Calculate the percent ionization of a .31 molar solution of acetic acid. The ionization constant of acetic acid is 1.8 x 10^-5
acetic acid is HAc.
.........HAc ==> H^+ + Ac^-
I..l....0.31.....0......0
C........-x......x......x
E......0.31-x....x......x
Substitute the E line into the Ka expression and solve for x.
Then % ion = [(x)/0.31]*100 = ?
To calculate the percent ionization of a solution, you need to know the ionization constant (Ka) and the initial concentration of the acid.
In this case, the ionization constant (Ka) of acetic acid is given as 1.8 x 10^-5 and the initial concentration of the acid is 0.31 M.
The percent ionization can be calculated using the following formula:
Percent Ionization = (concentration of ionized acid / initial concentration of acid) x 100
However, in this case, we don't have the concentration of the ionized acid. To calculate it, we can assume x as the concentration of the ionized acid (since we do not know the exact value of it).
Acetic acid (CH3COOH) ionizes according to the following equation:
CH3COOH ⇌ CH3COO- + H+
Using the given ionization constant (Ka), we can write an expression for the equilibrium of the reaction:
Ka = [CH3COO-][H+] / [CH3COOH]
Since the concentration of CH3COO- is equal to the concentration of H+ (since one CH3COOH molecule produces one CH3COO- ion and one H+ ion), we can write: [CH3COO-] = [H+]
Substituting these values into the expression for Ka, we have:
Ka = [H+]^2 / [CH3COOH]
We also know that the initial concentration of the acid (CH3COOH) is 0.31 M.
Thus:
Ka = x^2 / 0.31
Now, we can solve this equation for the concentration of H+ (or CH3COO-), which will give us the concentration of ionized acid:
x^2 = Ka * 0.31
x = √(Ka * 0.31)
Finally, we can calculate the percent ionization:
Percent Ionization = (x / 0.31) x 100
Substitute the value of x with √(Ka * 0.31) and solve to get the final answer.