1. Chemistry Two SJBoylan Friday, October 25, 2019 Page 1 of 50
Chemistry
Two
Stephen Joseph Boylan
August 2019
2. Chemistry Two SJBoylan Friday, October 25, 2019 Page 2 of 50
Chapter 10 Solutions - A solution is a homogenous mixture of a solvent and one or more solutes. The solvent is
the major constituent of the solution. The solute is the minor constituent of the solution. A constituent is a
chemical species of the solution. Solubility is the amount of solute dissolved in an amount of solvent.
Molarity is concentration in units of moles per liter. Molarity may change with a change in temperature due to a
change in density.
Molality is concentration in units of moles solute per kilogram of solvent. Typically, molality will not change
with a change in temperature.
Colligative properties of a solution depend on the concentration of the solute. Some colligative properties are
freezing point depression, boiling point elevation osmotic pressure and vapor pressure lowering..
Freezing point depression is a decrease in the freezing point. The freezing point is the temperature and pressure
where the liquid phase and the solid phase of the solution exist in equilibrium. The first equation for freezing
point depression is delta Tf = Tf0 - Tf. delta Tf is the freezing point depression. Tf0 is the freezing point of
the pure solvent. And Tf is the freezing point of the solution.
The other equation for freezing point depression is delta Tf = i x Kf x m. m is the molality, moles solute per
kilograms solvent. Kf is the molal freezing point constant. For water Kf = 1.86 OC/m. oC is temperature in
degrees Celsius.
i is the vant Hoff factor. For molecules, i equals 1. For 1:1 salts, i equals 2. For 1:2 salts and 2:1 salts, i = 3.
Boiling point elevation is an increase in the boiling point. The boiling point is the temperature and pressure
where the vapor phase and the liquid phase of the solutions exist in equilibrium. The first equation for boiling
point elevation is delta Tb = Tb - Tb0. delta Tb is the boiling point elevation. Tb is the boiling point of the
solution. And Tb0 is the boiling point of the pure solvent.
The other equation for boiling point elevation is delta Tb = i x Kb x m. m is the molality, moles solute per
kilograms solvent. Kb is the molal boiling point constant. For water Kb = 0.52 OC/m. oC is temperature in
degrees Celsius.
Osmotic pressure is the pressure developed between two different solutions across a semipermeable membrane.
Vapor pressure lowering is shown by Raoult’s law. Raoult’s law assumes an ideal liquid solution. Raoult’s Law
states the partial pressure, Pi, of constituent i above a solution is equal to the mole fraction, xi, of constituent i in
solution times vapor pressure, Poi, of pure constituent i.
The Raoult’s Law equation is Pi = xi Poi.
Henry’s law is used to model low concentrations of a volatile solute. Henry’s Law states the concentration, Cg,
of the solute in solution is proportional to the pressure, Pi, of the solute above the solution. The equation for
Henry’s Law is Cg = k Pi. k is the proportionality constant called the Henry’s Law constant. Henry’s Law is
usually applicable to dilute solutions.
3. Chemistry Two SJBoylan Friday, October 25, 2019 Page 3 of 50
Molarity – A flask contains 3.785 liters of solution. The solution contains 0.354 moles of copper(II) chloride,
CuCl2 . The density of the solution is 1.021 grams per milliliter..
A. Calculate the molarity of copper(II) chloride in the solution.
B. Calculate the molarity of the chloride ion in the solution.
C. Calculate the molality of the copper(II) chloride in the solution.
Molality - A bottle contains a solution of 5.829 kilograms of water and 0.298 moles of cobalt(II) nitrate,
Co(NO3)2. The density of the solution is 1.054 grams per milliliter.
A. Calculate the molality of the cobalt(II) nitrate in the solution.
B. Calculate the molality of the nitrate ion in the solution.
C. Calculate the molarity of the cobalt(II) nitrate in the solution.
4. Chemistry Two SJBoylan Friday, October 25, 2019 Page 4 of 50
Parts per million, ppm – A sample of seawater has a reported value of 8.12 parts per million of strontium. The
density of the seawater sample is 1.025 grams per milliliter. An aquarium is filled with 4556 liters of seawater.
Calculate the amount of strontium in the aquarium. Report the amount in units of grams strontium.
Parts per billion, ppb – A sample of electrical transformer oil has a reported value of 561.2 parts per billion
polychlorinated biphenyl, PCB. The density of the electrical transformer oil is 1.025 grams per milliliter. A 500
KVA transformer is filled with 764.6 liters of the same transformer oil as the sample. Calculate the amount of
polychlorinated biphenyl in the transformer. Report the amount in units of grams polychlorinated biphenyl.
Mass Percent Solution – A bottle is labeled 32.8 mass percent hydrochloric acid, HCl. The density of the acid is
1.1892 grams per milliliter.
A. Select a basis for this analysis ________________________________________________________
B. Calculate the mass percent of water in the acid.
____________________________________________________________________________________
C. Calculate the molality of the acid.
____________________________________________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________
D. Calculate the molarity of the acid..
____________________________________________________________________________________
____________________________________________________________________________________
____________________________________________________________________________________
5. Chemistry Two SJBoylan Friday, October 25, 2019 Page 5 of 50
Henry’s Law 1 – The concentration of oxygen in a sample of water is 0.0031 M. The partial pressure of the
water vapor above the water sample is 2.4 atm. Calculate the henry’s Law constant.
Henry’s Law 2 – The Henry’s Law constant of nitrogen over an aqueous solution is 6.2x10-4(mol/L)/atm. The
partial pressure of nitrogen above the solution is 0.78 atm. Calculate the concentration of nitrogen in the
aqueous solution.
Raoult’s Law – A solution is made of 52.3 grams of glucose, C6H12O6, in 298 grams of water. The vapor
pressure of pure water at the same conditions is 23.76 mmHg.
A. Calculate the molar mass of the solute glucose.
B. Calculate the moles of solute.
C. Calculate the moles of solvent.
D. Calculate the mole fraction of solute.
E. Calculate the vapor pressure of the water over the solution.
6. Chemistry Two SJBoylan Friday, October 25, 2019 Page 6 of 50
Boiling Point Elevation – An aqueous solution contains 0.85 m sugar. The boiling point of water at the same
conditions is 98.2 oC. Calculate the boiling point of the aqueous solution.
Freezing Point Depression –A. Estimate the freezing point of a solution at 0.150 m aqueous solution of sugar.
Tf0 = 0.00 C
_________________________________________________________________________________
__________________________________________________________________________________
B. Estimate the freezing point of a solution at 0.150 m aqueous solution of potassium nitrate. Tf0 = 0.00 C
__________________________________________________________________________________
____________________________________________________________________________________
C. Estimate the freezing point of a solution at 0.150 m aqueous solution of chromium (II) nitrate. Tf0 = 0.00 C
____________________________________________________________________________________
______________________________________________________________________________________
Osmotic pressure – 2.03 grams of a solute mixes with 285 milliliters of water. Assume the density of water is
1.000 grams per milliliter. The osmotic pressure of the solution is 0.856 atm at 25 oC. Determine the molar mass
of the solute.
7. Chemistry Two SJBoylan Friday, October 25, 2019 Page 7 of 50
Chapter 11 Reaction Kinetics - Reaction kinetics is the science describing how fast a chemical reaction
proceeds. In chemical process design, engineers use reaction kinetics to calculate the size of the reactor and
process tanks. A reactor is a process vessel where the chemical reaction takes place. Fast reactions will use
smaller reactors. Slow reactions will use larger reactors.
The average rate of disappearance of a reactant is calculated from the concentrations and time by the following
equation
Average rate = - ( Concentration 2 - Concentration 1 ) / ( Time 2 - Time 1) (1)
The average rate of appearance of a product is calculated from the concentrations and time by the following
equation.
Average rate = ( Concentration 2 - Concentration 1 ) / ( Time 2 - Time 1) (2)
Note that in the equation for the disappearance (1) there is a negative sign, while in the equation for the
appearance there is no negative sign.
