USF General Chemistry II Exam 3

Start Ch. 17- To deal with something quantitatively (quantity) with an equilibrium, you must know what?

the equilibrium constant, Ksp

How much is a nearly insoluble ionic compound?

Usually only half a gram at most.

The Solubility-Product Constant, Ksp, governs what?

Dissolving of a reaction.

Ksp = [M+][X-]

How to write solubility-product expression EXAMPLE

Have a compound that will disassociate-

Hg2Cl2 (s)

Write the chemical equation-

HgCl2 (s) <-> 2Hg+ (aq) + 2Cl- (aq)

The Ksp is given in the products-

Ksp = [Hg+]^2[Cl-]^2

What is the difference between Hg2Cl2 and HgCl2?

Hg2Cl2 is atomic- [Hg+]^2

HgCl2 is molecular-

[Hg^2+]

The Ksp of a slightly soluble ionic compound is expressed in terms of what?

Molar concentrations of ions in the saturated solution.

What is molar solubility?

Number of moles of compound that dissolve to give ONE liter of saturated solution.

Solving for Ksp given Solubility- Exactly 0.133 mg of AgBr will dissolve in 1.00 L of water. What is the value of Ksp for AgBr?

1. Chemical equation-

AgBr (s) <-> Ag+ (aq) + Br- (aq)

2. Ksp = [Ag+][Br-]

3. Change solubility into Molarities-

0.133 x 10^-3 g / 1.00 L x 1 mol / 187.772 g = 7.083 x 10^-7

4. Ice Chart-

Since Ag+ and Br- are one mole each, they both = x

5. Apply solubility found to Ksp equation-

Ksp =

(7.083 x 10^-7)^2

6. Final answer-

Ksp = 5.02 x 10^-13

Exam Note regarding Ksp values

When we are looking for the solubility of a compound, we will be given the Ksp value, or we will look it up on a table.

When the Ksp is known, we can find what?

The molar solubility- or the amount of a compound that dissolves in a solution.

Solving for Solubility given Ksp- Hg2Cl2 has a Ksp equal to

1.3 x 10^-18. What is the solubility of Hg2Cl2 in grams per liter?

1. Chemical equation- Hg2Cl2 (s) <-> Hg2^2+ (aq) + 2Cl- (aq)

2. Ksp = [Hg2^2+][Cl-]^2

3. Ice Chart-

Since we only have one Hg2^2+, = x. We have two Cl-, = 2x.

4. Plug in to Ksp equation-

Ksp = x(2x)^2 (reason this is to the second power is because we have two moles of Cl-)

Simplifly the expression to get

Ksp = 4x^3

4. Set equal to Ksp value given-

1.3 x 10^-18 = 4x^3

5. Manipulate the equation so x is alone-

x = 6.88 x 10^-7 M

6. Change to g/L-

6.88 x 10^-7 mol/ 1L x 472.086g / 1 mol =

3.2 x 10^-4 g/L

Exam Note regarding Molecular vs Atomic state

For problems where we are given the Ksp to find solubility, we can decipher if it will be the molecular state or the atomic state based on the Ksp table George gives us.

Ksp and Solubility have a (direct/indirect) relationship.

direct; the higher the Ksp, the more soluble it is.

What effect does the presence of a common ion effect have on solubility? In terms of

MX (s) <-> M+(aq) = X-(aq)

We can use Le Chatelier's Principle to predict that a reaction will shift in the reverse direction when M+ or X- is added.

Solving for Solubility of a compound using Common Ion Effect- What is the molar solubility of silver chloride in 1.0L of a solution that contains 2.0 x 10^-2 mol of HCl?

1. Use table to find Ksp for AgCl-

1.8 x 10^-10

2. Construct an Ice Chart-

AgCl (s) <-> Ag+ (aq) + Cl- (aq)

Ag+ = x

Cl- = 0.020M + x (HCL disassociates into Cl-)

3. Ksp = [Ag+][Cl-], plug in values found-

1.8 x 10^-10 = x(0.020 + x)

4. Simplifying assumption-

Since the Ksp is so small, we can assume-

1.8 x 10^-10 = 0.020x

5. Solve Equation-

x = 9.0 x 10^-9 M

The reaction quotient, Q, is used to determine what?

Whether a precipitation will occur or not.

We look for ________ conditions to determine the value of Q = [M+][X-]

Initial

The definition of Q is what?

The products at any instance between turning into an aqueous solution to the reaction reaching equilibrium.

Determine whether a precipitate will occur or not- Ca3(PO4)2 has a Ksp of 1.0 x 10^-26. A sample of urine contaisn 1.0 x 10^-3 M Ca^2+ and 1.0 x 10^-8 M PO4^-3

1. Chemical equation-

Ca3(PO4)2 <-> 3Ca^2+ + 2PO4^-3

2. Ksp = [Ca^2+]^3[PO4^-3]^2

3. Solve for Qc-

Qc = (1.0 x 10^-3)^3 (1.0 x 10^-8)^2

Qc = 1.0 x 10^-25

4. Compare Qc and Ksp- Ksp < Qc

A precipitate WILL form.

