Calculate [H3O+] and [OH-] in 0.22M KOH(aq)
To calculate the concentrations of [H3O+] and [OH-] in a solution of KOH, we need to understand the dissociation of KOH in water.
KOH, a strong base, dissociates completely in water to form K+ ions and OH- ions. The equation for this dissociation is:
KOH(aq) -> K+(aq) + OH-(aq)
Given that the initial concentration of KOH is 0.22 M, we can use this information to find the concentration of OH-.
Since KOH dissociates in a 1:1 ratio, the concentration of OH- will be equal to the concentration of KOH.
Therefore, [OH-] = 0.22 M.
To find the concentration of [H3O+], we need to know that water also provides a source of H+ ions (protons) through self-ionization:
H2O(l) -> H+(aq) + OH-(aq)
In pure water, the concentration of H+ ions is equal to the concentration of OH- ions and is 1 x 10^-7 M (at 25°C). However, in our case, we have added a solution of KOH, which will affect the concentration of H3O+.
To find the concentration of [H3O+], we can use the equation:
[H3O+][OH-] = 1 x 10^-14
Substituting the known concentration of [OH-] = 0.22 M:
[H3O+](0.22) = 1 x 10^-14
[H3O+] = (1 x 10^-14)/(0.22)
[H3O+] ≈ 4.55 x 10^-14 M
Therefore, the concentration of [H3O+] is approximately 4.55 x 10^-14 M.
Final results:
[H3O+] ≈ 4.55 x 10^-14 M
[OH-] = 0.22 M
To calculate the concentrations of [H3O+] and [OH-] in a solution of KOH, we need to use the concept of dissociation of water and the autoionization of water.
First, let's understand that KOH is a strong base and dissociates completely in water. It will produce K+ ions and OH- ions.
The balanced equation for the dissociation of KOH is:
KOH(aq) → K+(aq) + OH-(aq)
Since KOH is a strong base, we can assume that the concentration of OH- in the solution will be the same as the concentration of KOH. Therefore, [OH-] = 0.22 M.
To calculate [H3O+], we need to use the autoionization of water. In water, a small fraction of water molecules will self-ionize to produce H3O+ ions and OH- ions. The balanced equation for this reaction is:
H2O(l) ↔ H3O+(aq) + OH-(aq)
Since the concentration of H3O+ and OH- ions from the autoionization of water are equal, we can assume that [H3O+] = [OH-] = x (let's assume it is x).
Using the concept of the ion product of water (Kw = [H3O+][OH-]), we know that at 25°C, Kw is equal to 1.0 x 10^-14.
So, we can write this equation as:
x * x = 1.0 x 10^-14
Taking the square root of both sides:
x = square root of (1.0 x 10^-14)
By calculating this square root, we can get the value of x, which will be the concentration of both [H3O+] and [OH-].
Now let's calculate it:
x = square root (1.0 x 10^-14) = 1.0 x 10^-7
Therefore, the concentrations of [H3O+] and [OH-] in the 0.22 M KOH solution will both be equal to 1.0 x 10^-7 M.
KOH is a strong base; therefore, (OH^-) = 0.22M
Then (H^+)(OH^-) = Kw = 1E-14. Solve for (H^+).