The rate law represents how the reaction proceeds in time. Consider a reaction when chemical A reacts with
chemical B to produce products. The reaction equation for this reaction is:
A + B => products (3)
For each reactions there is a rate law that shows how the reaction proceeds in time. Typically the rate law has
the form
Rate = k [A]n [B]m (4)
Where k is the rate constant. [A] is the concentration if A. n is the order of reaction with respect to A. [B] is the
concentration of B. m is the order of reaction with respect to B. The overall order of reaction, o, is calculated by
adding n and m.
n + m = o (5)
The rate law can be used to calculate the rate constant as follows
k = Rate / ( [A]n [B]m ) (6)
Table 11A – Rate Law Models
Order Rate Expression
Differential
Rate Expression
Integrated
Straight Line
Linear Plot
Zero Order Rate = k [A] – [A]0 = - k t [A] versus t
First Order Rate = k [A] ln( [A] / [ A ]0) = - k t ln [A] versus t
Second Order Rate = k [A]2 1/ [A]0 – 1/[A] = - k t 1 / [A] versus t
Second Order in A and B Rate = k [A][B]
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Reaction Rate Relations - Write the reaction rate relations for the following reactions
A. N2O5 (g) => 2 NO2 (g) + 0.5 O2 (g)
________________________________________________________________________________
B. SO2Cl2 (g) => SO2 (g) + Cl2 (g)
________________________________________________________________________________
C. 2 N2H4 (l) + N2O4 (l) => 3 N2 (g) + 4 H2O (g)
________________________________________________________________________________
D. 2 Sb (s) + I2 (s) => 2 SbI3 (s)
________________________________________________________________________________
E. 4 PH3 (g) + 8 O2 (g) => P4O10 (s) + 6 H2O (g)
________________________________________________________________________________
F. CH3CHO (g) => CH4 (g) + CO (g)
________________________________________________________________________________
9. Chemistry Two SJBoylan Friday, October 25, 2019 Page 9 of 50
Reaction Order Determined by Ratio Method - The gas phase decomposition of acetaldehyde, CH3CHO, to
methane, CH4, and carbon monoxide is shown by the reaction equation:
CH3CHO (g) = > CH4 (g) + CO (g)
Experiments provided the following initial rate data
Table – Decomposition of Acetaldehyde
Point Number 1 2 3 4
[CH3CHO]
Mol/L
0.20 0.30 0.40 0.50
Rate,
(mol/L)/s
0.34 0.76 1.4 2.1
A. Determine the order of the reaction with respect to the initial acetaldehyde concentration using the ratio
method.
B. Determine the rate constant.
C. Write the rate expression for the decomposition of acetaldehyde.
Reaction Mechanisms
A. Write the rate expression for the elementary steps
NO3 (g) + CO (g) = > NO2 (g) + CO2 (g) (1)
_________________________________________________________________________
I2 (g) = > 2 I (g) (2)
_______________________________________________________________
NO (g) + O2 (g) = > NO3 (g) (3)
_____________________________________________________________________________
For the reaction H2 (g) + I2 (g) => 2 HI (g), the experimental rate expression is rate = k[H2][I2]. Show that
this is consistent with the reaction mechanism:
I2 (g) < = > 2I (g) (fast) (4)
H2 (g) + I (g) + I (g) => 2 HI (g) (slow) (5)
10. Chemistry Two SJBoylan Friday, October 25, 2019 Page 10 of 50
Model Discrimination and Parameter Estimation Sulfuryl Chloride Experiment - The reactants are SO2Cl2 and
the products are SO2 and Cl2.
A. Write the reaction equation
_________________________________________________________________________________
B. Complete the table: At time 0 min the concentration of SO2Cl2 is 0.0100 M At time 10 min the concentration
of SO2Cl2 is 0.00987 M. At time 30 min the concentration of SO2Cl2 is 0.00962 M At time 70 min the
concentration of SO2Cl2 is 0.00913 M
Table One Reactions
Point Time [SO2Cl2] [SO2] [Cl2] ln[SO2Cl2] ln[SO2Cl2]
1
2
3
4
C Make three graphs for this experiment
Plot [SO2Cl2] versus time ln [SO2Cl2] versus time ln [SO2Cl2] versus time
D Which graph has the straightest line _______________________________________________________.
E What is the order of this reaction. ___________________________________________
F. Write the rate equation
____________________________________________________________
G Calculate the reaction constant k
___________________________________________________________
G Calculate the half-life.
_______________________________________________________________
11. Chemistry Two SJBoylan Friday, October 25, 2019 Page 11 of 50
Arrhenius Equation and Influence of Temperature on Reaction Rate - The following data were obtained from
the article First Principles Theory for the H + CH4 => H2 + CH3 reaction by t. Wu, H. Werner, and U.
Manthe in Science page 2227, volume 306, December 2004.
Table - Arrhenius Equation and Influence of Temperature on Reaction Rate
Point Number Rate Constant
K (cm3/sec)
Temperature K Ln K Reciprocal
Temperature
1/ T K
1 2.0 x 10-16 500
2 8.5 x 10-18 400
3 5.0 x 10-19 333
A. Complete the Table
B. Plot ln versus reciprocal temperature
C. Determine the activation energy, Ea, in units of kJ/mol.
D. Calculate the frequency factor, A, for the reaction
12. Chemistry Two SJBoylan Friday, October 25, 2019 Page 12 of 50
Chapter 12 Gaseous Equilibrium – Terms and definitions
Law of mass action
Equilibrium
Steady state
Transient process
Open system
Closed system
Equilibrium constant K
Activities
Activity coefficients
Kp Kc Ka Kactivity Kb Kw Kf
Coefficient rule K’ = Kn
Reciprocal rule K” = 1/K
Rule of multiple equilibriums K3 = K1 + K2
Homogeneous equilibrium
Heterogeneous equilibrium
Large equilibrium constant favors more products
Small equilibrium constant favors more reactants
Reaction quotient
RICE table
Effects of concentration
Effects of pressure
Effects of temperature
Le Chatelier’s principle
Haber process for production of ammonia gas
Van’t Hoff equation ln(K2/K1) = (delta H / R)x(1/T1 – 1/T2)
Gibbs – Helmholts equation delta G = delta H – T delta S
13. Chemistry Two SJBoylan Friday, October 25, 2019 Page 13 of 50
Chapter 12 Gaseous Equilibrium
A. Determine the equilibrium constant for
4 NH3 (g) + 5 O2 (g) < = > 4 NO (g) + 6 H2O (g)
______________________________________________________________________________
B. Determine the equilibrium constant for
CH4 (g) + H2O (l) < = > CO (g) + 3 H2 (g)
______________________________________________________________________________
C. Determine the equilibrium constant for
2 NO3
- (aq) + 8 H+ (aq) + 3 Cu (s) < = > 2 NO (g) + 3 Cu2+ (aq) + 4 H2O (l)
______________________________________________________________________________
D. Determine the equilibrium constant for
BaCO3 (s) < = > BaO (s) + CO2 (g)
______________________________________________________________________________
E. Determine the reaction equation from K = ( PNH3
4 x PO2
5 ) / ( PNO
4 x PH2O
6 )
14. Chemistry Two SJBoylan Friday, October 25, 2019 Page 14 of 50
Chapter 12 Gaseous Equilibrium
___ (23) Calculate the equilibrium constant Kp for the formation of methyl alcohol, CH3OH, from carbon
monoxide, CO, and hydrogen, H2 . At equilibrium, the partial pressures of the gases are CH3OH = 0.0512 atm,
CO = 0.814 atm, H2 = 0.274 atm. The reaction equation is
CO (g) + 2 H2 (g) < = > CH3OH (g)
A. 0.00472
B. 0.230
C. 0.838
D. 51
(24) The reaction equation for the formation of methyl alcohol, CH3OH, from carbon monoxide, CO, and
hydrogen, H2 , is
CO (g) + 2 H2 (g) < = > CH3OH (g)
At equilibrium the partial pressures of the gases are PCO = 0.818 atm, PH2 = 0.187 atm, PCH3OH = 0.0514 atm.
