Gotta Love Equilibrium.

Because my mind is about to explode but I shouldn't be slacking off because my grade depends heavily on this exam. Let's talk equilibrium.

Important things to remember when dealing with equilibrium problems:

• Large k = mostly product (product favored)
• Small k = mostly reactants (reactant favored)
• Q>K is proceeds towards reactants
• Q <>
• Acids are H+ donors
• Bases are H+ acceptors
• Quadratic Formula: x = -b +/- Square root of b^2 - 4ac divided by 2a
• Factors that affect equilibrium: 1) Change in [ ] of reactants/products 2) Change in presssure 3) Change in temperature. Catalysts do not affect equilibrium
  
• Kw = Ka*Kb
• pH and pOH have a difference of 14
• pH = -log [H3o+]
• pOH = -log [OH-]
• pKa = -logKa
• pH = pKa + log(A-/HA)

I've always had difficulty with predicting which way equilibrium reactions would shift. But now that I've really looked into and broken it down. It's pretty simple. Let's take a look at this sample problem.

• Consider the following reaction and predict which way the reaction will shift:

HNO2 + ClO4- <---> HClO4 + NO2-

First, we have to figure out which ones are the acids and which ones are the bases. As we recall from our notes above, acids are H+ donors. Which of those things in the rxn donate an H+ proton? HNO2 does because ClO4- doesn't even have an H+ proton. So if HNO2 is an acid, then that means NO2- in the products is the conjugate base.

Now that means Clo4- is a base and HClO4 is the conjugate acid. Now we have our two acids which are:

HNO2 and HClO4
Acid and Conjugate Acid

Looking at the Ka value, because Ka values determine acidity which determine acid strength. The higher the Ka value, the stronger the acidity is. The Ka of HNO2 is 7.2x10^-4. HClO4 is Perchloric acid which is a very strong acid and has a huge Ka value. Thus, we confirm that on the reactants side, the acidity is weaker (HNO2) and the acidity on the products side is much stronger (Perchloric acid).

Since the acidity strength is stronger on the products, the reaction would shift towards the reactants. Because in equilibrium reactions, the reaction will always shift from strong to weak. Therefore, the rxn proceeds towards the reactants side.

And that, my friends is how you predict which way an equilibrium reaction will shift.

Now, let's move onto a more in depth problem (and also because solving for pH is fun). Here we go:

• You are given a beaker containing 500.0mL of 0.10M HF. Determine the pH of the solution and justify any assumption you are making in the solving of this problem.

Okay here we go. First step, don't let the 500.0mL throw you off because that's actually for part B of the problem. We already have the initial [ ] of HF so we're cool. Let's write an equilibrium reaction and fill out an ICE table (Initial, Change, Equi.)

HF (aq) +H2O (l) <---> F- (aq) + H3O+ (aq)
0.10 n/a 0 0
-x +x +x
0.10-x x x

Alright. Now that we have our table. We need a Kc equation.

Kc = [F-][H3O+] / [HF]

We don't include water in the expression because it is a liquid. Liquids and solids are not included in the expression. Now filling in the blanks we get:

Kc = x^2 / (0.10-x)

This looks like messy work we're going to assume that x is a small value and therefore negligible. Consulting an almighty Ka value chart, we look up the Ka value for HF reaction which is 6.6x10^-4. Now we make the equation:

6.6x10^-4 = x^2 / 0.10

Digging into our brains for that simple algebraic problem solving, x therefore has a value of: 8.124X10^-3. Because we assumed that x was small, let's check it by dividing our x value by the initial concentration of HF which was 0.10M.

8.124X10^-3/0.10M * 100% = 8.124%

Aw crap. 8.124% means that x isn't small enough to be negligible. X has be less than 5% in order for us to void it. That means we gotta bust out the quadratic. In order to use the quadratic we need to have our equation in the format of: ax^2 + bx +c = 0. After using the stupid quadratic formula, we find that x is either 0.0078 or a negative number. Can't have a negative [ ] so we assume that x is 0.0078M.

Now our brain is about to die so we don't even know what we're trying to solve anymore. Oh right, we need the pH. Looking back at our ice table, the [H3O+] = x which means at equilibrium the concentration is 0.0078. Using our nifty knowledge, we know that we can use the equation:

pH = -log [H3O+]
pH = 2.10790 ~ 2.11

And now after all that, we have figured out the pH. Yay. Now my brain is about to explode. That is all for Tevina's 101 in Equilibrium work. PAYCE OUT!