Wanted: Chemistry Help

Factual Questions
1

No, I DO NOT want someone to do my homework for me. I want someone to HELP me so that I can understand it. A little background:

I’m a junior in High School, taking Chemistry 3-4 which is College-introductory level. In the beginning of the year, it was fine. Basically a rehashing of Chem 1-2, maybe in a bit more detail. Now, though, it has gotten difficult. One of the smartest people I know IRL (also a junior) has a C in the class. I’m not doing that well. We’re the only two juniors in the class.

Right now, we are working on Equilibrium. Acid-base equilibrium to be specific. So, if anyone, anyone at all could help me figure out what’s going on, I would REALLY appreciate the help. Either email me or post here and I’ll email you or I’m on AIM most all the time I’m online as Garf226. Thanks in advance for any help I receive.

2

We’re on the same thing (college Chem 2), but I don’t understand it either.

We were, actually…finals are this week.

3

The state that exists when acidic and basic ions in solution exactly neutralize
each other.

http://www.fsj.ualberta.ca/chimie/learning_tools/Acid_title.html

Or Better:

http://www.chem.uncc.edu/faculty/murphy/1252/Chapter16A/sld001.htm

4

handy,
Thanks for the links, but I know what equilibrium IS, I just want someone to help me understand how the problems work. The problems we are working on now are more complex than those in the slideshow thing (2nd link) and I don’t understand what to plug in to the equation where.

5

Err, it would help if you post some info on the problem…just to get an idea of where to begin–like whether the pKa of the acid is known, or if we’re trying to determine it, what the data given is, etc.

6

Oh, and handy (although I know you’ll not be back to read this):
your definition of acid-base equilibrium is outta whack.

An equilibrium is achieved (in a simple acid-base problem) when the system reaches a state such that there is no further change towards either the protonated or deprotonated form of a compound–which is not to say that it doesn’t happen, but that the rate in one direction exactly equals the rate of the reverse case.

Very few acid-base equilibria are actually neutral. For example, if I take a weak acid and dump it in some water, there is an equilibrium that will result in the water being (duh) acidic–but still at equilibrium.

Okay, enough of Chem 102…

7

Not sure what type of problems you’re trying to tackle, Garf, but the general types of problems involved in Acid/Base equilibria involve:

  1. Determining the [H+] (or pH); Determining the [OH-] (or pOH), where denotes “the concentration of”, when given equilibrium constants

  2. Determining the equilibrium constant when given concentration values.

Problems of type 2 are fairly straightforward:

K=(product of the concentrations of ions in solution at equilibrium) / (concentration of acid at equilibrium).

Problems of type 1 are a bit trickier, but the general methodology is:
Given: HA <–> H+ + A-
K(for HA)
Starting concentration of HA (let’s call this value C)

Find: [HA], [H+], [A-] at equilibrium
Solved as follows:

Let [H+] = [A-] = x (where we are trying to solve for x)

K(HA) = [H+][A-]/[HA] = (x)(x)/(C - x)

This yields the quadratic equation K(HA)*(C-x) = x^2, which can then be solved for x.

Sometimes, to speed up the calculation, the approximation (C-x) ~= C is used, since x is usually small in magnitude when compared to C. Then, the approximate solution becomes:

x = sqrt(K(HA)*C) Using this approximation, you then have to go back and check to ensure that the assumption (C-x) ~= C is valid (i.e., that x really is small when compared to C). Note also that you can only use this approximation if you are not asked to find [HA], since, with the approximation, you are essentially ignoring changes in [HA].

There are variants on this type of problem. One variant is the buffer solution problem where starting values for both HA and A- concentrations are given. For these, you merely factor in the additional A- concentration (let’s call this B) in your calculations as in:

K(HA) = (x)(x+B)/(C-x)

Another type of problem involves multiprotic acids, that is, acids which deprotonate over multiple steps. For instance:

H[sub]3[/sub]A <–> H[sub]2[/sub]A- + H+
H[sub]2[/sub]A- <–> HA-- + H+
etc…

These are solved by using systems of simultaneous equations. The math starts to get really messy if you can’t use the (C-x) ~= C approximation.

Of course, for base equilibria, make appropriate substitutions to the terms, but the math is the same.

Hopefully, some of this is helpful… post back, and let us know what types of problems you’re trying to solve.