Calculate the equilibrium constant, Kp for this system.
________________________________________________________________________________
(25) Equilibrium shift from a change in amount. – The reaction equation for the equilibrium of cobalt (II) ionsin
a solution containing ammonia is:
Co(H2O)6
2+ (aq) + 4 NH3 (aq) < + > Co(NH3)4
2+ (aq) + 6 H2O
25.1 Describe the effects of adding ammonia to the solution.
_________________________________________________________________________
25.2 Describe the effects of adding water to the solution.
_________________________________________________________________________
25.3 Describe the effects of removing cobalt(II) ammonia complex ion from the solution.
__________________________________________________________________________
15. Chemistry Two SJBoylan Friday, October 25, 2019 Page 15 of 50
Gaseous Chemical Equilibrium – The following shift conversion reaction is used to manufacturer hydrogen gas.
At a temperature of 850 K the equilibrium constant is K = 0.46
CO2 (g) + H2 (g) < = > CO (g) + H2O (g)
(24) Write the equilibrium constant expression
_______________________________________________________________________
The initial conditions are P CO2 = 1.85 atm P H2 = 1.00 atm P CO = 0.00 atm P H2O = 0.00 atm
Complete the following table:
CO2 H2 < = > CO H2O
Initial P atm (25)
Change P atm (26)
Equilibrium P
atm
(27)
(28) From the table substitute in the equilibrium expression.
______________________________________________________________________
(29) Rearrange the equilibrium expression into quadratic form.
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
______________________________________________________________________
(29) Calculate x = __________________________________________________
(31) The equilibrium partial pressure of CO2 is _____________________________
(32) The equilibrium partial pressure of H2 is _______________________________
(33) The equilibrium partial pressure of CO is ______________________________
(34) The equilibrium partial pressure of H2O is ______________________________
(35) Verify your partial pressures by substitution into the equilibrium expression and calculating an equilibrium
constant.
______________________________________________________________________________
16. Chemistry Two SJBoylan Friday, October 25, 2019 Page 16 of 50
Phosphine gas is highly toxic and has a compound formula of COCl2. Phosphine gas decomposes to carbon
monoxide and chlorine gas according to the following reaction equation.
COCl2 (g) = > CO (g) + Cl2 (g) (1)
The follow data has been reported for the equilibrium constant. (A. Lord and H. O. Pritchard J. Chem.
Thermodynamics, 2, 187 )
Table – Equilibrium Constant for the Dissociation of Phosphine Gas
Point Number Temperature Kelvin Equilibrium Constant, Kp
1 635.7 0.01950
2 670.4 0.04414
3 686.0 0.07575
4 722.2 0.1971
5 760.2 0.5183
A. Plot equilibrium constant versus temperature. Extrapolate the results to 298 K.
B. Estimate the standard change in enthalpy.
C. Estimate the standard change in entropy.
D. Estimate the standard change in Gibbs free energy.
LeChateliers Principle – Any system in chemical equilibrium, as a result of the variation in one of the factors
determining the equilibrium, undergoes a change such that, if this change had occurred by itself, it would have
introduced a variation of the factor considered in the opposite direction.
Consider the reaction:
Co(H2O)6
2+ (aq) + 4 Cl- (aq) < = > CoCl4
2- (aq) + 6 H2O (1)
A. What changes will result from increasing the amount of chloride ion?
_____________________________________________________________________________
B. What changes will result from increasing the amount of water?
______________________________________________________________________________
C. What changes will result from removing cobalt (II) chloride ion from the system?
_______________________________________________________________________________
17. Chemistry Two SJBoylan Friday, October 25, 2019 Page 17 of 50
Chapter 13 Equilibrium Acids and Bases
An Arrhenius acid is a substance that produces excess hydrogen ion, H+, in water.
An Arrhenius base is a substance that produces excess hydroxide ion, OH- , in water.
A Bronsted-Lowry acid is a proton donor, H+ . A Bronsted-Lowry base is a proton acceptor.
A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor.
A conjugate base is the species formed when a proton, H+, is removed.
A conjugate acid is the species formed when a proton, H+, is added.
An amphiprotic species can either accept a proton, H+, or donate a proton, H+.
A polyprotic acid is an acid that can ionize to provide more than one hydrogen ion.
[ H+ ] is the symbol for the molar concentration of hydrogen ion.
[ OH- ] is the symbol for the molar concentration of hydroxide ion.
In neutral solution [ H+ ] = [ OH- ] . At 25 C the pH of a neutral solution is 7.0.
In an acid solution [ H+ ] > [ OH- ] . At 25 C the pH of an acid solution less than 7.0.
In a basic [ H+ ] < [ OH- ] . At 25 C the pH of an basic solution greater than 7.0.
The water reaction is H2O + H2O < = > H3O+ (aq) + OH- (aq)
The water reaction can also be shown as H2O < = > H+ (aq) + OH- (aq)
pH is the power of the hydrogen ion. pOH is the power of the hydroxide ion. The following equation apply.
pH = - log10 [ H+ ] pOH = - log10 [ OH- ] log10 is the base 10 logarithm.
These are also shown as pH = - log [ H+ ] pOH = - log [ OH- ]
The inverses are [ H+ ] = 10-pH [ OH- ] = 10-pOH
The ion product of water is Kw = [ H+ ] x [ OH- ] .
At 25 C, Kw = 1.0 x 10-14 pH + pOH = 14.0 and [ H+ ] x [ OH- ] = 1.0 x 10-14
Weak acid equations HB (aq) <=> H+ (aq) + B- (aq) Ka = [ H+ ] [ B- ] / [ HB ] p Ka = - logKa
Henderson-Hasselbalch Equation for Buffer Systems
pH = pKa + log(B- / HB ) pKa = -log(Ka)
add acid pH = pKa + log((B- - acid) / (HB + acid)) add base pH = pKa + log((B- + base) / (HB - base))
weak acid strong base titration monoprotic VACID MACID = VBASE MBASE pKa = -log(Ka)
START POINT pH = - log sqrt( Ka MACID ) MIDPOINT pH = pKa
EQUILVALENCE POINT pH = - log sqrt ( Kw Ka VTOTAL / (VACID MACID))
18. Chemistry Two SJBoylan Friday, October 25, 2019 Page 18 of 50
Chapter 13 Equilibrium Acids and Bases – Complete the table.
Row [ H+ ] [ OH- ] pH pOH Acidic or Basic
(a) A = 2.8 x 10 -8 B C D E
(b) F G = 2.9 x 10 -3 H I J
(c) K L M = 4.73 N O
(d) P Q R S = 3.18 T
B ________________________________________________________________________
C ________________________________________________________________________
D ________________________________________________________________________
E ________________________________________________________________________
F ________________________________________________________________________
H ________________________________________________________________________
I ________________________________________________________________________
J ________________________________________________________________________
K ________________________________________________________________________
L ________________________________________________________________________
N ________________________________________________________________________
O ________________________________________________________________________
P________________________________________________________________________
Q ________________________________________________________________________
R ________________________________________________________________________
T________________________________________________________________________
19. Chemistry Two SJBoylan Friday, October 25, 2019 Page 19 of 50
Bronsted-Lowry acids and bases
(1) Draw a circle around the acid on the reactant side..
(2) Draw a circle around the conjugate base on the product side.
(3) Draw a line connecting the acid with its conjugate base.
(4) Label the acid and label the conjugate base.
(5) Draw a line under the base on the reactant side..
(6) Draw a line under the conjugate acid on the product side.
(7) Draw a line connecting the base with its conjugate acid.
(8) Label the base and label the conjugate acid.
