Fraction Mole (x): Consider a solution that contains 1.00 mol (46.1 g) of ethanol, C2H5OH in
9.00 mol (162 g) of water.
What are the mole fractions?
Ethanol Xethanol = 1/(1 + 9) = 1/10
Water Xwater = 9/(1 + 9) = 9/10
What percent: what is the weight % of alcohol in this alcohol water mixture?
Weight % = [46.1/(46.1 + 162)] * 100 = 22.15%
Using Henrys Law: the Henrys Law constant in water at 25 is 4.48e10-5 M/mm Hg. What is the
concentration of CO2 in water when the partial pressure is 0.33 atm?
Sg = kPg
k = 4.4848e10-5 M/mm Hg
Pg = 0.33 atm = 250.8 mm Hg
Sg = (4.4848e10-5 M/mm Hg) * (250.8 mm Hg) = .011235 M = 1.1e10-2 M
Using Raoults Law: Suppose 651 g of ethylene glycol, HOCH2CH2OH, is dissolved in 1.50 kg
of water. What is the vapor pressure of the water over the solution at 90 C = 525.8 mm Hg
Psolvent = XsolventPsolvent
Xsolvent = mole fraction
651 g/(62 g/mol) = 10.5 mol
1500 g/(18 g/mol) = 83.33 mol
Xsolvent = 83.33/(83.33 + 10.5) = 0.888
Psolvent = Xsolvent(525.8) = 466.96
Boiling Point Elevation: What quantity (in grams) of ethylene glycol, HOCH2CH2OH, must be
added to 125 g of water to raise the boiling point by 1.0 C? (molal boiling point elevation
constant for water: 0.512 C/m)
Practice Test 1
Practice Test 2
Freezing Point Depression. In the Northern United States, summer cottages are usually closed up
for the winter. When doing so, the owners winterize the plumbing by putting antifreeze in the
toilet tanks, for example. Will adding 525 g of HOCH2CH2OH to 3.00 kg of water ensure that the
water will not freeze at -25 C? (molal freezing point depression concstant for water. -1.86 C/m)
Tf = Kfm
525 g/ (62.1 g/mol) = 8.45 mol glycol
m = (mol of solute)/(kg of solvent) = 8.45 mol glycol / 3 kg H2O = 2.82 m
Tf = Kfm = (-1.86 C/m)(2.82 m) = -5.24 C/m
Practice Test 3
1) The osmotic pressure of a solution is directly proportional to both the ______ and the
_____.
= MRT = cRT
concentration (molarity)
temperature
2) Osmosis of solvent from one solution to another can continue until both solutions are
hypotonic/hypertonic/isotonic(circle the correct one), which means they have the same
_____.
isotonic
concentration
3) A so-called protozoan (a single-celled animal) that normally lives in the ocean is placed
in fresh water. Will it shrivel or burst? Explain briefly.
Burst, because lots of fresh water will flow into the cell.
4) Pure antifreeze freezes at -11.5 C. A solution made by adding a small amount of water to
antifreeze has a lower/higher (circle the correct one) freezing point than pure antifreeze.
Lower, solvent = antifreeze, solute = water
5) Assign a type of colloid (foam, aerosol, gel, emulsion) to each of the following so-called
colloidal dispersions:
a) clouds and fog _____
b) shaving cream, whipped cream _____
c) marshmallow _____
d) mayonnaise, face cream _____
e) cheese, butter _____
a) aerosol
b) foam
c) foam
d) emulsion
e) gel
6) You make a saturated solution of sugar in water at 25 C. No solid is present in the beaker
holding the solution. What can be done to increase the amount of dissolved sugar in the
solution?
a) Add more solid sugar
b) Raise the temperature of the solution.
c) Add some sugar and raise the temperature of the solution.
