What volume (mL) of .20 M HCL needs to be added to 1.23 L of .10 M acetic acid to reduce percent ionization to 1.00%?
I would do this.
........HAc ==> H^+ + Ac^-
I.......0.1......0.....0
C........-x......x.....x
E.....0.1-x......x......x
Plug the E line into the Ka expression and solve for x = (H^+). I obtained 1.34E-3M for (H^+)
.........HAc ==> H^+ + Ac^-
What you want to do now is to add H^+ from the HCl so that the equilibrium will shift to the left and that decreases the ionization. So how much HCl must we add.
1.8E-5 = (H^+)(Ac^-)/(HAc)
If you want (Ac^-)(HAc) to be 0.01, then (H^+) must be what? That's
(H^+)(0.01) = K = 1.8E-5
and H^+ = 0.0018M
It was already 0.00134 so we must add how much? That's 0.0018-0.00134 = 4.6E-4M. You want 1.23 L of that so 4.6E-4M x 1.23 L = 5.66E-4 mols and since M = mol/L then L = mols/M =- 5.66E-4/0.2MHCl = 2.83E-3 L or 2.83 mL of the 0.2M stuff. You can check that out to see if it gives a solution that is 1% ionized.
That 0.0018M H^+ will be slightly diluted by 0.00180 x (1230/1232.8) = ?
The 0.1M HAc will be diluted by the same proportion of 0.1 x (1230/1232.8) = ?
Plug those values into the Ka expression and solve for (Ac^-), then
% ion = 100*(Ac^-)/(HAc). I came up with 0.99% which is about as close as I can get.
I made a typo part way down the page. Instead of retyping the whole things, since it is so long, here is the line in bold. I show it as multiplied and it should be divided.
If you want (Ac^-)/(HAc) to be 0.01, then (H^+) must be what?
I re-thought that problem (all night long) and there is nothing wrong with the first part; i.e., (H^+) must be 0.0018 to make it work and you must add 2.83 mL of the 0.2 M acid to get that. When I tried to check it to see if all that worked the best I could do was 0.99% and that wasn't good enough. Here is how it is checked.
...........HAc ==> H^+ + Ac^-
I..........0.1...0.00134..0.00134
add..............4.6E-4MHCL.....
C...........+x..............-x
E.........0.1+x....0.0018..0.00134-x
Set up Ka = (H^+)(Ac^-)/(HAc) and solve for x = 3.37E-4M. That makes Ac^- = 0.00134-3.37E-4 = 0.001
Then % ion = (0.001/0.1)*100 = 1.00%.
So it checks out.
To solve this problem, we need to first calculate the initial percent ionization of acetic acid. Then, we can determine how much hydrochloric acid (HCl) needs to be added to reduce the percent ionization to 1.00%.
Step 1: Calculate the initial percent ionization of acetic acid.
To do this, we need to know the initial concentration of acetic acid and its ionization constant. The ionization constant for acetic acid (CH3COOH) is known as the Ka value, which is 1.8 x 10^-5 at 25°C.
Given:
Initial concentration of acetic acid (CH3COOH) = 0.10 M
Using the Ka expression for acetic acid:
Ka = [H+][CH3COO-] / [CH3COOH]
At equilibrium, the concentration of undissociated acetic acid [CH3COOH] will be (0.10 - x) M, where x is the concentration of dissociated acetic acid.
Let's assume x is small compared to the initial concentration of acetic acid, so we can approximate [CH3COOH] as 0.10 M.
Simplifying the Ka expression:
1.8 x 10^-5 = (x)(x) / 0.10
Since x is small compared to 0.10, we can assume the change in concentration due to dissociation (x) is negligible compared to the initial concentration (0.10). Thus, we can approximate (0.10 - x) as 0.10 in the denominator.
1.8 x 10^-5 = x^2 / 0.10
Simplifying further:
x^2 = (1.8 x 10^-5)(0.10)
x^2 = 1.8 x 10^-6
x = √(1.8 x 10^-6)
Calculating x:
x ≈ 1.34 x 10^-3 M
To calculate the initial percent ionization (I), we can use the formula:
I = (x / initial concentration) * 100
Plugging in the values:
I = (1.34 x 10^-3 / 0.10) * 100
I ≈ 1.34%
Therefore, the initial percent ionization of acetic acid is approximately 1.34%.
Step 2: Calculate the volume of HCl needed to reduce the percent ionization to 1.00%.
We want to reduce the percent ionization of acetic acid to 1.00%. Let's assume the final concentration of acetic acid after adding HCl is the same as the initial concentration, which is 0.10 M.
Let V be the volume of HCl needed in milliliters (mL).
The concentration of HCl is 0.20 M, and its reaction with acetic acid is as follows:
HCl + CH3COOH ⟶ CH3COOH2+ + Cl-
Since the final concentration of acetic acid should remain the same (0.10 M) and the percent ionization should be reduced to 1.00%, we can set up the equation:
(1.00 / 100) = (x / Vml) * 100
Simplifying:
1.00 = x / V
Rearranging the equation to solve for V:
V = x
Plugging in the value of x:
V = 1.34 x 10^-3 mL
Therefore, approximately 1.34 x 10^-3 mL (or 1.34 μL) of 0.20 M HCl needs to be added to 1.23 L of 0.10 M acetic acid to reduce the percent ionization to 1.00%.