What is the value of [OH-] in a 0.015 M CH3COOH solution? Ka = 1.8 x 10-5
I started by writing the equation:
CH3COOH + H2O <---> H3O+ + CH3COO-
Ka= [H3O+][CH3COO-]/[CH3COOH]= 1.8 E-5
... Now what? I don't understand where OH relates to this ...
Oh, don't worry, I understand the confusion! So here's the thing: in this equation, CH3COOH is an acid, right? And acids tend to donate protons (H+) to water, forming H3O+. Now, since H3O+ and OH- ions are linked in water, the presence of H3O+ means there must be some OH- floating around too. It's like a happy little family of ions!
But wait, you might ask, why isn't OH- explicitly mentioned in the equation? Well, CH3COOH is a "weak" acid, meaning it doesn't fully dissociate into H3O+ and CH3COO-. However, there's a neat little relationship to calculate OH- in a solution: Kw = [H3O+][OH-]. And since you know the value of Ka and the equation involves H3O+ and CH3COO-, you can use that to find OH-.
So, go ahead and use the Ka value to solve for [H3O+]. Then, using Kw = [H3O+][OH-], you can find the concentration of OH-. Just remember, Kw is always 1.0 x 10^-14 at 25 degrees Celsius.
Hope that helps you find that sneaky OH- hiding in the equation, my friend!
In the equation you wrote, CH3COOH acts as an acid and donates a proton (H+) to water, forming H3O+ (hydronium ion) and CH3COO- (acetate ion). The concentration of H3O+ indirectly affects the concentration of OH- in the solution.
To find the concentration of OH- in the solution, we can use the fact that the concentration of H3O+ multiplied by the concentration of OH- is always equal to the equilibrium constant for the ionization of water, which is known as Kw. At 25°C, Kw equals 1.0 x 10^-14.
Therefore, [H3O+][OH-] = 1.0 x 10^-14
Given that the concentration of H3O+ in the solution can be determined from the value of Ka, we can rearrange the equation to solve for [OH-]:
[OH-] = (1.0 x 10^-14) / [H3O+]
Now, we need to calculate the concentration of H3O+ in the solution using the dissociation constant (Ka) and the concentration of CH3COOH:
Ka = ([H3O+][CH3COO-]) / [CH3COOH]
1.8 x 10^-5 = ([H3O+] * [CH3COO-]) / (0.015)
Simplifying the equation, we have:
[H3O+] * [CH3COO-] = 1.8 x 10^-5 * 0.015
[H3O+] = (1.8 x 10^-5 * 0.015) / [CH3COO-]
Now, substituting this value for [H3O+] into the equation [OH-] = (1.0 x 10^-14) / [H3O+], we can calculate the value of [OH-].
To find the value of [OH-] in the solution, you need to consider the relationship between H3O+ (hydronium ion) and OH- (hydroxide ion) in water. In a neutral solution, the concentration of H3O+ is equal to the concentration of OH-, so [H3O+] = [OH-] = x, where x is the concentration of both ions.
However, in an acidic solution like the one with CH3COOH, the concentration of H3O+ is greater than the concentration of OH-. Therefore, we need to apply the concept of the acid dissociation constant (Ka) to determine the concentration of OH- indirectly.
Given that Ka = [H3O+][CH3COO-]/[CH3COOH] = 1.8 x 10^-5, we can rearrange the equation to find [H3O+] in terms of [CH3COO-] and [CH3COOH]:
[H3O+] = (Ka x [CH3COOH]) / [CH3COO-]
Since we're given the concentration of CH3COOH (0.015 M), we can substitute that value into the equation. However, we still need to determine [CH3COO-].
The solution to this problem involves the concept of the ionization of a weak acid (CH3COOH) in water. When CH3COOH ionizes, it produces H3O+ and CH3COO-. The extent of ionization depends on the value of Ka.
Since CH3COOH is a weak acid, it doesn't completely dissociate into ions. To determine the value of [CH3COO-], we need to make an assumption that x (the change in concentration) is small compared to the initial concentration of CH3COOH (0.015 M). This assumption is valid because CH3COOH is a weak acid.
Let's say that x is the concentration of [H3O+], [CH3COO-], and [OH-] (they are all the same because the solution is neutral). The initial concentration of CH3COOH is 0.015 M, and given that the acid-ionization equation is CH3COOH + H2O <---> H3O+ + CH3COO-, we can establish the following equilibrium expression:
Ka = [H3O+][CH3COO-]/[CH3COOH]
1.8 x 10^-5 = (x)(x) / (0.015 - x)
Solving this equation for x will give us the value of [H3O+] and also [OH-] in the solution. Once we have x, we can substitute that value into either [H3O+] or [OH-] to find the concentration.