If you add 171.15g of sugar (C12H22O11) to 500 g of water, over what range of temperature ( C)
will the water be liquid? Kb = 0.5 C/m and Kf = -2 C/m.
mol = 171.15 g sugar/ (342.5 g/mol sugar) = 0.5 mol
0.5 mol / 0.5 kg = 1 m
Tf = Kfm = (-2 C/m)(1 m) = -2 C
Tb = Kbm = (0.5 C/m) (1m) = 0.5 C
If you add 29.22 g of NaCl to 500 g of water, over what range of temperature ( C) will the water
be liquid? Kb = 0.5 C/m and Kf = -2 C/m. Assume a vant Hoff factor, i, of 1.8.
mol = 29.22 g NaCl/ (58.44 g/ mol NaCl) = 0.5 mol NaCl
0.5 mol NaCl/ 0.5 kg H20 = 1m
Tf = Kfmi = (-2 C/m)(1m)(1.8) = -3.6 C
Tb = Kbmi = (0.5 C/m)(1m)(1.8) = 0.9 C
Practice Test 4
1.) Osmosis, the flow across a _____ membrane is the solvent flowing from the less/more
concentrated solution toward the higher/lower concentration. (circle the correct word
combination.)
Semipermiable
less
higher
2.) How is the rate of a reaction defined?
Reaction rate = (change in concentration) / (change in time) = (-[A])/(t) = ([B])/(t)
3.) What are the relative rates of appearance or disappearance of each product and reactant,
respectively, in the following reaction? (a, b, and c are the so-called stochiometric
coefficients): aA _bA cC
-(1/a)( [A]/t) = -(1/b)( [B]/t) = (1/c)( [C]/t)
4.) A reaction has the following experimental rate equation
Rate = k[A][B]
If the concentration of A is doubled and the concentration of B is halved, what happens to
the reaction rate?
5.) Consider the exothermic combustion of coal. Which of the following answer(s) could
increase the rate of reaction?
a) using smaller pieces of coal
b) increasing the concentration of oxygen
c) lowering the temperature
6.) Rates increases/decreases with increasing concentration of reactants.
Increase
7.) Rates increase/decrease at higher temperatures.
Increase
Practice Test 5
1.) What is the rate equation for a first-order reaction: R products?
Reaction rate = -( [R]/t) = k[R]1 = k[R]
2.) How is the integrated rate equation defined for a first order reaction R products?
ln ([R]t/[R]0) = -kt
3.) What is the unit for the rate constants of a first-order reaction?
4.) How is the integrated rate equation defined for a second-order reaction products?
1/[R] = kt + 1/[R]0
5.) What is the reaction order if rate = k[A]n[B]m?
n + m = reaction order
6.) The rate equation for a reaction A + B C is
Rate = k[A][B]
and the rate constant is 1.9 L/mol hr. Determine the initial rate of the reaction when [A] =
4 x 10-4 mol/L and [B] = 2 x 10-4 mol/L. Show your work for full credit.
Rate = k[A][B] = 1.52 x 10-4 mol/L hr (plug and chug)
7.) How is the half-life defined for a first order reaction?
t1/2 = 0.69/k
Arrange the following aqueous solutions in order of (i) increasing vapor pressure of water and
(ii) increasing boiling point.
a) 0.35 m HOCH2CH2OH (nonvolatile,nonelectrolyte)
b) 0.50 m sugar
c) 0.20 m KBr (a strong electrolyte)
d) 0.20 m Na2SO4 (a strong electrolyte
(i)
D<B<C<A
(ii)
A<C<B<D
After five half-life periods for a first-order reaction, what fraction of reactant remains?
1 half life: 0.5
2 half life: 0.25
3 half life: 0.125
4 half life: 0.0625
5 half life: 0.03125
For a reaction A + B C + D, how would you prove a first-order or second-order rate
dependence for [A]?
Plot ln[A] vs. t
Plot 1/[A] vs. t
See which one will give you a straight line.
The following statements relate to the reaction with the following rate law: Rate = k[H2][I2]. Fill
in the blanks.
H2 (g) + I2 (g) 2HI (g)
a) This is a _____-order reaction overall.
second-order
b) Raising the temperature will cause the value of k to _____.
increase
c) If the concentrations of both reactants are doubled, the rate will _____.
increase by a factor of 4
d) Adding a catalyst in the reaction will cause the initial rate to _____.
speed up
Raising the temperature lowers the activation energy?