When a problem gives you the concentrations of two samples that are MIXED, make sure to calculate the ____ initial concentrations.

new

Determine whether a precipitate will occur or not using new initial concentration- Exactly 0.400L of 0.50M Pb^2+ and 1.60L of 2.50 x 10^-2 M Cl- are MIXED TOGETHER to form a 2.00L solution. Calculate Qc and predict whether PbCl2 will precipitate. Ksp for PbCl2 is

1.6 x 10^-5.

1. Calculate the Molarities of [Pb^2+] and [Cl-]-

(0.50M)(0.400L)

/(2.00L) = 0.100 M [Pb^2+]

(2.5 x 10^-2M)(1.60L)

/(2.00L) = 0.0200M [Cl-]

2. Chemical reaction-

PbCl2 (s) <-> Pb^2+ (aq) + Cl- (aq)

3. Ksp - PbCl2 = 1.6 x 10^-5

4. Solve for Qc-

Qc = (0.100)(0.0200)^2 = 4.00 x 10^-5

5. Compare Qc and Ksp- Ksp < Qc

A precipitate WILL form.

Fractional Precipitation is the technique of what?

Separating the weakest ion (least soluble compound) from a solution by adding a reactant to precipitate with the first ion, and so on; like a conveyer belt.

In fractional precipitation, if solubility is close between the compound and the reactant added, what happens?

They BOTH precipitate.

True or False: Between the compound and the reactant added, whichever has the smaller Ksp will precipitate sooner.

True

Fractional Precipitation- which precipitate will form first?- A solution contains 1.0 x 10^-4 M Cu+ and 2.0 x 10^-3 M Pb^2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp = 1.4 x 10^-8) or CuI (Ksp - 5.3 x 10^-12) precipitate first? Specify the concentration of I- necessary to begin precipitation of each salt.

1. Ksp expression for PbI2

1.4 x 10^-8n = Ksp = [Pb^2+][I-]^2

2. [Pb^2+] is known, solve for [I-]-

1.4 x 10^-8 = (2.0 x 10^-3) [I-]^2

[I-] = 2.6 x 10^-3 M

Any I- in excess of the calculated concentration will cause solid PbI2 to form.

3. Ksp expression for CuI

5.3 x 10^-12 = (1.0 x 10^-4) [I-]

[I-] = 5.3 x 10^-8

A concentration of I- in excess of this calculated concentration will cause CuI to form.

4. Which will precipitate first?

CuI will precipitate first since the I- required is less.

5. What is the concentration of Cu+ ion remaining just as the PbI2 begins to precipitate?

Ksp(PbI2) = [I-] = 2.6 x 10^-8

Ksp(CuI) = 5.3 x 10^-12 = [Cu+](2.6 x 10^-3)

[Cu+] = 2.04 x 10^-9 M

When a salt contains the conjugate base of a weak acid, the pH will affect what?

The solubility of the salt.

According to Le Chatelier's Principle, if you add an acid-base pair (Buffer), it will suppress the salt's solubility. What has to happen to the reaction?

It must reach equilibrium again, resulting in a change of pH.

As the acid concentration increases, X- reacts with the H3O+, forming MX and reducing the X- concentration. What is the result?

More MX dissolves, increasing the solubility.

What is a complex?

Species formed in chemical reactions that are very resistant to breaking apart. Complexes hinder sparingly soluble compounds from dissolving, making them less soluble.

KNOW THESE- Metal ions that form complex ions- hint- seven

Ag+, Cd2+, Cu2+, Fe2+, Fe3+, Ni2+, Zn2+

Ligands (define)

Lewis bases that have a net negative charge through availability of a lone pair of electrons. Includes CN-, NH3, S2O3^-2, and OH-

What is the complex-ion formation equilibrium?

An equilibrium driven process that favors products. Free cations and free ligands usually don't exist as free, due to equilibrium lying to the right (favors products).

Kf, formation constant

Kf = (products)/

(reactants)

True or False: equilibrium favors the complex ion.

True; this is because the Kf values are quite large.

Kf for complex ions- What is the concentration of Ag+ (aq) ion in 0.00010 M AgNO3 that is also 1.0 M CN-? Kf for Ag(CN)2^- is 5.6 x 10^18

1. First, we assume Ag+ forms the entire complex ion.

2. Calculate [Ag+] using the disassociation equilibrium.

3. Chemical Reaction-

Ag+ (aq) + 2CN- (aq) <-> Ag(CN)2^- (aq)

4. Ice Chart (Formation)-

Initially we have 0.00010 M Ag+, but that will be given to the complex ion. We started with 1.0 M 2CN-, but since we are moving the 0.00010 M Ag+, it must be applied to the 2CN- as well. 1.0 M - 2(0.00010) = 0.9998 is our new value.

