Posted by
**Jodi** on
.

A student prepared a .10M solution of acidic acid. Acidic acid has a Ka of 1.75 x 10-3. What are the hydronium ion concentration and the PH of the solution?

I think the PH is 1.76 because I hit -log(1.75 x 10-3) on my calculator and that's what I got thanks in advance for your help

Is this acetic acid? And is Ka 1.75 x 10^-5? Whatever, I will use HA for the acid and we will use the value of Ka you posted. The ionization of HA is

HA ==> H^+ + A^-

Ka = (H^+)(A^-)/(HA)

(H^+) = y

(A^-_ = y

(HA) = 0.1 -y

Plug these values into the Ka expression and solve for (H^+). Then pH = -log(H^+).

Post your work if you get stuck. I found 1.9 for the pH.

I don't understand what you posted I cant even attempt this.My problem reads

A student prepared a 0.10M solution of acetic acid,CH3COOH. Acetic acid has a Ka of 1.75 x 10^-5. What are the Hydronium concentration and the PH of the solution. sorry for the incorrect posting

You should have been able to follow my post but here goes again.

Acetic acid is a WEAK acid. The definition of a weak acid is that it does not ionize completely; that is, it ionizes less than 100%. Thus, CH3COOH, acetic acid, produces the following when it ionizes.

CH3COOH + H2O ==> H3O^+ + CH3COO^-

If we call the hydronium concentration (at equilibrium) Y, designated by (H3O^+), then the CH3COO^- concentration is Y, also, and we designate that (CH3COO^-). If we started with 0.1 M CH3COOH, then it must be 0.1 - Y at equilibrium. It was 0.1 M before ionization and after ionization it is 0.1 - the amount ionized (Y) or 0.1 - Y.

The mathemtatical expression for Ka is

Ka = (H3O^+)(CH3COO^-)/(CH3COOH)

Now, plug Y in for (H3O^+) and Y for (CH3COO^-) and 0.1 - Y for (CH3COOH) and solve for Y. That give you the (H3O^+) which the problem asks for first. The second part is to calculate pH and that is obtained by pH = - log(H3O^+).

The answer for (H3O^+) = 0.00132 M and the answer for pH is 2.88