HNO2 (aq) + OH-
(aq) < = > NO2
-
(aq) + H2O (1)
HCHO2 (aq) + H2O < = > CHO2
-
(aq) + H3O+
(aq) (2)
HC2H3O2 (aq) + HS-
(aq) < = > C2H3O2
-
(aq) + H2S (aq) (3)
CN-
(aq) + H2O < = > HCN (aq) + OH-
(aq) (4)
HCO3
-
(aq) + H3O+
(aq) < = > H2CO3 (aq) + H2O (5)
20. Chemistry Two SJBoylan Friday, October 25, 2019 Page 20 of 50
A. Polyprotic Acid - Phosphoric acid, H3PO4, is a polyprotic acid. The reaction equations for the ionization of
phosphoric acid are:
H3PO4 (aq) <=> H+ (aq) + H2PO4
- (aq)
H2PO4
- (aq) <=> H+ (aq) + HPO4
2- (aq)
HPO4
- (aq) <=> H+ (aq) + PO4
3- (aq)
The following chart shows the percent formation of each ion versus pH for phosphoric acid, H3PO4 .
At pH = 1.5, the percent formation of H3PO4 is high: At pH = 3.0, the percent formation of H3PO4 is medium.
At pH = 5.0, the percent formation of H3PO4 is low. At pH above 4.0, the percent formation of H3PO4 is low.
At a pH of 9.5, the percent formation of HPO4
2- is: A. high B. medium C. low D. none of these
B . Calculate the pH for a weak acid – Nitrous acid, HNO2 , is a weak acid with an equilibrium constant of
Ka = 6.0 x 10-4. Make a solution to 0.100 M nitrous acid. Calculate the pH.
Reaction Equation ______________________________________________________________________
I Initial ________________________________________________________________________________
C Change ______________________________________________________________________________
E Equilibrium ___________________________________________________________________________
21. Chemistry Two SJBoylan Friday, October 25, 2019 Page 21 of 50
Calculate the Equilibrium Constant for a Weak Acid – Prepare a solution of pure aspirin and water. Aspirin has
the compound formula of HC9H7O4 and is a weak organic acid. Add 1.26 gram of acid to water. Bring the final
volume to 350.0 mL. The pH of the final solution of 2.60.
A. Draw a two flask diagram of this process.
B. Complete the RICE table for this reaction.
R Reaction ____________________________________________________________________________
I Initial ________________________________________________________________________________
C Change ______________________________________________________________________________
E Equilibrium ___________________________________________________________________________
C. Calculate the molar mass of aspirin.
_______________________________________________________________________________________
D. Calculate the moles of aspirin.
________________________________________________________________________________________
E. Calculate the initial molarity of the acid.
_______________________________________________________________________________________
F. Calculate the equilibrium hydrogen ion concentration from the pH.
______________________________________________________________________________________
G. Calculate the value of the weak acid equilibrium constant, Ka.
22. Chemistry Two SJBoylan Friday, October 25, 2019 Page 22 of 50
Dissolving Salts in Water – Write the reaction equations for dissolving the following compounds in water. State
whether the solution is acidic, basic or neutral.
Zinc nitrate Zn(NO3)2 __________________________
Potassium perchlorate KClO4 _____________________
Sodium phosphate Na3PO4 _________________________
Ammonium fluoride NH4F ________________________________
Sodium hydrogen carbonate NaHCO3 ________________________________________
Table – Typical Salt Dissolution
Spectator Ions Acidic Ions Basic Ions
Cation Lithium, sodium, potassium,
calcium, strontium, barium
Ammonium, aluminum,
magnesium, transition metal ions
Anion Chloride, bromide, iodide,
nitrate, perchlorate
Acetate ion, carbonate ion,
fluoride ion, phosphate ion
Table – Acid Base Equilibrium Constants
Weak Acids Ka Weak Acids Ka Weak Acids Ka Weak Acids Ka
HF Ka = 6.9x10-4 F- Kb = 1.4x10-11 HCO3
- Ka=4.7x10-11 CO3
2- Kb = 2.1x10-4
HC2H3O2 Ka =1.8x10-5 C2H3O2
1 Kb = 5.6x10-10 H2CO3 Ka = 4.4x10-7 HCO3
- Kb=2.3x10-8
NH4
+ Ka = 5.6x10-10 NH3 Kb = 1.8x10-5
H3PO4 Ka = 7.1x10-3 H2PO4
- Kb = 1.4x10-12
H2PO4
- Ka = 6.2x10-8 HPO4
2- Kb = 1.6x10-7
HPO4
2- Ka = 4.5x10-13 PO4
3- Kb = 2.2x102-
HClO Ka = 2.8x10-8 ClO- Kb = 3.6x10-7
23. Chemistry Two SJBoylan Friday, October 25, 2019 Page 23 of 50
Weak Acid Strong Base Titration - Formic acid, HCHO2, is titrated with potassium hydroxide, KOH. This is a
weak acid strong base titration. There are 50.00 mL of formic acid at 0.157 molarity. The molarity of the
potassium hydroxide is 0.235 M. The equilibrium constant for formic acid is Ka= 1.9x10-4.
1. Complete the following diagram with the information provided above.
2. Write the reaction equations for this reaction
______________________________________________________________
______________________________________________________________
________________________________________________________________
________________________________________________________________________
3. Calculate the volume of potassium hydroxide added at the equivalence point.
4. Calculate the total volume at the equivalence point.
5. Calculate the volume added at the midpoint.
6. Calculate the power of the acid equilibrium constant.
7. Calculate the pH before adding any base pH = _________
8. Calculate the pH after adding 8.40 mL of base. pH = ______________
9. Midpoint Calculate the pH at the mid-point. pH = ________________
24. Chemistry Two SJBoylan Friday, October 25, 2019 Page 24 of 50
10. Calculate the pH after adding 24.2 mL of base. pH = ________________
11. Equivalence Point Calculate the pH at the equivalence point. pH = ________________
12. After the equivalence point, calculate the total volume _______ and pH after adding 56.1 mL of base.
pH = _______________
13. Complete the following table
Table One Weak Acid Strong Base Titration
Point Number Description pH Volume of base added,
mL
1 7. At Start
2 8. After first addition
3 9. Midpoint
4 10. After Mid-Point
5 11. Equivalence Point
6 12. After equivalence point
14. Make a titration curve. Plot pH versus volume of base added
15. On the titration curve, identify the midpoint, pKa and the equivalence point.
25. Chemistry Two SJBoylan Friday, October 25, 2019 Page 25 of 50
Buffer System – Reaction Equations and Graph A buffer solution is made by mixing acetic acid, HC2H3O2, and
sodium acetate, NaC2H3O2, in water.
1 Write the reaction equation for the dissolution of sodium acetate in water.
_________________________________________________________________
2 Write the reaction equation for the equilibrium of acetic acid in water.
_________________________________________________________________
3 Write the equation for the acid equilibrium constant of acetic acid.
_______________________________________________________________
Hydrochloric acid, HCl, is added to the buffer solution. The pH decreases.
4 Write the reaction equation for the absorption of hydrochloric acid gas in water.
____________________________________________________________________
5 Write the reaction equation for the reaction of hydrochloric acid with acetate ion.
__________________________________________________________________
Another buffer solution is made with acetic acid and sodium acetate. Sodium hydroxide, NaOH, is added to this
buffer solution. The pH of the solution increases.
6 Write the reaction equation for the dissolution of sodium hydroxide in water.
____________________________________________________________________
7 Write the reaction equation for the reaction of sodium hydroxide with acetic acid.
_______________________________________________________________
8 Write the following labels on the chart
26. Chemistry Two SJBoylan Friday, October 25, 2019 Page 26 of 50
Buffer System Calculations and Diagram – Acetic Acid Sodium Acetate Water
A buffer solution is made of 0.456 L of 0.500 M acetic acid, HC2H3O2 and 0.400 L: of 0.429 M sodium acetate,
NaC2H3O2. Assume the volumes are additive. The acid ion product is Ka = 1.8 x 10-5
A. As you do the calculations fill in the blanks in the diagram.
B. Calculate the pH of the buffer solution.
_______________________________________________________________________________________
C. Add Acid - Add 0.050 mL of 1.00 M HCl to the buffer solution at start. Calculate the pH after the addition.
__________________________________________________________________________
D. Add Base - Add 0.050 mL of 1.00 M NaOH to the buffer solution at start. Calculate the pH after the
addition.