Consider the following aqueous solutions: (i) 0.20 m HOCH2CH2OH; (ii) 0.10 m CaCl2; (iii) 0.12
m KBr; and (iv)) 0.12 m Na2SO4.
a) Which solution has the highest boiling point?
b) Which solution has the lowest freezing point?
c) Which solution has the highest water vapor pressure?
Half-Life (t1/2) and first-order reactions: Sucrose decomposes to fructose and glucose in acid
solution as a first-order process (k = 0.208 h-1 at 25 C). What amount of time is required for
87.5% of the initial time concentration of sucrose to decompose?
100% 87.5% 12.5% remaining fraction
0.125 = [A]t/[S]0 = remaining fraction
Half-life
1
2
3
t1/2 = 0.693/k = 0.693/0.208 h-1 = 3.33
three half-lifes: (3)(3.33 h) 10
Fraction
0.50 (50%)
0.25 (25%)
0.125 (12.5%)
Determining a rate equation: The rate of the reaction CH3CO2CH3 (aq) + OH- (aq) CH3CO2
(aq) +CH3OH (aq) was studied at 25 C. Data were collected for several experiemtns.
Experiment
#
1
2
3
Initial Concentration
[CH3CO2CH3]
0.05 M
0.05 M
0.10 M
-
[OH ]
0.05 M
0.10 M
0.10 M
Initial Rate
(mol/L s)
0.00034
0.00068
0.00136
What is the rate equation? What is the rate constant?
Rate [OH-]1
Rate [CH3CO2CH3]1
Rate = k[CH3CO2CH3]1 [OH-]1
#1 Rate = 0.00034 mol/L s = k (0.05 mol/L) (0.05 mol/L)
k = (0.00034 mol/L s) / ((0.05 mol/L) (0.05 mol/L)) = 0.14 L/mol s
The following statements relate to the reaction with the following rate law: Rate = k[H2][I2]
H2(g) + I2(g) 2HI(g)
Determine which of the following statements are true
a) This is a second order reaction overall.
TRUE
b) Raising the temperature will cause the value of k to decrease.
FALSE
c) Raising the temperature lowers the activation energy for this reaction.
FALSE
d) If the concentrations of both reactants are doubled, the rate will double.
FALSE
e) Adding a catalyst in the reaction will cause the initial rate to increase.
TRUE
A reaction has the following experiemental rate equation. Rate = k[A]2[B]. If the concentration
of A is doubled and the concentration of B is halved, what happens to the reaction rate?
Reaction rate doubles
Identify which of the following statements are correct / incorrect.
a) Rates increase with increasing concentration of reactants because there are more
collisions between reactant molecules.
CORRECT
b) At higher temperatures a larger fraction of molecules have enough energy to get over the
activation energy barrier.
CORRECT
c) Catalyzed and uncatalyzed reactions have identical mechanisms.
INCORRECT
Suppose you have two aqueous solutions separated by a semipermeable membrane. One contains
5.84 g NaCl dissolved in 100 mL solution, and the other contains 8.88 g of KNO3 dissolved in
100 mL solution. In which direction will the solvent flow?
(5.84 g NaCl)/(58.4 g/mol) = 0.1 mol in 100 mL of H2O
1.0 mol/L
(8.88 KNO3 g )/(101.1 g/mol) = 0.088 mol in 100 mL of H2O
0.88 mol/L
The solvent flows from less concentrated to more concentrated solutions, therefore the KNO3
flows to the side of NaCL
Give the relative rates for the disappearance of reactants and formation of products for the
following reaction:
4PH3 (g) P4 (g) + 6H2 (g)
aA + bB cC + dD
-(1/a)([A]/t) = -(1/b)([B]/t) = (1/c)([C]/t) = (1/d)([D]/t)
-(1/4)([PH3]/t) = ([P4]/t) = (1/6)([H2]/t)
What are the relative rates of appearance or disappearance of each product and reactant
respectively, in the decomposition of nitrosyl chloride, NOCl?