5. Ice Chart (Disassociation)-

Flip the reaction and give the initial numbers. Find out the change. This will be 0.00010 -x for Ag(CN)2^-, x for Ag+, and 0.9998 + 2x for 2CN-.

6. Finding Kd value-

Use the inverse of Kf.

1/5.6 x 10^18 = 1.8 x 10^-19

7. Plug in numbers-

1.8 x 10^-19 = x(0.9998 +2x)^2 / (0.00010 - x)

8. Get rid of extra x's (No quadratics)

1.8 x 10^-19 = x(0.9998)^2 / 0.00010

9. Solve-

x = 1.8 x 10^-23 = [Ag+]

Cations can be separated into groups according to what?

Precipitation properties

The qualitative analysis scheme for the separation of metal ions uses _______ ________ to separate Co2+, Fe2+, Ni2+, and Zn2 (Analytical Group III).

sulfide solubility

What is the quantitative analysis scheme?

To separate and identify ions into known analytical groups.

Start of Ch. 18- Thermodynamics (define)

The study of heat and other forms of energy involved in chemical or physical processes.

What are the standard state conditions? Why are they needed?

For a pure substance, 1 atm pressure.

For a substance in solution, 1 M solution.

These are needed to harmonize experiments and compare data.

What is the First Law of Thermodynamics?

Energy cannot be created or destroyed but can be converted from one form to another.

Internal energy, U, is the sum of what two energies that make up a system?

Potential and Kinetic energy

The two types of energy exchange between the system and its surroundings are...

Heat, q

Work, w

The First Law of Thermodynamics describes the universe as what?

The sum of all energy in a system and its surroundings.

Sign Convention for q- When heat is EVOLVED by a system, q is __________, and when heat is ABSORBED by a system, q is _________.

negative; positive

Sign Convention for w- When a system expands, change in volume (DV) is positive, so w is _________. When the system contracts, change in volume (DV) is negative, so w is _________. This is a (direct/indirect) relationship.

negative; positive; indirect

Sign Convention for w- When the system does work on the surroundings, the internal energy (decreases/increases), and when the surroundings do work on the system, the internal energy (decreases/increases).

decreases; increases

Calculate the amount of work done given a change in volume at constant pressure- Equation

w = (-P)(Delta V)

Heat isn't made, it is __________.

liberated

Calculate change in enthalpy at constant pressure- Equation

q (Heat) x p (pressure) = DH (change in enthalpy)

First Law of Thermodynamics- Equation

DU (Change in Energy of Universe) = DH (change in enthalpy) - P (pressure) x DV (change in volume)

What does "spontaneous" mean?

A process that occurs on its own and will do so until equilibrium is reached.

What is entropy?

A measure of how dispersed the energy of a system is in the different possible ways a system can contain energy; a state of disorder.

Entropy supports _____ ________.

heat transfer

Entropy Change - Equation

DS (Change in entropy) = Sf - Si

What is a state function? Give examples.

When one variable depends on two others to determine the state of a substance. Examples are Entropy and Gibbs Free Energy.

When physical is going from solid to liquid to gas, what is happening to the entropy?

The entropy is increasing.

Whichever side (reactants or products) has more moles has more what?

Entropy

Why does a gas have more entropy than a solid or liquid?

This is due to the spontaneity of the state it is in. Gas is more spread out, which indicates more spontaneity, hence more entropy.

What is the Second Law of Thermodynamics?

The total entropy of a system and its surroundings ALWAYS increases for a spontaneous process; Entropy supports spontaneity.

Using the Second Law of Thermodynamics, state whether a reaction is spontaneous or not- Equation

Find DS (change in entropy)- DS = entropy created + q/T

Find spontaneity-

DS > q/T

When molecules spread out more, greater energy will disperse, for a greater chance at what?

A spontaneous reaction

Fusion is the opposite of __________, and vaporization is the opposite of _________.

melting; condensation

Use the Second Law of Thermodynamics to predict whether or not a chemical reaction is spontaneous- Acetone, CH3OCH3, has a standard formation (DHf) of -247.6 kJ/mol at 25 C. The same quantity for the vapor is -216.6 kJ/mol. What is the entropy change when 1 mol liquid acetone vaporizes at 25 C?

1. Find DH-

216.6 kJ/mol - -247.6 kJ/mol = DH = 31.0 kJ

Make sure to account for moles

2. Plug numbers into equation DS = DH/T

3. 31,000 J / 298K = 104 J/K = DS

For a reaction to be spontaneous, the entropy of a system and its surroundings must (increase/decrease)

increase

What is the Third Law of Thermodynamics?