__________________________________________________________________________
E. Determine the limits of the buffer solution
__________________________________________________________________________
__________________________________________________________________________
27. Chemistry Two SJBoylan Friday, October 25, 2019 Page 27 of 50
Precipitation Definitions Refer to IUPAC Goldbook for definitions
Solid ___________________________________________________________________
Crystal_________________________________________________________________
Dissolution_________________________________________________________________
Cation_________________________________________________________________
Anion_________________________________________________________________
Centrifuge_________________________________________________________________
Equilibrium_________________________________________________________________
Phase_________________________________________________________________
Reaction Equation_________________________________________________________________
Solubility molar_________________________________________________________________
Solubility mass_________________________________________________________________
Solubility Product Constant Ksp__________________________________________________________
Activity_________________________________________________________________
Activity coefficient_________________________________________________________________
Object oriented analysis_________________________________________________________________
Variables _________________________________________________________________
Relations_________________________________________________________________
Algorithm_________________________________________________________________
Solubility Chart_________________________________________________________________
Reaction Quotient_________________________________________________________________
Common_________________________________________________________________
Common Ion_________________________________________________________________
Complex Ion_________________________________________________________________
Central Ion_________________________________________________________________
Ligand_________________________________________________________________
Lewis Acid_________________________________________________________________
Lewis Base_________________________________________________________________
Precipitation _________________________________________________________________
Precipitate_________________________________________________________________
Gravity cloud rain_________________________________________________________________
Filter_________________________________________________________________
Anhydrous _________________________________________________________________
Metastable hydrate _________________________________________________________________
Common ion effect_________________________________________________________________
Complex ion effect_________________________________________________________________
Insoluble_________________________________________________________________
Moderately soluble_________________________________________________________________
Soluble_________________________________________________________________
28. Chemistry Two SJBoylan Friday, October 25, 2019 Page 28 of 50
Solubility Calculation Common Ion - Strontium Carbonate SrCO3 Ksp = 7.0x10-10
1) Write the equilibrium reaction equation:
___________________________________________________________________________________
2) What is the value of the cation to anion ratio? ____________________________________________
3) Write the expression for the solubility product
___________________________________________________________________________________
4) Calculate the concentration of the anion when [Sr2+]=7.9x10 -5
_____________________________________________________________________
5) Calculate the concentration of the cation when [CO3
2-]=8.3x10 -4
_____________________________________________________________________
6) Write the equation for the molar solubility
___________________________________________________________________________________
7) Write the equation for the mass solubility _______________________________________________
8) Calculate the molar mass
___________________________________________________________________________________
9) Calculate the mass solubility in pure water
___________________________________________________________________________________
10) Calculate the mass solubility when the common ion is [CO3
2-]=0.010
_____________________________________________________________________
Solubility Calculation Common Ion - Lead Chloride PbCl2 Ksp = 1.7x10-5
1) Write the equilibrium reaction equation:
___________________________________________________________________________________
2) What is the value of the cation to anion ratio? ____________________________________________
3) Write the expression for the solubility product
___________________________________________________________________________________
4) Calculate the concentration of the anion when [Pb2+]=
__________________________________________________________________________________
5) Calculate the concentration of the cation when [Cl-]=
___________________________________________________________________________________
6) Write the equation for the molar solubility
___________________________________________________________________________________
7) Write the equation for the mass solubility _______________________________________________
8) Calculate the molar mass
___________________________________________________________________________________
9) Calculate the mass solubility in pure water
___________________________________________________________________________________
10) Calculate the mass solubility when the common ion is [Cl-]=0.0010
___________________________________________________________________________________
29. Chemistry Two SJBoylan Friday, October 25, 2019 Page 29 of 50
Precipitation Solubility Chart
A. Write the reaction equation for solid calcium iodate, Ca(IO3)2 , in equilibrium with its ions in water.
_____________________________________________________________________________
B. Write the equation for the solubility product of solid calcium iodate, Ca(IO3)2 , in equilibrium with its ions in
water.
______________________________________________________________________________________
C. Solve the equation for the solubility product for the concentration of the calcium ion.
________________________________________________________________________________
D. Solve the equation for the solubility product for the concentration of the iodate ion.
__________________________________________________________________________________
E. Complete Table One
Table One – Solubility of Calcium Iodate Ksp = 6.47x10-6
Point Number Calcium Ion Concentration, mol/L Iodate Ion Concentration, mol/L
1 0.0120
2 0.0140
3 0.0160
4 0.0186
5 0.0200
6 0.0220
7 0.0240
8 0.1000
F. Make a graph of Calcium Ion Concentration versus Iodate Ion Concentration.
G. Show on your graph where Q>Ksp, Q=Ksp, Q<Ksp. Show where precipitation can occur, dissolution can
occur, saturated solution, unsaturated solution and supersaturated solution.
30. Chemistry Two SJBoylan Friday, October 25, 2019 Page 30 of 50
Precipitation Mixing Two Solutions – A starting solution of 252 mL of 0.052 M calcium nitrate is mixed with a
solution of 378 mL of 0.049 M sodium carbonate. The mixture becomes opaque small crystals and then the
crystals fall to the bottom of the solution. Assume the volumes are additive. Ksp CaCO3 = 4.7x10-9
A. Draw a three flask diagram showing this precipitation.
B. Do a qualitative analysis for precipitation. Write the formula for the precipitate.
C. Write the equilibrium equation and the solubility product expression for the precipitate.
________________________________________________________________________________
D. Calculate the volume of mixture in liters.
_______________________________________________________________________________
E. Calculate the molarity of the calcium ion in the mixture.
_______________________________________________________________________________
F. Calculate the molarity of the carbonate ion in the mixture.
_______________________________________________________________________________
G. Calculate the reaction quotient, Q, for the solution in the mixture
_______________________________________________________________________________
H. Compare Q to K. Does the reaction quotient indicate that a precipitation will take place?
_______________________________________________________________________________
I. Calculate the molarity of calcium ion in the final solution.
_______________________________________________________________________________
J. Calculate the molarity of carbonate ion in the final solution.
_______________________________________________________________________________
K. Calculate the molarity of sodium ion in the final solution.
________________________________________________________________________________
L. Calculate the molarity of nitrate ion in the final solution.
________________________________________________________________________________
CaCO3, 0.630, 0.021, 0.029, 6.6x10-4 , yes, 5.9x10-7 , 8x10-3 , 4.72x10-9 , 0.059, 0.042
31. Chemistry Two SJBoylan Friday, October 25, 2019 Page 31 of 50
Complex Ions and Solubility – Zinc(II) hydroxide is in equilibrium with its ion in water according to
Zn(OH)2 (s) < = > Zn2+ (aq) + 2 OH- (aq) Ksp = 5.0 x 10-17 (1)
Zinc(II) ion forms a complex ion according to
Zn2+ (aq) + 4 NH3 (aq) < = > Zn(NH3)4
2+ (aq) Kf = 1.0 x 109 (2)
A. Combine equations (1) and (2) to make equation (3). Calculate the equilibrium constant.
_____________________________________________________________________________________ (3)
__________________________________________________________________________________
B. Draw a diagram with two beakers showing zinc hydroxide solid equilibrium, add ammonia and complex ion
equilibrium.
C. Estimate the molar solubility of zinc(II) hydroxide in pure water.
__________________________________________________________________________________
D. Estimate the molar solubility of zinc(II) hydroxide in a 0.10 M ammonia solution.
___________________________________________________________________________________
E. Calculate the ratio of zinc(II) hydroxide solubility in the ammonia solution to zinc(II) hydroxide solubility in
pure water.
_______________________________________________________________________________
F. Is the solubility of zinc(II) hydroxide in the ammonia solution greater than or less than its solubility in pure
water and why?
