Hi! So I solved this problem but got the wrong answer because I assumed x was small enough to ignore. However, achieve says that I need to use the quadratic formula. I keep getting the wrong answer for this and was hoping someone could explain what they did/ what they got as the values for the quadratic formula. Right now I have 8.40*10^-4 = x^2/0.1044-x. Thanks!
This was the question: Muscles produce lactic acid, CH3CH(OH)COOH(aq) , during exercise. Calculate the percent ionization (deprotonation), pH, and pOH of a 0.1044 M solution of lactic acid. The acid‑dissociation (or ionization) constant, Ka , of this acid is 8.40×10−4 .
Week 2 #3 Achieve
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 108
- Joined: Fri Sep 24, 2021 5:24 am
Re: Week 2 #3 Achieve
Since X cannot be negligible, i got x^2+0.00893x-9.91*10-4 = 0.
After using the quadratic equation, you should get x= [H+]=0.027.
And since your are asked to calculate the percent ionization of this weak acid, you do [H+]/[HA]= (0.027)/(0.111) *100% = 24.6% ionization.
Hope you found this helpful :)
After using the quadratic equation, you should get x= [H+]=0.027.
And since your are asked to calculate the percent ionization of this weak acid, you do [H+]/[HA]= (0.027)/(0.111) *100% = 24.6% ionization.
Hope you found this helpful :)
-
- Posts: 15
- Joined: Fri Sep 24, 2021 5:59 am
Re: Week 2 #3 Achieve
Usually after doing the quadratic formula, if one of the x values is positive and the other is negative, I use the positive one (you can't have a negative concentration!) Then you just plug that x value in to solve for the blank concentrations from your ICE table.
-
- Posts: 110
- Joined: Fri Sep 24, 2021 5:29 am
Re: Week 2 #3 Achieve
Hi Marleena! I believe that we usually all get different numbers for these Achieve questions so I'm not sure how much my personal values would help, but I think I can assist you in setting up the quadratic formula. Given the values you listed above, I believe that you can manipulate the equation to make it easier to solve using the quadratic formula. First, you could multiply the left side of the equation, your K value, by (0.1044 - x), which should give you 0.000087696 - 0.00084x = x^2. Moving x^2 to the other side of the equation should make the entire thing equal to 0 and make it easy to plug values of A,B, and C into the quadratic formula (A=1, B= 0.00084, C= - 0.000087696). From there, you should be able to solve for x (use the positive value) and answer the question.
Hope this helps!
Hope this helps!
Return to “Equilibrium Constants & Calculating Concentrations”
Who is online
Users browsing this forum: No registered users and 18 guests