2NOCl (g) 2NO (g) + Cl2 (g)
aA + bB cC + dD
-(1/a)([A]/t) = -(1/b)([B]/t) = (1/c)([C]/t) = (1/d)([D]/t)
-(1/2)([NOCl]/t) = (1/2)([NO]/t) = ([Cl2]/t)
Write the equilibrium constant expressions for the following reactions.
a) N2 (g) + 3H2 (g) 2NH3 (g)
Kc = Keq = [NH3]2/[H2][H2]3
b) H2CO3 (aq) + H2O HCO3 (aq) +H3O+ (aq)
Kc = Keq = [HCO3-][H3O+]/[H2CO3]
Write the equilibrium constant expressions for the following reactions.
a) S (s) + O2 (g) SO2 (g)
Keq = [SO2]/[O2]
b) NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)
Keq = [NH4+][ OH-]/[NH3]
Write the equilibrium constant expression for each of the following reactions in terms of
concentrations.
a) PCl5 (g) PCl3 (g) + Cl (g)
keq = kc = [PCl3][Cl2]/[PCl5]
b) CO2 (g) + C (s) 2CO (g)
heterogeneous equilibrium
keq = kc = [CO]2/[CO2]
c) CH3CO3H (aq) + H2O (l) CH3CO2 (aq) + H3O+ (aq)
The Equilibrium Constant: Are the following reactions product or reactant favored?
a) Cu(NH3)42+ Cu2+ (aq) + 4NH3 (aq) K = 1.5*10-13
K < 1 : reactant favored
b) Cd(NH3)42+ Cd2+ (aq) + 4NH3 (aq) K = 1.0*10-7
K < 1: reactant favored
The reaction quotient: The brown gas nitrogen dioxide, NO2, can exist in equilibrium with the
colorless gas N2O4. K = 170 at 298 K for the reactions
2NO2 (g) N2O4 (g)
Suppose that, at a specific time, the concentration of NO2 is 0.015 M and the concentration of
N2O4 is 0.025 M. Is Q larger than, smaller than, or equal to K? If the system is not at
equilibrium, in which direction will the reaction proceed to achieve equilibrium?
Q = [N2O4]/[NO2]2 = 0.025/(0.015)2 = 110
Q<K
More product will be produced
System will shift to the right (the product side) The formation of ammonia from its elements is
an important industrial process.
N2 (g) 2H2 (g) 2NH3 (g)
a) How does the equilibrium composition change when extra H2 is added? When extra NH3
is added?
Shift to product side, shift to reactant side
b) What is the effect on the equilibrium when the volume of the system is decreased?
Shift to product side
Increasing pressure
Reactants: 4 particles/moles
Products: 2 particles/moles
Write the formula and give the name of the conjugate base of each of the following acids.
a) H20 H+ + OHb) HSO4- H+ + SO42c) HF H+ + FWrite the formula and give the name of the conjugate acid of each of the following bases.
a) NH3
b) HSO4c) D
The reaction quotient: A mixture of SO2, O2, and SO3 at 1000 K contains the gases at the
following concentrations: [SO2] = 5.0 X 10-3 mol/L, [O2] = 1.9 X 10-3 mol/L, and [SO3] = 6.9 X
10-3 mol/L. Is the reaction at equilibrium? If not, which way will it proceed to reach equilibrium?
2SO2 (g) + O2 (g) 2SO3 (g)
K = 279
Product favored, because K >> 1
Q = [SO3]2/[SO2]2[O2] =(6.9 X 10-3)/[(5.0 X 10-3)2( 1.9 X 10-3)]
Q > K, so reaction proceeds to the left to convert products into reactants
Disturbing a chemical equilibrium: Dinitrogen trioxide decomposes to NO and NO2 in an
endothermic process (H = 40.5 kJ/mol).
(H)E + N2O3 NO (g) + NO2 (g)
Predict the effect of the following changes on the position of the equilibrium, that is, state which
way the equilibrium will shift (left, right, or no change) when each of the following changes is
made.
a) Adding more N2O3 (g)
right
b) Adding more NO2 (g)
left
c) Increasing the volume of the reaction flask
lower pressure, right
d) Lowering the temperature
left
A solution of hydrochloric acid has [HCl] = 4.0 X 10-3 M. What are the concentrations of H3O+
and OH- in this solution at 25 C?