____________________________________________________________________________________
___________________________________________________________________________________
32. Chemistry Two SJBoylan Friday, October 25, 2019 Page 32 of 50
Chapter 16 Spontaneity
Spontaneity Spontaneous nonspontaneous
Mass Molarity Volume Pressure
Temperature distance Position Energy
Enthalpy Delta change Differential change Summation
Multiplication product Standard state Heat transfer Work
Equilibrium constant Reaction rate constant Temperature Kelvin Reaction Quotient
Time Entropy Specific entropy Amagats free energy
Gibbs free energy First Law Thermodynamics Second Law Thermodynamics
Third Law Thermodynamics
delta H = sum( delta H0 products ) - sum(delta H0 reactants )
delta S = sum( S products ) - sum( S reactants )
Gibbs-Helmholtz Equation delta G0 = delta H0 - T delta S
Table - Thermodynamic Properties
Delta H0 S0 Delta G0
CO2 (g) -393.5 0.2136 -394.2
CO (g) -110.5 0.1976 -137.2
O2 (g) 0.0 0.2050 0.0
CuCl2 .(s)
Cu2+ (aq) 64.8 -0.0996 65.5
Cl- (aq) -167.2 0.0565 -131.2
H2O (g) -241.8 0.1887 -228.2
H2O (l) -282.8 0.0699 -237.2
Ca2+ (aq) -542.8 -0.0531 -553.6
SO4
2+ (aq) -909.3 0.0201 -744.5
CaSO4 (s) -1434.1 0.1067 -1321.8 Ksp=7.1x10-5
Guidelines to predict the Sign of the Entropy
Less Entropy < More Entropy
Less Random < More Random
More Order < Less Order
Less Molecules < More Molecules
Small Volume < Large Volume
High Pressure < Low Pressure
Low Temperature < High Temperature
Liquid < Solid
Solid < Liquid
Less Entropy < More Entropy
33. Chemistry Two SJBoylan Friday, October 25, 2019 Page 33 of 50
Thermo Chemistry - Evaluate the following reaction:
CO2 (g) < = > CO (g) + 0.5 O2 (g) (1)
A. Identify the type of reaction
______________________________________________________________________________________
B. Determine the sign of the predicted change of entropy.
_____________________________________________________________________________________
C. Determine the calculated change of entropy.
______________________________________________________________________________________
D. Compare the sigh of the predicted change of entropy to the sign of the calculated change of entropy.
______________________________________________________________________________________
E. Calculate the change of enthalpy.
______________________________________________________________________________________
F. Is the reaction endothermic or exothermic?
______________________________________________________________________________________
G. Calculate the change in Gibbs free energy.
______________________________________________________________________________________
H. Is the reaction spontaneous or nonspontaneous?
______________________________________________________________________________________
I. Is the reaction exergonic or endergonic?
______________________________________________________________________________________
34. Chemistry Two SJBoylan Friday, October 25, 2019 Page 34 of 50
Thermo Chemistry - Evaluate the following reaction:
CaSO4 (s) = > Ca2+ (aq) + SO4
2- (aq) (2)
A. Identify the type of reaction
______________________________________________________________________________________
B. Determine the sign of the predicted change of entropy.
_____________________________________________________________________________________
C. Determine the calculated change of entropy.
______________________________________________________________________________________
D. Compare the sigh of the predicted change of entropy to the sign of the calculated change of entropy.
______________________________________________________________________________________
E. Calculate the change of enthalpy.
______________________________________________________________________________________
F. Is the reaction endothermic or exothermic?
______________________________________________________________________________________
G. Calculate the change in Gibbs free energy.
______________________________________________________________________________________
H. Is the reaction spontaneous or nonspontaneous?
______________________________________________________________________________________
I. Is the reaction exergonic or endergonic?
______________________________________________________________________________________
35. Chemistry Two SJBoylan Friday, October 25, 2019 Page 35 of 50
Table 1601 - Gibbs-Helmholtz Equation delta G0 = delta H0 - T delta S
Condition Delta Gibbs
free energy
Delta enthalpy Temperature
absolute
Delta entropy
1
+ + Low + nonspontaneous
2
- + High + spontaneous
3
+ + Low - nonspontaneous
4
+ + High - nonspontaneous
5
- - Low + spontaneous
6
- - High + spontaneous
7
- - Low - spontaneous
8
+ - High - nonspontaneous
___ (1) For a chemical reaction, the change in enthalpy is positive, the temperature is very high and the change
in entropy is positive. Will the reaction be spontaneous and nonspontaneous?
_______________________________________________________________________
___ (2) For a chemical reaction, the change in enthalpy is negative, the temperature is very low and the change
in entropy is positive. Will the reaction be spontaneous and nonspontaneous?
_______________________________________________________________________
___ (3) For a chemical reaction, the change in enthalpy is – 56.8 kJ per mole, the temperature is 500.0 K and the
change in entropy is 0.2014 kJ per mole K. Calculate the change in Gibb’s Free Energy.
_______________________________________________________________________
___ (4) For a chemical reaction, the change in enthalpy is – 56.8 kJ per mole, the temperature is 854.2 K and the
change in Gibb’s free energy is – 69.8 kJ per mole, kJ per mole K. Calculate the change in entropy in units of
kJ per mole K.
_______________________________________________________________________
36. Chemistry Two SJBoylan Friday, October 25, 2019 Page 36 of 50
Chapter 16 Spontaneity
From https://courses.lumenlearning.com/chemistryformajors/chapter/spontaneity/
1. What is a spontaneous reaction?
2. What is a nonspontaneous reaction?
3. Indicate whether the following processes are spontaneous or nonspontaneous.
a. Liquid water freezing at a temperature below its freezing point
b. Liquid water freezing at a temperature above its freezing point
c. The combustion of gasoline
d. A ball thrown into the air
e. A raindrop falling to the ground
f. Iron rusting in a moist atmosphere
4. A helium-filled balloon spontaneously deflates overnight as He atoms diffuse through the wall of
the balloon. Describe the redistribution of matter and/or energy that accompanies this process.
5. Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of
these plastics in air to form carbon dioxide and water is a spontaneous process; however,
plastic materials tend to persist in the environment. Explain.
37. Chemistry Two SJBoylan Friday, October 25, 2019 Page 37 of 50
Electrochemistry Definitions – Write the definitions for the following terms.
