HW - #2

[Alexa Marinelli-1B]

The Ka of a monoprotic weak acid is 0.00431. What is the percent ionization of a 0.123 M solution of this acid?
Hi, I have tried this problem a few times and it keeps telling me that it is incorrect but I am unsure why. I set up my ICE table so that the equilibrium concentration for HA: 0.123 - x, H: x, and A: x. That resulted in me having Ka = [x][x]/[0.123-x] = 0.00431 --> x2 + 0.00431x - 5.3*10^-5. The x I ended up getting was 0.0054. I then divided that by 0.123 * 100 to get the percent ionization of 4.39%. If someone can please let me know where I went wrong with my math I would appreciate it!

[Ryan Blaydon 1K]

I am not entirely sure but it looks like when you were multiplying 0.00431 times 0.123 you got a different value than I did. I got 5.3E-4 not to the -5. I would try and see if this works, I hope this helps!

[Jaylin Hsu 1C]

Hello, so it seems like your steps are correct, including your ice tablet and equilibrium constant setup. However, please check you did not make any careless errors when solving the quadratic formula.

[Holland Smith 3C]

After attempting the problem right now, I can confirm that Ryan is correct. the quadratic formula should be x^2 + .00431x - 5.2*10^-4 (not -5). After following your steps, it looks like it was just the mathematical error, everything else looks correct. Hope this helps!

[Caroline 2A]

Hi!

First, start by assembling an ICE table. Because the value of M is more than 1000 times greater than the Ka, we are able to assume that the concentration change x is sufficiently small compared to the initial concentration. We can check this by dividing the value of X by the initial concentration and multiplying that value by 100%. If x is less than 5% of the concentration, then our assumption is valid (and it is valid in this case).

From there, we can find the pH by using the equation pH= -log[H+] where x is equal to H+. This should give you your final answer.

Hope this helps :)