HCl + H2O H3O+ + Cl[H3O+] = [acid] = [HCl] = 4.0 X 10-3
Kw = [H3O+][OH-] = 1.0 X 10-14 at 25 C
[OH-] = Kw/[H3O+] = 1.0 X 10-14/4.0 X 10-3 = 2.5 X 10-12 M
What is the pH of a 0.0012 M NaOH solution at 25 C?
NaOH + H2O Na+ + OH[OH-] = [NaOH] = 0.0012 M
pOH = -log[OH-] = -log(0.0012) = 2.92
14 2.92 = 11.08 pH
What are the hydroxide and hydronium ion concentration in a 0.0012 M solution of NaOH?
NaOH strong base
NaOH --100% H2O-> Na+ (aq) + OH- (aq)
[NaOH] = [OH-] = 0.0012 M
Kw = [H3O+][OH-] = 1.0x10-14 at 25 C
[H3O+] = 1.0x10-14/0.0012 = 8.3x10-12
Equilibrium Constants for Acids: Several acids are listed here with their respective equilibrium
constants
C2H5OH
ke = 1.3x10-10
HCO2H
ke = 1.8x10-4
HPO42-
ke = 3.6x10-13
CH3CO2H
ke = 1.8x10-5
HCO2H > CH3CO2H > C3H5OH > HPO42Which is the strongest acid? Which is the weakest acid?
HCO2H strongest
HPO42- - weakest
A 0.10 M solution of a weak acid has a pH of 2.43. What is the value of ka for the acid?
HA + H2O H3O+ + AReactant favored k << 1
ka = [H3O+][A-]/[HA]; [H3O+] = [A-]
pH = -log[H3O+] = 2.43
[H3O+] = 10-pH = 10-2.43 = 3.7x10-3 M
FINISH TYPING QUESTIONS
Practice Test
1) What is the definition of the pH value?
pH = -log[H3O+]
2) How are the pH and the pOH value related?
pH + pOH = 14
3) In weak acids/bases, the acid or base strength is determined by the _____ dissociation
(ionization).
percent
4) An anion that is the conjugate base of a strong acid (in aqueous solution) will/will not
affect the pH value (circle the correct choice!).
HNO3/NO3will not
5) An anion that is the conjugate base of a weak acid (in aqueous solution) will
increase/decrease (circle one!) the pH value.
CH3CO2H
CH3CO3- + H2O CH3CO2H + OHOH- is a strong base, pH >> 1
6) Cations of the strong Arrhenius bases (in aqueous solution) will/will not affect the pH
value (circle the correct choice!).
Will not affect the pH value
Na+, Cl+, Ca2+, Ba2+
7) Decide which one(s) will give rise to a neutral solution in water:
a) NH4Cl
b) NaHCO3
c) NaNO3
d) KBr
a) slightly acidic
b) slightly basic
c) neutral
d) neutral
8) What is effect the on the pH of adding NH4Cl to a solution of NH3 (aq)? The pH will go
up, go down, or not change? Circle the correct answer!
La Chatelier Principle
NH3 (aq) + H2O NH4+ (aq) + OH- (aq), the reaction moves to the left, consumes OH- so it
makes it less basic (more acidic)
pH will go up
9) How is a Bronsted-Lowry acid defined?
proton donor
10) A Bronsted-Lowry base is defined as a _____ and must have _____.
proton acceptor
lone pair/nonbonding pair
11) An Arrhenius base is a substance that, when dissolved in water, _____.
increase the concentration of OH- ions
12) Define the term Lewis acid:
electron pair acceptor
13) List the following acids in order of increasing acid strength:
HBr, HCl, HI, HF
_____ < _____ < _____ < _____
HCl < HF < HBr < HI
14) Circle the one(s) of the following that are weak acids:
a) acetic acid
b) sulfuric acid
c) formic acid
d) hydrochloric acid
e) perchloric acid
f) citric acid
acetic acid, formic acid, citric acid
15) Circle the one(s) of the following that are weak bases.
