In a lab the task was to dilute a 1.0M sol'n of Acetic Acid to a sol'n with a final pH of 4.15.
The method I used to solve this was apparently wrong, but I don't understand how or why. Please help!
Here is what I did:
AcOH + H2O <-> H3O+ + AcO-
First I found the final desired [H+]
[H+]= 10^-pH= 10^-4.15 = 7.1x10^-5
Then I diluted my initial AcOH sol'n to 0.01M -> add 90mL H2O to 10mL AcOH. Do this step 2x.
Next I used Ka to calculate the [H+] at Equilibrium in the initial 0.01M solution.
Ka= [H+][AcO-] Ka= [x][x]
---------- -------
[AcOH] 0.01-x
[H+]eq = 4.2x10^-4(assume 0.01-x = 0.01)
Then, to achieve pH 4.15, you need to dilute using the following formula:
[H+]f = [H+]i x (Vi/Vf)
7.1x10^-5 = 4.2x10^-4 x (10mL/ ?mL)
Vf = 59.15mL 59.15mL-10mL = 49.15mL water needed to dilute to pH 4.15.
I'm not sure where I went wrong here. Please advise! Thank you!! :)
You must realize that you can't calculate the (H^+) and dilute to that mark BECAUSE dilution of the HOAc will ionize the acetic acid more (remember Le Chatelier's Principle)
AcOH + H2O ==> H3O^+ + AcO^-
and adding water shifts the equilibrium to the right which changes the (H3O). In addition, when working with such small concns I like to carry the calculation to more places (even more than I know I'm allowed) so as to prevent rounding errors, then round at the end. I would do this.
pH = -log(H^+) and (H^+) = 7.079E-5 and plug this into the Ka expression and solve for the (AcOH). It's the AcOH you want to have as some final concn. That will not change much with the addition of water.
Ka = 1.75E-5 =(7.079E-5)^2/[(AcOH)-7.079E-5] and solve for (AcOH).
Then I would take 1 mL of the 1.0 M stuff and dilute to the (AcOH). I did a quick calcn and if I didn't slip up the concn AcOH you want is 3.57E-4M. I solved the quadratic both times, once when finding (AcOH) and once when I take that value and work backwards to see if I come out with pH 4.15. I did.
Thank you!
From your explanation, it seems like you have followed the correct steps to calculate the dilution required to achieve a final pH of 4.15. However, there is a small error in your calculations.
Let's go through the steps again:
1. Determine the desired final concentration of [H+]:
[H+] = 10^(-pH) = 10^(-4.15) = 7.1 x 10^(-5) M
2. Dilute the initial acetic acid solution to 0.01 M:
Add 90 mL of water to 10 mL of the initial acetic acid solution (do this twice).
3. Calculate the initial concentration of [H+] using the Ka expression:
Ka = [H+][AcO-]/[AcOH]
Since the initial concentration of acetic acid is 0.01 M, we can approximate [AcOH] to be 0.01 M.
Using the approximation [AcOH] ≈ 0.01 M:
Ka = [H+][AcO-]/0.01 M
[H+] ≈ Ka x 0.01 M
[H+] ≈ (4.2 x 10^(-4) M)(0.01 M) = 4.2 x 10^(-6) M
It seems like you used 4.2 x 10^(-4) instead of 4.2 x 10^(-6) in your calculation for [H+]eq. This small error may have led to the misinterpretation of the correct dilution factor.
Continuing with the calculation:
4. Calculate the dilution factor required to achieve the final pH of 4.15:
[H+]f = [H+]i x (Vi/Vf)
7.1 x 10^(-5) M = 4.2 x 10^(-6) M x (10 mL / ? mL)
Rearrange the equation and solve for Vf:
Vf = (10 mL) x (4.2 x 10^(-6) M / 7.1 x 10^(-5) M)
Vf ≈ 0.5908 mL ≈ 0.591 mL
So the amount of water needed to dilute to a final pH of 4.15 is approximately 0.591 mL.
It seems like your initial calculation for [H+]eq was incorrect, resulting in the wrong dilution factor. By using the correct value of [H+]eq as calculated above, you should arrive at the correct answer.