If a solution is 0.25 M KOH...how can i figure out the hydronium concentration and hydroxide concentration?

i'm confused..because isn't this a salt?

In solution it would separate into ions. As this is a strong base it would be fully dissociated in solution i.e K+ ions and OH- ions.

To determine the hydronium (H3O+) concentration and the hydroxide (OH-) concentration in a solution of 0.25 M KOH, you need to consider the dissociation of the salt in water.

KOH, or potassium hydroxide, is a strong base and completely dissociates into potassium ions (K+) and hydroxide ions (OH-) in water. Therefore, the concentration of hydroxide ions in the solution will be equal to the concentration of KOH.

In this case, the concentration of hydroxide ions (OH-) is 0.25 M.

Since water is a neutral solvent, the hydronium ion concentration (H3O+) can be calculated using the equation Kw = [H3O+][OH-], where Kw (the ionization constant of water) is 1.0 x 10^-14 at 25°C.

Since we know the concentration of hydroxide ions (OH-) is 0.25 M, we can substitute this value into the equation:

1.0 x 10^-14 = [H3O+][0.25]

Now, divide both sides of the equation by 0.25:

[H3O+] = (1.0 x 10^-14) / 0.25

Calculating this gives the hydronium ion concentration in the solution.

So, the hydroxide concentration is 0.25 M, and the hydronium concentration can be determined using the given equation.