Reduction ___________________________________________________________________________
Oxidation____________________________________________________________________________
Electrode____________________________________________________________________________
Cathode_____________________________________________________________________________
Anode______________________________________________________________________________
Redox reaction________________________________________________________________________
Half reaction__________________________________________________________________________
Voltaic cell___________________________________________________________________________
Electrolytic cell________________________________________________________________________
Galvanic cell__________________________________________________________________________
Reducing agent_________________________________________________________________________
Oxidizing agent_________________________________________________________________________
Abbreviated cell notation_________________________________________________________________
Salt bridge____________________________________________________________________________
Cell_________________________________________________________________________________
Cell voltage___________________________________________________________________________
Standard cell voltage____________________________________________________________________
Standard reduction voltage_______________________________________________________________
Standard oxidation voltage_______________________________________________________________
Nernst equation________________________________________________________________________
Gibbs free energy to reaction quotient equation_______________________________________________
Potential to reaction quotient relation_______________________________________________________
Activity______________________________
Activity coefficient______________________________
Table E100 – Electrochemistry
ELECTRICITY FARADAY CONSTANT CHEMISTRY
Current Ampere A
Charge Coulomb C
Power Watt W
Energy joule J
Potential E Volt V
Resistance Ohm
time
F = 9.648x104 Coulomb
Equals
1 mole electron
Mass m gram g
Moles M
Molar Mass MM
Balanced reaction equation
Time
rate
Charge coulomb C 1 C = 1 A x s = 1 J / V Current ampere A 1 A = 1 C / s
Potential volt V 1 V = 1 J / s Power watt W 1 W = 1 J / s
Energy joule J 1 J = 1 V C 1 kWh = 3.600x106 J
38. Chemistry Two SJBoylan Friday, October 25, 2019 Page 38 of 50
Standard Potentials, Aqueous Solutions 25 C Acidic [H+ ] = 1 M Eo red (V)
1 Li+(aq) + e- => Li(s) -3.040
2 K+(aq) + e- => K(s) -2.936
3 Ba2+(aq) + 2e- => Ba (s) -2.906
4 Ca2+(aq) + 2e- => Ca (s) -2.869
5 Na+(aq) + e- => Na (s) -2.714
6 Mg2+(aq) + 2e- => Mg (s) -2.357
7 Al3+(aq) + 3 e- => Al(s) -1.68
8 Mn2+(aq) + 2 e- => Mn (s) -1.182
9 Zn2+(aq) + 2e- => Zn (s) -0.762
10 Cr3+(aq) + 3e- => Cr (s) -0.744
11 Fe2+(aq) + 2e- => Fe (s) -0.409
12 Cr3+(aq) + 3e- => Cr (s) -0.408
13 Cd2+(aq) + 2e- => Cd (s) -0.402
14 PbSO4 (aq) + 2e- => Pb(s) + SO4
2- (aq) -0.356
15 Tl+(aq) + e- => Tl (s) -0.336
16 Co2+(aq) + 2e- => Co (s) -0.282
17 Ni2+(aq) + 2e- => Ni (s) -0.236
18 AgI(s) + e- => Ag(s) + I- (aq) -0.152
19 Sn2+(aq) + 2e- => Sn (s) -0.141
20 Pb2+(aq) + 2e- => Pb (s) -0.127
21 2H+(aq) + 2e- => H2 (g) 0.000
22 AgBr(s) + e- => Ag(s) + Br- (aq) 0.073
23 S(s) + 2H+ (aq) + 2e- => H2S (aq) 0.144
24 Sn4+(aq) + 2e- => Sn2+ (aq) 0.154
25 SO4
2- (aq) + 4H+(aq) + 2e- => SO2 (g) + H2O 0.155
26 Cu2+(aq) + e- => Cu+ (aq) 0.161
27 Cu2+(aq) + 2e- => Cu (s) 0.339
28 Cu+(aq) + e- => Cu(s) 0.518
29 I2 (s) + 2e- => 2I- (aq) 0.534
30 Fe3+(aq) + e- => Fe2+ (s) 0.769
31 Hg2
2+ (aq) + 2e- => 2Hg (l) 0.796
32 Ag+(aq) + e- => Ag(s) 0.799
33 2Hg2+(aq) + 2e- => Hg2
2+ (aq) 0.908
34 NO3
- (aq) + 4H+ (aq) + 3e - => NO(g) + 2H2O 0.964
35 AuCl4
-(aq) + 3e- => Au(s) + 4Cl- (aq) 1.001
36 Br2 (l) + 2e- => 2 Br- (aq) 1.077
37 O2 (g) + 4H+ (aq) + 4 e- => 2H2O 1.229
38 MnO2 (s) + 4H+ (aq) + 2 e- => Mn2+ (aq) + 2H2O 1.229
39 Cr2O7
2- (aq) + 14H+ (aq) + 6e- => 2Cr3+(aq) + 7H2O 1.33
40 Cl2 (g) + 2e- => 2 Cl- (aq) 1.360
41 ClO3
- (aq) + 6H+ (aq) + 5 e- => 0.5Cl2 (g) + 3H2O 1.458
42 Au3+ (aq) + 3 e- => Au (s) 1.498
43 MnO4
-(aq) + 8H+ + 5e- => Mn2+ (aq) + 4H2O 1.512
44 PbO2 (s) + SO4
2- (aq) + 4H+ (aq) 2e- => PbSO4(s) + 2H2O 1.687
45 H2O2 (aq) + 2H+ (aq) + 2e- => 2H2O 1.763
46 Co3+(aq) + e- => Co2+ (aq) 1.953
47 F2 (g) + 2e- => 2 F - (aq) 2.889
1. Which species is the strongest oxidizing agent? _______________________________
2. Which species is the strongest reducing agent? _______________________________
3. Which species will reduce zinc(II) ion but not aluminum ion? _________________
4. Which species will oxidize C 3+ but not bromide ion? ___________________________
39. Chemistry Two SJBoylan Friday, October 25, 2019 Page 39 of 50
Electrochemistry Voltaic Cell Two Beakers –
1. Write the reduction half reaction and standard potential for lead.
___________________________________________________________________________________
2. Write the reduction half reaction and standard potential for copper.
___________________________________________________________________________________
3. Which half reaction has the lowest standard potential. _________________________
4. Write the lowest potential reaction as an oxidation reaction.
___________________________________________________________________________________
5. Combine the reduction reaction with the oxidation reaction to get the combined reaction.
___________________________________________________________________________________
6. Write the abbreviated form of this voltaic cell.
___________________________________________________________________________________
7. Draw a diagram of the copper lead voltaic cell. Label the anode, cathode, oxidation, reduction, Cu, Cu2+, Pb,
Pb2+, salt bridge, K+ , NO3
- . For each wire show an electron and the direction the electron is moving. Show a
voltmeter that displays the standard potential in volts.
9. Calculate delta G for the standard cell in units of kJ per mol.
___________________________________________________________________________________
10. Calculate the equilibrium constant K for the standard cell.
___________________________________________________________________________________
11. The concentration of copper ion is 0.150 M. The concentration of lead ion is 0.1 M. Use the Nernst equation
to estimate the potential of the voltaic cell.
___________________________________________________________________________________
40. Chemistry Two SJBoylan Friday, October 25, 2019 Page 40 of 50
Electrolytic Cell – Chromium metal is electroplated from a solution of potassium dichromate by the following
reaction:
Cr2O7
- (aq) + 14 H+ (aq) + 12 e- => 2 Cr (s) + 7 H2O (1)
1. Write the atomic mass of the element chromium. __________________________
2. From the reaction equation, write the ratio of moles electrons to moles of chromium. ______________-
3. Draw a diagram of the electroplating unit. Label your diagram.
4. Grams Plated A current of 6.00amperes is run for 48 minutes through the electroplating solution.
Estimate the mass of chromium metal in grams that will be plated. _______________
5. Metal Plating Time A current of 6.00 amperes is run through the electroplating solution. The volume of the
solution is 0.215 liters. The molarity of the solution is 1.25 moles potassium dichromate per liter.
Estimate the time in seconds to completely convert the chromium ion to chromium metal. ______________
6. Electrical Energy for Mass of Metal A potential of 4.5 volts is applied to a plating solution. Estimate the
electrical energy in kilowatt-hours needed to recover 1.00 grams of chromium metal from the chromium
solution.
41. Chemistry Two SJBoylan Friday, October 25, 2019 Page 41 of 50
Chapter 17 Electrochemistry
A. Write the anode reaction, cathode reaction and overall reaction for a Leclanche cell.
_______________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
B. Write the anode reaction, cathode reaction and overall reaction for a lead storage battery.
_______________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
C. Write the anode reaction, cathode reaction and overall reaction for a lithium ion battery.
Cathode (+) LiCoO2 <= discharge charge => Li1-xCoO2 + x Li+ + xe-
Anode(-) Cn + xLi+ + x e- <= discharge charge => CnLix
Overall: LiCoO2 + Cn <= discharge charge => CnLix + Li1-xCoO2
Nernst Equation – A voltaic cell has the following reaction:
MnO2 (s) + 4 H+ 9aq) + 2 Br- (aq) = > Br2(l) + Mn2+ (aq) + 2 H2O (2)
A. Calculate the standard voltage
_______________________________________________________________________________
B. Write the Nernst equation for this reaction.
_______________________________________________________________________________
C. Estimate the voltage when [Mn2+] = 0.50 M, [ Br-] = 0.73 M, and pH = 4.72
_____________________________________________________________________________________
42. Chemistry Two SJBoylan Friday, October 25, 2019 Page 42 of 50
Chapter 18 Nuclear Chemistry
Define the following terms
Alpha particle
Beta particle
Gamma radiation
K electron capture
Neutron
Deuterium
Tritium
Proton
Nuclear equation
Balance nuclear charge (atomic number, Z) protons
Balance nuclear mass (mass number, A) protons and neutrons
Neutron to proton ratio
Stability
Unstability
Isotope
Nuclide
Nucleon
Even number of nucleons are more stable
Ridge of stability
Bismuth Z = 83
Magic numbers 2, 8, 20, 28, 50, 82, 126
Radioactivity
43. Chemistry Two SJBoylan Friday, October 25, 2019 Page 43 of 50
Nuclear Reaction Equations
A. Write a balanced nuclear reaction equation for the fission of U-235. There is one neutron on the reactant
side. There are four neutrons on the product side. One of the products is Zn-72.
