Halp with AP Chemistry Homework

[Zarrexaij]

I've been assigned some homework I'm completely lost with. We've just started covering Acid-Base Equilibrium. We've just gotten into calculating Ka, Kb, pH, pOH, [OH-], and [H3O+]. Now we'll dealing with weak acids, and we've barely touched on it. I honestly can't get past one part of this problem:

.100 molar solution of CLCH2COOH is 11% ionized (disassociated). Calculate [CLCH2COO-], [H+], [CLCH2COO], and Ka.

This as far as I've gotten, and I'm not sure if this is right:

.1 M x .11 = .011

Ka= [CLCH2COO-][H+]/[CLCH2COOH]

Ka= [.011][.011]/[.1]
= .00121

Now where do I go from there, assuming I'm doing this right?

[RandomLogic]

Write the overall equation for the problem. Acid-Base dissociation is really no different from your standard equilibrium problems. IE write either your ICE or BCE equations (initial/before change equilibrium). You'll use your Ka when you calculate the delta X value for change in mols.

Assuming you understand the top part, then what you do is solve for delta X. In this case you get lucky with a 1 to 1 ratio for all of it. but you have the initial value of CLCH2COOH as .1 mols and initially you have 0 mols of both CLCH2COO and H. So you lose (-x) on the left and +x for both on the right. Ka should be given in the problem for you as you can't calculate it with the information you are given*, but you know that it as an equation it looks like this [CLCH2COO][H]/[CLCH2COOH] = Ka. well you plug in your x values (x)*(x)/(.011-x)=Ka which like I said should be given. You solve for x and then calculate the Equilibrium (E part) concentrations of your solution. your ph is then -log([H+]) and pOH is 14 - log([H+]) if you are required to calculate that.

I can't format it any better, otherwise I could write out the equation and show you. But I think that should answer your question.

Also your initial concentrations are off, if you have .1 M (molar solution) you use that, if you have .1 mols in 1 L of water for example then you have to calculate the previous value.

[Zarrexaij]

Just for a little reference... we completely skipped the equilibrium chapter because we're really running out of time, and we have ten more chapters to cover before May.

We went over some basic equilibrium stuff for a day. The delta thing makes no sense to me, but the rest does.

I got the .011 from multiplaying its molarity by the percent ionization to get how much of it actually dissassociated.

[RandomLogic]

Just for a little reference... we completely skipped the equilibrium chapter because we're really running out of time, and we have ten more chapters to cover before May.

How the hell did you skip that?! Thats how you learn acid-base dissociations!

I'll see if I can find the ICE/BCE example to show you.

Found one. http://www.chem1.com/acad/webtext/chemeq/Eq-05.html Take a look specifically at example 3. It uses pressures for gases, but the same is valid for aqueous and such solutions. You'll notice the ICE part on the side but they have it listed as initial pressure I, change C, equilibrium pressure E and that is what generally is referred to as the ICE equation.

Problem 4 on that same page shows you how to do it when you have initial concentrations on both sides just ignore the normalized gas stuff at the bottom.

[Zarrexaij]

How the hell did you skip that?! Thats how you learn acid-base dissociations!We learned about acid-base dissociation a long time ago. My teacher doesn't really like to follow the book all that well (which is why I'll be looking over it soon).

[RandomLogic]

We learned about acid-base dissociation a long time ago. My teacher doesn't really like to follow the book all that well (which is why I'll be looking over it soon).

Check out the linked problem. I think it should help. Post again if you need any further questions answered.

[Zarrexaij]

Check out the linked problem. I think it should help. Post again if you need any further questions answered.
Uh, help? It just confused the hell out of me.

I guess I'll ask for help tomorrow. :smallsigh:

[RandomLogic]

Uh, help? It just confused the hell out of me.

I guess I'll ask for help tomorrow. :smallsigh:

Really? anything specific?

Basically what you do is write out the balanced chemical reaction. Then in a column to the left or the reaction write Initial, Change, Equilibrium. Now under each ion you write what you have. Initial for your problem is .1 mols of CLCH2COOH, 0 mols of both CLCH2COO and H. Now the change part is the end result of the balanced chemical equation. Since its 1 to 1 to 1 its not a big deal so you'll subtract x (some unknown you are trying to solve for) from the left as that amount will be the dissociated part, and you'll gain +x for both ions on the right. (Which makes sense, if you lost 1 mol of stuff on the right, and it split into 2 ions you would have an equal amount of both at equilibrium, right?) So then at equilibrium you put the intial + change and write that there, for CLCH2COOH you have (.1 - x) and for both CLCH2COO and H you have (x). So now this is where Ka comes in, its defined as the concentrations of the stuff on the right, divided by the concentrations of the stuff on the left. so its [CLCH2COO]*[H]/[CLCH2COOH]. So you plug in the concentrations that you have into the equation (x)*(x)/(.1-x) = Ka. Ka should be given as a constant in the problem. So basically you have (x)*(x)/(.1-x)=constant and you solve for x using the quadratic formula (probably the easiest way, unless you have a TI-89) and you've found the amount dissociated from the the original substance. Now because that is the amount gained on the right side your equilibrium concentrations of both [CLCH2COO] and [H] are the x value you found. If you had something like a 1 to 2 ratio you would multiply by the balanced chemical equation ratio, or if you had some initial quantity you would add that as well.

[Zarrexaij]

Okay, I don't know why you think you can't calculate Ka, since our example problem for that sort of stuff was this:

A .2 M solution of a monoprotic acid HX has a pH of 2.2. Is HX strong or weak? If weak, calculate Ka.

Here's the steps we went through:

1. Set up an equation

HX + H2O -> X- + H3O+

2. Given information

.2 molar acid and 0 and 0 starting molar of the hydronium and X ions.

3. Calculate reaction

{table=head]HX|X-|H3O+
.2-Z|Z|Z[/table]

4. Use the pH to find the amount of HX dissociated.

pH = - log [H3O+]
-2.2 = log H3O+
log = .00631

5. Replace Z with the found value.

{table=head]HX|X-|H3O+
.1936|.00631|.00631[/table]

6. Find Ka given that Ka = [X-][H3O+]/[HX]

(.00631)(.00631)/.1936 (amount of HX not dissociated)

Ka = .000206

Edit: Okay, figuring out how I got that, I managed to get the problem done after testing if I could find the same value for that sample problem by using the percentage ionization instead of pH. Was making it harder than it was.

[RandomLogic]

Yeah I know you can directly calculate Ka, but that is at a specific temperature and you have to have the reaction at equilibrium already. Usually you are given the constant itself and have to calculate the part dissociated.

Good to hear you figured it out. Also good luck on the AP test, its the hardest one besides one of the English tests.