Systems of Quadratic Equations

[FireWall]

What is the pH of a solution that contains 0.250 M acetic acid (K = 1.8 * 10^-5) and 0.150 formic acid (K = 1.8 * 10^-4)? The easy part is setting up the system (i.e. the actual chemistry), the hard part is solving it.

Alright, it turns out that you can use simple subsitition to solve the system if you apply a shortcut to the original problem.

1.8 * 10^-5 = x(x + y) / 0.250

1.8 * 10^-4 = y(x + y) / 0.150

If you start out like that, you can solve for y in the first equation, and then plug it back in to the second equation.

Multiply out the 0.250:

4.5 * 10^-5 = x(x + y)

4.5 * 10^-5 / x = x + y

4.5 * 10^-5 / x - x = y

Etc. Turns out to be more simple than I expected. Thanks to Streetwolf for the help.

[kretz]

do your own homework

[FireWall]

do your own homework

Instead of pointing out the obvious--specifically, that you are a worthless fucking moron--I will simply state:

1. My chemistry professor doesn't assign homework.
2. Even if she did, she wouldn't collect homework on the last day of class, which is tomorrow.
3. Problems like that aren't even going to be on the final exam.
4. I understand the chemistry, which is all that she cares about.

Do me a favor and stop posting.

[Streetwolf]

Explained it on AIM.

For those who are curious, since they are weak acids they will not dissociate much. Thus, instead of 0.250 - x and 0.150 - y in the denominators, you can just put them as 0.250 and 0.150 and those serve as good approximations. The solutions I got (x = 0.0008 and y = 0.0048) were close to the actual answers, by only a factor of 0.00001 and 0.0001 respectively.

[zany]

I actually had several problems like this on my final exam. That was fun at 8 in the morning :/