a) A weak acid, HX, is 1.3% ionized in .20 M solution. What percent of HX is ionized in a .030 M solution? b) Does the percent ionization increase or decrease upon dilution? c) Does the H3O+ concentration of the above weak acid increase or decrease?

Please help, struggling like crazy.

HX <-> H+ + X-

Concentration = mol/vol
Since we only know mol and not concentration, we can just use mol since it is directly proportional to concentration
HX H+ X-
I 0.2 0 0
C -y +y +y
E 0.2-y y y

Since y is very small(weak acid) Equilibrium mol for HX is 0.2

Ionization % = [H+]eq/[HX]i *100%
For 0.2M, [H+] = 1.3% / 100% * 0.2= 2.6*10^-5
y=2.6*10^-5 M

Now we can calculate Ka of HX
Ka = [H+][X-] / [HX]
=(2.6*10^-5)^2/0.2
=1.3*10^-4

ICE table for 0.03M
HX H+ X-
I 0.03 0 0
C -z +z +z
E 0.03-z z z
(Once again, HX at equi = 0.03)

Using Ka = [H+][X-] / [HX]
1.3*10^-4 = z^2/0.03
z=1.97*10^-3 = [H+]

Ionization % for 0.03M =1.97*10^-3/0.03 *100%
=6.58%

I'm not too sure if this is correct, haven't done this is awhile and quite rusty
(Do check the calculations too)

a) To find the percent of HX ionized in a 0.030 M solution, we can use the concept of percent ionization.

Percent ionization (% ionization) is given by the equation:

% ionization = (concentration of ionized HX / initial concentration of HX) x 100

We know that in a 0.20 M solution, the weak acid HX is 1.3% ionized. This means that:

% ionization = 1.3

Let's substitute the known values into the equation and solve for the concentration of ionized HX in the 0.20 M solution:

1.3 = (concentration of ionized HX / 0.20) x 100

Dividing both sides of the equation by 100:

0.013 = concentration of ionized HX / 0.20

Multiplying both sides of the equation by 0.20:

0.013 x 0.20 = concentration of ionized HX

0.0026 M = concentration of ionized HX in 0.20 M solution

Now, let's find the percent ionization in a 0.030 M solution. Using the same equation:

% ionization = (concentration of ionized HX / initial concentration of HX) x 100

% ionization = (concentration of ionized HX / 0.030) x 100

Substituting the concentration of ionized HX we found earlier:

% ionization = (0.0026 M / 0.030) x 100

% ionization = 8.67

Therefore, in a 0.030 M solution, the percent of HX that is ionized is approximately 8.67%.

b) From the calculations in part a, we can observe that the percent ionization decreases upon dilution. In the 0.20 M solution, the percent ionization was 1.3%, but in the 0.030 M solution, it decreased to 8.67%.

c) The H3O+ concentration in the weak acid solution will decrease upon dilution. This is because the weak acid ionizes to a lesser extent with lower concentration. As the weak acid becomes more diluted, there are fewer HX molecules to ionize, resulting in a lower concentration of H3O+ ions.

To determine the answers to these questions, we need to use the concept of percent ionization and the relationship between concentration and percent ionization.

a) To find the percent of HX ionized in a 0.030 M solution, we can use the relationship between the percent ionization and the concentration of the weak acid. The percent ionization is defined as the ratio of the concentration of ionized HX to the initial concentration of HX, multiplied by 100.

Given that the weak acid HX is 1.3% ionized in a 0.20 M solution, we can represent this as follows:
Percent ionization = (concentration of ionized HX / initial concentration of HX) * 100 = 1.3%

Let's assume x represents the concentration of ionized HX in a 0.20 M solution. Since the percent ionization is given as 1.3%, we can write the equation as follows:
(0.013 * 0.20) / 0.20 * 100 = (x / 0.030) * 100

Simplifying the equation, we get:
0.13 = (x / 0.030) * 100

Solving for x, the concentration of ionized HX in a 0.030 M solution, we have:
0.13 = (x / 0.030)
x = 0.13 * 0.030
x = 0.0039 M

Therefore, in a 0.030 M solution, 0.0039 M of HX is ionized.

b) Now that we know the concentration of ionized HX in both the 0.20 M and 0.030 M solutions, we can compare the percent ionization.

In the 0.20 M solution, the percent ionization is 1.3%. In the 0.030 M solution, the percent ionization is:
(0.0039/0.030) * 100 = 13%

So, the percent ionization increases upon dilution from 1.3% to 13%.

c) The H3O+ concentration of a weak acid is related to its concentration and percent ionization. As we determined in part a, the concentration of ionized HX in the 0.030 M solution is 0.0039 M. Therefore, the H3O+ concentration of the weak acid in the 0.030 M solution is also 0.0039 M.

In conclusion:
a) The percent of HX ionized in a 0.030 M solution is 13%.
b) The percent ionization increases upon dilution.
c) The H3O+ concentration of the weak acid remains the same at 0.0039 M.