

After finding the value of x, we can substitute it back into the equilibrium concentration expressions to obtain the molar amounts of each species.

For example:
[HClO] = 1.0x10^-4 M - x
[H+] = x
[ClO-] = x

In addition, since HClO is a weak acid, it partially dissociates, producing H3O+ ions. Thus, the concentration of H3O+ ions will be the same as the concentration of H+ ions.

The concentration of OH- ions can be calculated as follows:
[OH-] = Kw / [H+]
where Kw is the ion product of water (1.0x10^-14 at 25°C).

The concentration of H2O can be considered to be approximately the same as the initial concentration of HClO (1.0x10^-4 M).

By following these steps, you can determine the relative molar amounts of the species HClO, OH-, H3O+, OCl-, and H2O in the given solution.

H2O, HClO, H3O+, OH-, OCl-