0.1 mol of HCl is added to 1 L of water.

What is [H+] and [OH-]?

Oh boy! It's acid time! By adding 0.1 mol of HCl to 1 L of water, we can assume complete dissociation of HCl. So for every mole of HCl, you'll get one mole of H+ ions. Therefore, the concentration of [H+] is 0.1 mol/L.

Now, since water remains mostly undissociated, we can assume the water concentration remains constant. In pure water, [H+] and [OH-] are equal at 1 x 10^-7 M. However, since we've added HCl, the reaction shifts towards the production of more H+. This means that the concentration of [OH-] will decrease.

However, calculating the exact concentration of [OH-] would require some fancy calculations and knowledge of the equilibrium constant. But since we have such a dilute solution, we can approximate that the [OH-] concentration will be negligible compared to the initial concentration in pure water.

So, to sum it up: [H+] is 0.1 M, and [OH-] is super tiny and can be considered negligible.

To determine the values of [H+] and [OH-], we need to consider the dissociation of water and the ionization of HCl.

1. Dissociation of water:
H2O ⇌ H+ + OH-

In pure water, the concentration of H+ ions and OH- ions are equal, resulting in a neutral solution. Therefore, [H+] = [OH-].

2. Ionization of HCl:
HCl → H+ + Cl-

When HCl is added to water, it completely dissociates into H+ and Cl- ions. Since the concentration of HCl is 0.1 mol and it fully dissociates, the concentration of H+ will also be 0.1 mol.

Therefore, [H+] = 0.1 mol and [OH-] = 0.1 mol.

To determine the concentration of hydrogen ion ([H+]) and hydroxide ion ([OH-]) in a solution, we need to consider the dissociation of water and the dissociation of the solute, which in this case is HCl.

First, let's look at the dissociation of water:
H2O ⇌ H+ + OH-

In pure water, at 25°C, the concentration of [H+] and [OH-] are equal and can be calculated using the equation Kw = [H+][OH-], where Kw is the ion product constant for water, which is equal to 1.00 x 10^-14 at 25°C.

Since [H+] = [OH-] in pure water, we can substitute [OH-] with [H+] in the equation above:
Kw = [H+][H+], or
1.00 x 10^-14 = [H+]^2

Taking the square root of both sides, we get:
[H+] = √(1.00 x 10^-14)

[H+] ≈ 1.00 x 10^-7 M

Now, let's consider the dissociation of HCl:
HCl → H+ + Cl-

Since we added 0.1 mol of HCl to 1 L of water, the concentration of HCl is 0.1 M.

Therefore, the concentration of [H+] in the resulting solution will be 0.1 M.

Since the concentration of [H+] is significantly higher than the concentration of [OH-] in this solution, we can neglect the contribution of [OH-] from the dissociation of water. Therefore, the concentration of [OH-] can be assumed to be negligible.

To summarize:
[H+] ≈ 0.1 M
[OH-] ≈ negligible

Technically this is a VERY tough probem and can't be answered with the information given. However, I assume that the person making up the problem meant that 0.1 mol HCl was added to enough water to make 1L of solution.

(H^+) = (HCl) since HCl is 100% ionized.
(H^+) = 0.1 mols/1L = 0.1 M.
Then to find OH, use
(H^+)(OH^-) = Kw = 1E-14.
You know Kw and H^+, solve for OH^-