A substance that is perfectly crystalline at zero kelvin, has an entropy of zero. Everything is compared to this perfect standard; however, there is ALWAYS ENERGY, so we can never truly reach zero.

Standard for gas?

Standard for liquid?

- 1 atm

- 1 M solution

Entropy Increases in Three Situations:

1. A molecule is broken into two or more smaller molecules

2. Increase in the number of moles of gas

3. Changes from solid to liquid to gas.

Predict the sign of DS for this reaction- Ba(OH)2 + 8H2O (s) + 2NH4NO3 (s) -> 2NH3 (g) + 10H2O (l) + Ba(NO3)2 (aq)

- 11 moles vs 13 moles

- Going from solids to liquids and gases

- DS is positive

Calculate the entropy change at 25 C for the following reaction- CH3Ch2OH (l) + O2 (g) -> CH3OOH (l) + H2O (l)

- Standard entropies given: CH3CH2OH (l)- 161 J/K x mol

O2 (g)- 205 J/K x mol

CH3OOH (l)- 160 J/K x mol

H2O (l)- 69.9 J/K x mol

1. Add reactant and product sides-

161 + 205 = 366 J/K

160 + 69.9 = 229.9 J/K

2. Subtract products - reactants for answer-

DS = 229.9 J/K - 366 J/K = -136 J/K

What are the two thermodynamic driving forces for a chemical reaction/process?

Entropy and enthalpy

Describe the change in free energy for a chemical reaction/process in terms of changes in enthalpy and entropy.

When enthalpy is high, Gibbs Free Energy is high; When entropy is high, Gibbs Free Energy is low. This can be deciphered by looking at the equation-

G = H - TS

Describe the relationship between change in free energy and the maximum amount of work that can be done on a system.

Spontaneous reactions can be used to perform useful work; a direct relationship

Ex: combustion of gasoline to move a vehicle.

Calculate change in Gibbs Free Energy, DG- Using standard enthalpies of formation, DHf, and the value of DS from the previous problem (-136 J/K), calculate DG for this reaction-

CH3CH2OH (l) + O2 (g) -> CH3COOH (l) + H2O (l)

1. Find the values of DHf-

-275 kJ/mol + 0 -> -484.5 kJ/mol + -285.8 kJ/mol

2. Solve for DH-

-495.3 kJ/mol

3. Use equation- DG = DH - T(DS)-

DG = -495.3 kJ/mol - 298K(-0.136 kJ/mol)

4. Solve for DG-

DG = -454.77 kJ/mol

When entropy is a large (positive/negative) value and enthalpy is a large (positive/negative) value, a spontaneous reaction will occur.

positive; negative

Calculate DG using the standard free energies of formation- CH3CH2OH (l) + O2 -> CH3COOH (l) + H2O (l)

1. Find the values of DG-

-174.8 kJ/mol + 0 -> -392.5 kJ/mol + -237.2 kJ/mol

2. Solve for DG-

-629.7 -(-174.8) = -454.9 = DG

Name the spontaneity of DG based on its value:

DG < -10 kJ: spontaneous

DG > +10 kJ: nonspontaneous

DG = (Between +10 and -10): equilibrium

Why is a chemical reaction spontaneous only when DG is below -10? Why is it nonspontaneous when it is above +10?

This means the reactants (initial state) have more free energy than the products (final state); products (final state) gave more free energy than the reactants (initial state).

Why is a chemical reaction at equilibrium when DG = 0?

Forward and backward reaction are occurring at the same rate.

If only gases are present, use _________. If only solutes in liquid solution are present, use _________.

pressure; molarities

Write the expression for thermodynamic equilibrium constant for these reactions-

a. N2O4 (g) <-> 2NO2 (g)

b. Zn (s) + 2H+ <-> Zn2+ (aq) + H2(g)

a.

K = P^2NO2/PN2O4

b.

K= [Zn^2+]PH2 / [H+]^2

Standard free energy change related to Q- Equation

DG = DG^0 + RT(lnQ)

Standard free energy change related to K and DG = 0- Equation

DG^0 = -RT(lnk)

Calculate the value of thermodynamic equilibrium constant at 25 C for the reaction N2O4 (g) <-> 2NO2 (g)- Standard free energy of formation at 25 C is 51.39 kJ/mol for No2 and 97/82 kJ/mol for N2O4.

1. Find DG-

DG = 2(51.30 kJ/mol) - 97.82 kJ/mol = 4.78 kJ

2. Manipulate the equation-

DG = -RT(lnK)

lnk = -DG/RT

3. Solve the equation-

lnk = -(4,780 J/mol)/(8.315 J/mol x K)(298K)

lnk = -1.929

e^-1.929 = 0.145 = K

What is the difference between DG and DG^0?

DG is free energy change of a reaction at any temperature; DG^0 is the change at standard conditions in 1 mole.