Notes:

The Ka1 for H2SO3 is 0.015
The Ka2 is 1.2 x 10-7
For H2CO3: Ka1 is 4.3 x 10-7 and Ka2 is 5.6 x 10-11

To understand how to do many of the Equilibrium problems you MUST watch video 13.2 in the APChem video folder.

Ksp = 4 ×10−38 for Fe(NO3)3

Notes on finding the pH of salts:

Problem #1: The Ka of formic acid (HCOOH) is 1.8 x 10¯4. What is the pH of 0.35 M solution of sodium formate (NaHCOO)?
Solution:
Calculate the Kb of the formate (HCOO¯) ion:
  • 1.00 x 10¯14 = (1.8 x 10¯4) (x)x = 1.00 x 10¯14 / 1.8 x 10¯4 = 5.56 x 10¯11
Calculate the [OH¯]:
  • 5.56 x 10¯11 = [(x) (x)] / (0.35 - x)neglect the minus x
  • x = 4.41 x 10¯6 M
Calculate pOH, then pH:
  • pOH = - log 4.41 x 10¯6 = 5.36pH = 14 - pOH = 14 - 5.36 = 8.64

Problem #2: Calculate the pH of a 0.36 M CH3COONa solution given the pKa of acetic acid is 4.745.
Solution:
Using the pKa, calculate the Ka of acetic acid:
  • Ka = 10¯pKa = 10¯4.745 = 1.80 x 10¯5
Using the Ka of the acid, find the Kb of the base (the CH3COONa):
  • 1.00 x 10¯14 = (1.80 x 10¯5) (x)x = 5.56 x 10¯10
Use the Kb expression to calculate the [OH¯]:
  • 5.56 x 10¯10 = [(x) (x)] / (0.36 - x)neglect the minus x
  • x = 1.415 x 10¯5 M (I threw in a guard digit.)
Calculate pOH, then pH:
  • pOH = - log 1.415 x 10¯5 = 4.85pH = 14 - pOH = 14 - 4.85 = 9.15
Comment on the solution to Example #3:
You can use the pKa to get the pKb
  • 4.745 + x = 14.000x = 9.255
Use the pKb to get the Kb:
  • Kb = 10¯pKb = 10¯9.255 = 5.56 x 10¯10
Go to rest of solution, at "Use the Kb expression," just above.

Problem #3: What is the pH of a 0.0510 molar solution of salt NaCN (the Ka for HCN is 6.166 x 10¯10)?
Solution:
CN¯ is the salt of a weak acid and hydrolyzes thusly:
  • CN¯ + H2O ⇔ HCN + OH¯
Determine the Kb for this reaction (using Kw = KaKb):
  • 1.00 x 10¯14 = (6.166 x 10¯10) (x)x = 1.6218 x 10¯5 (I kept a couple guard digits.)
Write the Kb expression and solve for [OH-]:
  • Kb = ([HCN] [OH¯)] / [CN¯]1.6218 x 10¯5 = [(x) (x)] / 0.0510
  • x = 9.0946 x 10¯4 M
Determine pOH (with negative log) and then pH (using pH + pOH = 14):
  • -log 9.0946 x 10¯4 = 3.0412pH = 14 - 3.0412 = 10.959 (to three sig figs)



When to use the quadratic Equation:

Because x > 5%of the original concentration, the assumption was not valid, and the full expression must be evaluated using the quadratic equation.

equilnote.png
When you know the K for two reactions and then add the reactions together to get the new K you multiply the two K's for the reactions added together.
Gibbs_Free_Energy.png
Enthalpy;
D H


Spontaneity

+ D S

+ D H


At all temperature
+ D S

- D H


At high temperature
- D S

- D H


At low temperature
- D S

+ D H


Not spontaneous
D G is dependent on pressure. The equation that links them together is:
D G = D G° + RTln(P)
  • D G is the free energy
  • D G° is the standard free energy, for a gas the standard state is always 1 atm.
  • R= 8.3145 J
  • T is the temperature in kelvin.
  • P is the partial pressure.
Free energy is also related to the equilibrium constant k and Q.
D G = D G° + RTln(Q)
At equilibrium, D G is 0 because at the product and reactant have equal free energy.
0 = D G° + RTln(Q)
D G° = -RTln(k).
Free energy is the max work obtained from the reaction.
Workmax= D G
Enthalpy

Aaah, yes, enthalpy. If you've just heard of enthalpy, you will have no idea what it is. When you're done with enthalpy, you'll still not have a clear good idea of what it is, but you will know how to do problems with it, and that's what is important.
The enthalpy of a system, H, is simply defined as:
H = E + PV
Enthalpy is equal to the total energy of the system, plus the pressure of the system times the volume of the system. It's sort of hard to grasp what enthalpy is, from that definition. Instead of enthalpy (H) itself, you will usually deal with a change of enthalpy (DH). And if pressure is constant, and the only work allowed to work on the system is through volume, then:
D H = q
Yep, you should think of enthalpy as sort of like heat. It's not heat exactly, but if those two conditions are met, then it is heat.
So if the change of enthalpy is increasing, that means it is gaining an increase of energy, and therefore is endothermic. If the change of enthalpy is decreasing, that means it is losing heat to the surroundings, or exothermic.
Final thought: +D H = Endo, -D H = Exo.
How do you find the change of H? It's the enthalpy of the final products, minus the enthalpy of the reactants. Or...
D H = D Hproducts - D Hreactants
So if the products have less energy than reactants, then D H will be negative, indicating energy was lost. And vice versa.


Entropy
Entropy, D S, is the degree of chaos in the universe. For example, when a glass cup breaks on the floor, the glass breaks into little pieces. The cup changes from a unified object to many smaller pieces.
D S is positive if things are going to more disorder.
D S is negative if things are going to less disorder.
The entropy of the universe is positive; the universe is becoming more disordered.
D Suniverse = D Ssystem + D Ssurrounding
The entropy for the three states of matter
Solid < Liquid < Gas
The molecules in the gas phase can be arranged in more ways than in the solid or liquid state. The gas phase also has less intermolecular forces to hold the molecules together.
Heat flow changes D Ssurrounding. If the system is exothermic, heat is released; the heat increases the disorder of the surrounding. If the system is endothermic, the surrounding loses heat. With less heat in the surrounding, the molecules move slower and entropy decreases.
D Ssurrounding = - D H/T
  • D H is the enthalpy of the system.
  • T is the temperature in kelvin.

entropy.png
For Entropy, Enthalpy try this web page:
http://www.chem1.com/acad/webtext/thermeq/TE4.html

Videos:

Watch these Videos and answer these questions with video 13.1 and 13.2:





Labs:

Must do Lab:







Do Lab 6 instead of 7.

Discussion


WebSite:

For Entropy, Enthalpy try this web page:
http://www.chem1.com/acad/webtext/thermeq/TE4.html