A student prepared a 0.10 M solutino of acetic acid, CH3COOH. Acetic acid has a Ka of 1.75 x 10^-5. What are the hydronium ion concentration and the pH of the solution?

Let's let CH3COOH be represented by HAc (H is the H of the COOH and Ac stands for the remaining part of the molecule.).

HAc ==> H^+ + Ac^-

Beginning concentrations:
(HAc) = 0.1
(H^+) = 0
(Ac^-) = 0

After ionization:
(H^+) = x
(Ac^-) = x
(HAc) = 0.1 - x

Write the Ka expression and plug the after ionization into it in the appropriate places, then solve for x. Following that, use pH = -log(H^+).

Post your work if you get stuck.

1.75 x 10^-5 = [x][x]/[0.10-x]?

I'm completely lost...

See my response to your later post above.

Repost if you still don't get it.

To find the hydronium ion concentration and the pH of the solution, we need to first calculate the degree of ionization (α) of acetic acid. The degree of ionization represents the fraction of acetic acid molecules that dissociate into hydronium ions (H3O+) and acetate ions (CH3COO-).

The equation for the dissociation of acetic acid is as follows:

CH3COOH ⇌ H3O+ + CH3COO-

The initial concentration of acetic acid (CH3COOH) is 0.10 M. At equilibrium, let's assume that x M of the acetic acid dissociates.

The concentration of hydronium ions (H3O+) and acetate ions (CH3COO-) formed will be x M each.

Using the equilibrium expression for Ka, which is equal to the ratio of the concentration of the product (H3O+) to the concentration of the reactant (CH3COOH), we can write:

Ka = [H3O+][CH3COO-] / [CH3COOH]

Substituting the known values, we have:

1.75 x 10^-5 = [x][x] / [0.10 - x]

Since the value of x will be very small compared to the initial concentration of acetic acid, we can make the approximation that 0.10 - x ≈ 0.10. This simplifies the equation to:

1.75 x 10^-5 = x^2 / 0.10

Rearranging the equation, we find:

x^2 = 1.75 x 10^-5 * 0.10

x^2 = 1.75 x 10^-6

Taking the square root of both sides, we obtain:

x ≈ 1.32 x 10^-3

This represents the concentration of hydronium ions (H3O+) and acetate ions (CH3COO-) formed.

Next, we can find the pH of the solution using the formula:

pH = -log[H3O+]

Plugging in the hydronium ion concentration, we have:

pH = -log(1.32 x 10^-3)

Calculating the log value, we find:

pH ≈ 2.88

Therefore, the hydronium ion concentration is approximately 1.32 x 10^-3 M and the pH of the solution is approximately 2.88.