Chemistry Final Review

Seite 1

K(p)

concentrations of products over reactants

ONLY GAS!!!!!!

Partial pressures- NOT concentration

homogeneous equilibrium= everything in the equilibrium mixture is present in the same phase.( to use Kp, everything must be a gas)

Just like Kc, Kp always has the same value (provided you don't change the temperature)

EQUILIBRIUM

Le Chatelier's Principle

Equilibrium concentrations are only effected by temperature

***Same number of moles on both sides means no change

***

Using an ICE Table with Concentrations Example

often carried out for weak acid titrations

Find the concentration of A- for the generic acid dissociation reaction:

HA(aq)+H2O(l)⇌A−(aq)+H3O(aq)

with [HA(aq)] initial=0.150M and Ka=1.6×10−2

1. Fill in given concentrations:

Reaction:HAA-H3O+I0.150 M0.000 M0.000 MC???E???

2. Calculate the change concentrations by using a variable 'x'Reaction:HAA-H3O+I0.150 M0.000 M0.000 MC-x M+x M

+x M

E???

The change in concentration is unknown, so the variable x is used to denote the change. x is the same for both products and reactants because equal stoichiometric amounts of A- and H3O+ are generated when HA dissociates in water

3. Calculate the concentrations at equilibrium

Reaction:HAA-H3O+I0.150 M0.000 M0.000 MC-x M+x M+x ME0.150 - x Mx Mx M

4. Use the ICE table to calculate concentrations with K​aThe expression for Ka is written by dividing the concentrations of the products by the concentrations of the reactants. Plugging in the values at equilibrium into the equation for Ka gives the following:

Ka=x20.150−x=1.6×10−2

To find the concentration x, rearrange this equation to its quadratic form, and then use the quadratic formula to find x:

(1.6×10−2)(0.150−x)x2+(1.6×10−2)x−(0.150)(1.6×10−2)==x20

solve for x and eliminate the negative version of x

K(c)

ACIDS AND BASES

Circle of pH

(for K constant)

Calculating Weak Acid and Weak Base pH

Find concentrations using what is given

Find K(a) or K(b) value if pK(a) or pK(b) is given

Make ICE table with concentrations

Solve for x using K value

Use x to find concentrations

Use concentration to find value needed in question (pH, [H+], [OH-], pOH)

Calculate Ksp of Precipitate Example

Calculate the solubility product constant for lead(II) chloride, if 50.0 mL of a saturated solution of lead(II) chloride was found to contain 0.2207 g of lead(II) chloride dissolved in it.First, write the equation for the dissolving of lead(II) chloride and the equilibrium expression for the dissolving process.

PbCl2(s) --> Pb2+(aq) + 2 Cl-(aq)Ksp = [Pb2+][Cl-]2

Second, convert the amount of dissolved lead(II) chloride into moles per liter.

(0.2207 g PbCl2)(1/50.0 mL solution)(1000 mL/1 L)(1 mol PbCl2/278.1 g PbCl2) = 0.0159 M PbCl2 Third, create an "ICE" table.

PbCl2 (s) Pb2+(aq) Cl-(aq) Initial Concentration All solid 0 0 Change in Concentration - 0.0159 M (dissolves) + 0.0159 M + 0.0318 M Equilibrium Concentration Less solid 0.0159 M 0.0318 M Fourth, substitute the equilibrium concentrations into the equilibrium expression and solve for Ksp.

Ksp = [0.0159][0.0318]2 = 1.61 x 10-5

Calculating Concentration in Titration using Molarity

M(a)V(a) = M(b)V(b)

Normality

calculate moles of acid or base and convert to moles of OH or H

moles of OH or H

L

ELECTROCHEMISTRY

Assigning Oxidation Numbers:

Free elements = 0

Single element ions= change on ion

Alkali metals = +1

Alkaline Earth Metals = +2

Fluorine = -1

Hydrogen = +1 (unless in binary with metal, then -1)

Oxygen = -2 (unless peroxide O(2) , then its -1

Calculate remaining charges

Balance in Acidic Medium:

Write unbalanced equation for the reaction in ionic form

Make two half reactions (use oxidation reactions and separate , then balance by adding electrons)

Balance the atoms (except for O and H)

Add H2O to balance O and H atoms

Add electrons to balance

Equalize number of electrons in both half reactions by multiplying

Add two half reactions together and cancel out things as necessary

Balance in Basic Medium:

Follow all steps in acidic medium

Add OH- to both sides of equations for every H+ that appears in the final equation

For every H and OH, make H20

Reduce the number of H20 if they cancel

Balance by Half Reaction Method:

Assign oxidation numbers

Write half reactions for elements that changed oxidation numbers

Add coefficients and balance elements

Add electrons to both reactions to balance charges

Multiply electrons to ensure they cancel out when reactions are added

Add coefficients back into original equation

Check overall balanced equation

If elements are present that did not change oxidation numbers, check that they are balanced

Calculate Cell Potential (Voltage):

Write two reduction half reactions

Look up cell potentials (from chart)

The reaction with a more negative value (in the reduction chart) is oxidation, so flip this reaction and write it as oxidation

Change the sign of the cell potential for the flipped reaction

Add the cell potentials together to get the total voltage

When the cell potential is positive it is spontaneous

Electrochemical Cell:

increase in size

concentration decreasing

positive electrode

decrease in size

concentration increasing

negative electrode

KINETICS

Rate Expression

reactants are negative

Reaction Order

units for rate constant, k

M/s

1/s

1/(M*s)

Catalyst

Increases reaction rate and rate constant by providing a lower activation energy needed

SOLUTIONS

Types of Acids and Bases

Bronsted Acid - will donate a H+

Bronsted Base- will accept a H+

*** can be both!!!

Equilibrium

Acids and Bases

ElectroChemistry

Kinetics

Solutions