Great Thanks to Tina Nye (A-5; 05-06) for much work on
graphics: making a rough idea reality!
A system in
equilibrium is like our ants up there! As long as the ants work at the same
speed, the piles of sand remain in equilibrium. Neither gets bigger or
smaller (It is important to note however that the piles are not the same size!).
The work that each ant does exactly offsets what the other is doing. Products
are being stacked up and the same rate that they are being taken away.
Reactants are being stacked up at exactly the same rate that they are
being taken away. If this were a chemical reaction, reactants would be colliding
to make product at the same rate that products would be colliding with
each other to make fresh reactant again! The reaction never really stops, but
the forward reaction and reverse reaction proceed and the same rate, so
we don't see a change in the size of the piles! Nifty Huh!!!!?
Chemical equilibrium is reached when the
rates for the forward and reverse chemical reactions are
equal for a chemical system. Or you could say when products are being made as
fast as they are breaking down to form reactants again. The concentrations of products and reactants are generally
To know what the concentrations are, you use the Equilibrium Constant
= [products]/[reactants]. Its a ratio : ) So Keq > 1 favors products, Keq
< 1 favor reactants.
For a general equation like:
wA + xB <===> yC + zD
the square brackets mean concentration, in molarity, if it is given in some
other unit you must calculate molarity. Coefficients in the chemical equation become
exponents in the Keq expression. Leave out any solids or pure liquids
since they have undefined molarities ; ).
If you spend time with the following notes, you'll master this topic in no time.
final concentrations and cheating while were doing it.
Keq expression, no numbers yet:
<===> 2 HI Given H2=I2=
0.200 M and Keq = 0.64
- ICE table:
Keq = [HI]2
Substitute in actual numbers
0.64 = (2x)2/ (0.200-x)
(0.200-x) Cheat by ignoring the -x in
denominator. This will not work well because Keq is not small (.64 to cheat
Keq should be less than 0.1), but
let's see what happens.
x= .08 now substitute back.
When Keq is sufficiently small, this is an okay
approximation. The quadratic is best though.
Le Chatelier's Principle
A chemical system, once it as equilibrium will respond to a stress or change in the environment by responding to reduce the stress.
A complete set of notes including familiar examples.
Chatelier's Principle notes.
Ionization of Acids and Bases: Ka & Kb
or strong bases always disassociate (ionize) 100%, we really aren't going see
much of them in this unit. Weak acids and weak bases on the other hand like to
stay together as molecules, they don't disassociate very much.
The measure of
how "strong" an acid is is given by Ka. The measure of how
"strong" a base is is given by Kb. A high Ka means that you get more
ionization so produce more H3O+ (hydronium) in a solution
with water. More hydronium means you have a "stronger" acid.
Ka = [products]/[reactants]
Kb = [products]/[reactants]
Acids when ionized, always separate at the H+, bases at the
OH-. Also remember, square brackets mean concentration in molarity.
If Ka is 3.5 x 10-8 what is the [H+] in a 0.050M solution of
Equation for ionization of hypochlorous acid: HClO <---> H+ +
2) Ka expression (no numbers): Ka = [H+][ClO-]/[HClO]
( Well will always leave out the -x term in the denominator. Our
justification is that Ka will always be small so x will be small and we can
ignore the quadratic without causing too much grief : )
3) Substitute and solve:
let x stand for both [H+]and [ClO-] then
x2/0.050 = 3.5 x 10-8.
x = 4.2 x 10-
& Kb Tutorial
The Solubility Product: Ksp
Ksp comes from
Keq, only this time our reactants are solid so they get left out. That is because
for a solid concentration is meaningless. I mean the solid is 100% solid but
there is no solvent so you are sort of dividing by 0 to try to get
Steps in solving problems
1. Write a balance dissolving equation.
Ex: Al(OH)3 <---> Al3+ + 3OH- (Leave the polyatomics
together, split the compound between the metal and the anion).
2. Write a Ksp expression (no numbers yet).
Ex: Ksp = [Al3+][OH-]3 (Coefficients become
powers, and are on the outside of the brackets. The charges are on the inside of
the brackets are just part of the ion's label, not really numbers at
3. NEW, NEW, NEW: Substitute X for ion according to mole ratio. If X =
Al3+ , then OH- = 3X because they are 1:3 in the balanced chemical equation.
Substitute into Ksp expression.
Ex: Ksp = (x)(3x)3 (Exponent is on outside of parenthesis, so
cube 3x first then multiply by x)
Then Ksp = 27x4
From here you could solve for X (molar solubility) given KSP or you could solve
for the numerical value of Ksp given X (molar solubility)
Common Ksp's are:
x2 ( Mole ratio 1:1)
4x3 (Mole ratio 1:2, or 2:1)
27x4 (Mole ratio 1:3, or 3:1)
108x5 (Mole ratio 2:3 or 3:2)
if you have some other solution for Ksp, check your algebra.
A thorough text heavy set of notes.
Predicting whether precipitation occurs boils down to one question. Is my ION PRODUCT
(Qip) greater than the KNOWN SOLUBILITY PRODUCT (KSP)?
Qip > Ksp then precipitation.
Ksp > Qip no precipitation.
The final concentration of
CaCl2 in a solution is 1.5 x10-5 M and the final concentration of NaF in a solution
AFTER mixing is 2.2
x10-2 M Will precipitation occur?
Ksp of CaF2 is 4.0 x10-11
Step 1: Write a balanced dissolution reaction.
(Hint: if you're not sure what will form, read the question carefully, the known Ksp of the compound must be given. That is what might or might not form depending on the ION PRODUCT).
Ex. CaF2 <--> Ca2+ + 2F-
Step 2: Write a Ksp expression but call it Qip
(it will have our data in it, from the concentrations that we put
together in a test tube.)
Ex:Qip = [Ca][F-]2
Step 3: Subsitute the ACTUAL VALUES of the concentrations of the ions that we actually have.
because we have no unknowns. We just want to see if our ion product is greater than the known
Ksp. If so, the solution can't hold that many ions and precipitation occurs.
Ex: Ca2+ = 1.5 x10-5 M and F- = 2.2 x10-2 M
(from the problem)
Kip= [Ca2+] [F-]2
Qip = (1.5 x10-5)(2.2 x10-2)2
Hint: raise to power 1st)
Qip = 7.26 x10-9
Ksp = 4.0 x10-11
Qip is 180x more than Ksp so we get a precipitate or GUNK in the bottom of test tube.
Remember negative exponents are fractions 10-9 is bigger than 10-11
Looks like Qip is greater than Ksp!!!