Week 2 Achieve #2

[Leilani Mareko-1F]

Hello I have a question regarding this problem from achieve week 2, number 2.

Question: The Ka of a monoprotic weak acid is 0.00264. What is the percent ionization of a 0.138 M solution of this acid?

I was able to set my equation up using the ICE table and ended up finding a percentage of 13.8% but when I submit my answer, it says that its invalid and that I should try using the quadratic formula. Can anyone help me understand how to set up this equation using the quadratic formula? I had just thought that 13.8% would be the answer and I thought the steps I took were correct.

Thank you.

[Jessica Oey]

The reason we don't have to use the quadratic formula in some problems is because the change x from the original concentration is negligible when Ka is smaller than 10^-4. Here, the Ka is not smaller than that so you can't drop the x.

[105988375]

We don't have to use the quadratic in some problems because the change in x from the original concentration of the reactant is so small that its irrelevant, and we can therefore ignore it. We can do this when the Ka is smaller than 10^-3. So in this case, our Ka value is way larger than 10 ^-3, so we can't disregard the x.

if you set up your ICE table correctly, you should get this final equation to solve:
0.00264 = x^2/0.138-x

You can solve it using the quadratic formula :) hope this helped!

[Vivian_Lau_2D]

Hello. We would have to use the quadratic equation for this problem because Ka isn't less than 1*10^-3--Ka isn't small. Because Ka is large, you can't use approximations and would have to use the quadratic equation. Thus, you Ka (0.00264) should equal x^2/(0.138-x). From there, you would set the equation to 0 and plug the a, b, and c values into the quadratic equation to solve for x. Hope this helps!

[Claudia Rajski 2F]

I am also stuck on this question but for a different reason. So, when I did this I didn't use the quadratic formula and it seemed right to me but I didn't get the answer right. I am a little confused on how you would use the quadratic formula and I have also attached my work if someone could look over it I would greatly appreciate it. Thank you!

[Hannah Mirmo 1F]

I am also stuck on this question but for a different reason. So, when I did this I didn't use the quadratic formula and it seemed right to me but I didn't get the answer right. I am a little confused on how you would use the quadratic formula and I have also attached my work if someone could look over it I would greatly appreciate it. Thank you!

Make sure you set up an ICE Table. And you should get [x^2/(.143-x) ] = .00432 , then solve for x using the quadratic formula!

[lanatruong]

To add on, after setting up the ice table for Ka = [H+][A-]/[HA] and getting .00432 = [x^2/(.143-x) ], you would solve for x using the quadratic formula. You can set this up by making this equation into the form of AX^2+BX+C=0 which would be X^2+0.00432X-0.00061776=0. You can then plug in the A, B, and C values into the quadratic equation to find the value of X which is equal to [H+] at equilibrium. Lastly, you can calculate the % ionization with the equation
([H+] at equilibrium/[HA] at initial) x 100. This should get you your final answer. Hope this helps!

[May Jiang 1E]

I followed all these steps numerous times and even checked my work, my x value was 0.0228 after solving for a solution of .143 M and with a K of 0.00432. However, I am still getting the wrong answer–I'm doing 0.0228/0.143 = 15.9%. I thought the equilibrium concentration of H+ was equal to x and the initial concentration is 0.143 M. Can someone help guide me along the right path? I can't seem to find what I'm doing wrong

[005716808]

I too am having difficulties with this question
the formula I was using was sqrt of ka/c and I keep getting it wrong. any strategies would me much appreciated thank you

[005716808]

Hi this is my work. can someone pls help me see where I'm going wrong. this question is driving me crazy lol

[victoria_sitter_1L]

Hi
The number 0.0061 is incorrect. When checking, I got 0.04255.
I think this might be the cause :)

[Ciabel Querol 3E]

Hi this is my work. can someone pls help me see where I'm going wrong. this question is driving me crazy lol

Hello! In this case, the Ka is not less than 1 x 10^-3, so you should not assume that the x being subtracted from the reactants is negligible. Therefore you should multiply .00431 by .104 - x, resulting in .000448 - .00431x = x^2. Continue from there with the steps you did previously. Hope this helps!

[305943084]

In this case, the Ka value is large enough that the change in product and reactant concentration is too significant to be ignored (you can tell by the resultant rate of dissociation you calculated: 13.8% is larger than 5%). That's why the solving of a quadratic equation must be involved to get an answer that is accurate enough.