_______________________________________________________________________________________
B. Write a balanced nuclear reaction equation for the fission of U-235. There is one neutron on the reactant side.
There are two neutrons on the product side. One of the products is Xe-135.
_______________________________________________________________________________________
C. Write a balanced nuclear reaction equation for the formation of Np-232 by alpha particle decay.
_______________________________________________________________________________________
D. Write a balanced nuclear reaction equation for the loss of an alpha particle from Th-230.
_______________________________________________________________________________________
E. Write a balanced nuclear reaction equation for the formation of Np-232 by alpha particle decay.
_______________________________________________________________________________________
F. Write a balanced nuclear reaction equation for the formation of Am-241 through beta emission.
_______________________________________________________________________________________
G. Write a balanced nuclear reaction equation for the formation of Ac-288 by beta emission.
_______________________________________________________________________________________
H. Write a balanced nuclear reaction equation for a positron emission from Pm-142.
_______________________________________________________________________________________
I. Write the balance nuclear reaction equation for Cf-149 bombarded by C-12. The products are one nuclide and
four neutrons.
___________________________________________________________________________________
44. Chemistry Two SJBoylan Friday, October 25, 2019 Page 44 of 50
Radioactive Decay Calculations
N = number of atoms k = first order rate constant t 1/2 = half life k = 0.693 / t1/2
A = activity Rate = k N ln(N / No) = - k t N = No exp( - k t) A = k N
NA = Avogadro’s number 0.6022x1024 atoms per mole Ci = Curie 3.700x1010 atoms per second
Bq = Becquerel 1 atom per second
The half-life of Rb-90 is 2.8 minutes.
A. Convert the half-life to units of seconds.
___________________________________________________________________________________-
B. Calculate the first order rate constant.
____________________________________________________________________________________
C. Sample AA3 contains 0.0000100 grams of Rb-90. Calculate the number of atoms in sample AA3.
_____________________________________________________________________________________
D. Calculate the rate of disintegration of Rb-90 in sample AA3 in units of atoms per second.
________________________________________________________________________________
E. Calculate the activity of Rb-90 in sample AA3 in units of Curie.
_______________________________________________________________________________
F. Calculate the activity of RB-90 in sample AA3 in units of Becquerel
_____________________________________________________________________________
G. Sample BB2 has a disintegration rate of 1.00x104 atoms per minute from Rb-90. Convert the disintegration
rate to atoms per second.
_______________________________________________________________________________
H. Calculate the number of atoms of Rb-90 in sample BB2.
______________________________________________________________________________
I. Calculate the mass in grams of Rb-90 in sample BB2.
__________________________________________________________________________________
45. Chemistry Two SJBoylan Friday, October 25, 2019 Page 45 of 50
Complete the table.
Table N207 – Natural Radioactive Decay Series 4n + 2 for Uranium 238
Name of
element
Reactant Half life Decay
mode
product Atomic
number
product
Neutron
number
product
Mass
number
product
n from
4n + 2
Neutron capture chain in supernova
explosion
U-238
1 U-238 4.51x109 year alpha Th-234
2 Th-234 24.1 day beta Pa-234
3 Pa-234 1.18 minute beta U-234
4 U-234 2.48x105 year alpha Th-230
5 Th-230 7.52x104year alpha Ra-226
6 Ra-226 1622 year alpha Rn-222
7 Rn-222 3.825 day alpha Po-218
8 Po-218 3.05 minute alpha Pb-214
9 Pb-214 26.8 minute beta Bi-214
10 Bi-214 19.7 minute beta Po-214
11 Po-214 1.6x10-4 second alpha Pb-210
12 Pb-210 22 year beta Bi-210
13 Bi-210 5.01 day beta Po-210
14 Po-210 138.4 day alpha Pb-206
Pb-206 Stable lead
isotope for 4n +
2 series
B. Calculate the change of energy for the following nuclear reactions.
Table - Atomic Mass Units amu
0n1 1.0086654 35Br87 86.920711 55Cs144 143.932077 62Sm160 159.93514
30Zn72 71.926858 37Rb90 89.914802 57La146 145.92579 92U235 235.0439299
92U235 + 0n1 = > 37Rb90 + 55Cs144 + 2 0n1 (1)
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92U235 + 0n1 = > 35Br87 + 57La146 + 3 0n1 (2)
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92U235 + 0n1 = > 30Zn72 + 62Sm160 + 4 0n1 (3)
46. Chemistry Two SJBoylan Friday, October 25, 2019 Page 46 of 50
Chapter 18 Nuclear Chemistry – Calculate the binding energy
A. Write the nuclear reaction equation for the separation of plutonium 239 into protons and neutrons.
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B. Calculate the mass defect of plutonium 239, in units of grams per mole.
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C. Calculate the binding energy of plutonium 239 in units of kJ per nucleon mole.
D. Write the nuclear reaction equation for the separation of helium 4 into protons and neutrons.
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E. Calculate the mass defect of helium 4 in units of grams per mole.
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F. Calculate the binding energy of helium 4 in units of kJ per nucleon.
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Table of Nuclear Masses on the C-12 Scale
Delta E = (9.00x1010 kilojoules / gram ) x ( delta m grams mass defect / A mass number)
Mass amu
Electron 0.00055
Neutron, n 1.00867
Hydrogen 1 , proton 1.00728
Helium 4 4.00150
Carbon 12 11.99671
Iron 56 55.92066
Cobalt 59 58.91837
Hafnium 179 178.9065
Uranium 235 234.9934
Uranium 238 Na
Plutonium 239 239.0006
47. Chemistry Two SJBoylan Friday, October 25, 2019 Page 47 of 50
Chapter 20 Chemistry of Metals
Reactions of Alkaline Metals and Alkaline Earth Metals
1) Write the reaction equation for the reaction of magnesium solid with chlorine gas. Name the product
compounds.
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2) Write the reaction equation for the reaction of lithium solid with sulfur vapor. Name the product compounds.
_______________________________________________________________________________
3) Write the reaction equation for the reaction of sodium solid with liquid water. Name the product compounds.
_______________________________________________________________________________
4) Write the reaction equation for the reaction of calcium solid with hydrochloric acid. Name the product
compounds.
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49. Chemistry Two SJBoylan Friday, October 25, 2019 Page 49 of 50
Organic Chemistry
Write the compound formula and draw structures for the following organic chemicals.
1) Alcohol – Ethanol _________________ 2) Ether – Dimethyl ether_________________
2) Aldehyde – Acetaldehyde_________________ 4) Ketone - Acetone_________________
5) Carboxylic acid – Acetic acid_________________ 6) Ester – Methyl acetate_________________
7) Amine – Ethylamine_________________ 8) Double bond - Ethylene_________________
9) Alkane – n-propane_________________ 10) Alkane branched – iso-butane_______________
50. Chemistry Two SJBoylan Friday, October 25, 2019 Page 50 of 50
Polymers
1) Write the reaction equation for the addition polymer reaction of ethylene.
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2) Write the reaction equation for the addition polymer reaction of vinyl chloride.
___________________________________________________________________________________
3) Write the reaction equation for the condensation polymer reaction of ethylene glycol with terephthalic acid to
make PET polyester
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4). Draw the structure of alpha-glucose.
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5) Write the reaction equation for the reaction of alpha-glucose with another alpha-glucose to make alpha-
maltose.
___________________________________________________________________________________
6) Write the reaction equation for the reaction of glycine plus alanine to make glycylalanine (Gly-Ala)
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