a) an aqueous solution of NH3
b) sodium hydroxide
c) the chloride anion in aqueous solution
d) Ba(OH)2
e) water
f) KOH
an aqueous solution of NH3, the chloride anion in aqueous solution, water
16) Circle the one(s) of the following that are amphiprotic:
a) HSO4
b) FeCl2
c) H2O
d) NaOH
e) HCO3
f) H2SO4
HSO4, H2O, HCO3
amphiprotic- can donate and accept protons
17) Circle the one(s) of the following that are effective buffer systems:
a) An aqueous solution containing a pair of a strong base and its conjugate acid
b) Household vinegar
c) Pure water
d) An aqueous solution containing table salt
e) An aqueous solution containing equimolar amounts of both fluoride ions and HF
f) An aqueous solution containing a pair of a weak acid and its conjugate base
g) Human blood
An aqueous solution containing equimolar amounts of both fluoride ions and HF, and
aqueous solution containing a pari of weak acid and its conjugate base, Human blood
Calcium fluoride dissolves to a slight extent in water. Calculate the Ksp value for CaF2 if the
calcium ion concentration has been found to be 2.3x10-4 mol/L.
CaF2 Ca2+ (aq) + 2F- (aq)
K = [Ca2+][F-]2
If [Ca2+] = 2.3x10-4 mol/L, then [F-] = 2[Ca2+] = 4.6x10-4 mol/L
K = (2.3x10-4)(4.6x10-4) = 4.9x10-11
K << 1 (reactant favored)
Practice Test
1) The effect of a catalyst on an equilibrium is to
a) increase the rate of the forward reaction only
b) increase the equation constant so that products are favored
c) slow the reverse reaction only
d) increase the rate at which equilibrium is achieved
e) shift the equilibrium to the right
2) The equilibrium constant for a gas phase reaction is Keq = 230 at 300 degrees Celsius. At
equilibrium
a) reactants predominate
b) roughly equal amounts of products and reactants are present
c) only reactants are present
d) products predominate
e) only products are present
3) The equilibrium constant expression depends on the _____ of the reaction.
a) stochiometry
b) temperature
c) mechanism
d) stochiometry and mechanism
e) The quantity of reactants and products initially present
4) At chemical equilibrium,
a) both the forward and reverse reaction are occurring, but at different rates
b) no reaction occurs because equilibrium has been reached
c) both the forward and reverse reaction are occurring at the same rate
d) only the forward reaction occurs
e) only the reverse reaction occurs
f) none of the above
5)
Write the equilibrium constant expression for the following reaction in terms of
concentrations.
PbCl2 (s) Pb2+ (aq) + 2Cl- (aq)
Keq = WHAT
Practice Test
1) Circle the one(s) of the following that are characteristics of a buffer system.
a) Presence of a catalyst.
b) Presence of a sufficient amount of a strong acid.
c) Presence of a sufficient amount of both a strong acid and a strong base.
d) Presence of a saturated solution of a salt that is the conjugated base of a weak acid.
e) Presence of a sufficient amount of both a weak acid and its conjugate base.
f) A useful buffer solution controls the pH at the desired value.
g) The pH value of a buffer solution changes only slightly after addition of a reasonable
amount of either a strong acid or base.
2) In a buffer solution, how are the pH and the pKa value related (formulate the equation)?
Henderson-Hasselbach equation
pH = pKa + log[base]/[acid]
3) Assume you dilute an aqueous buffer solution with pure water. Will the pH value change?
Explain briefly. Write the equation you are referring to.
No the pH value will not change. Water is neither a strong acid or strong base. Ratio of
base : acid wont change by diluting the buffer
4) Assume an aqueous buffer solution. If there is more of the conjugate base in the solution
than acid, then pH > pKa, or pH < pKa, pH = pKa? Circle the correct answer?
pH = pKa + log[base]/[acid]
pH > pKa
5) Formulate the equilibrium constant for the ion-product of water.
Kw = [OH-][H3O+] = 1.0x10-14 at 25 C
6) Circle the one(s) of the following that is/are amphiprotic
a) H2O
b) NaOH
c) H3PO4
amphiprotic because it can be either an electron acceptor or donor
7) Define the term Lewis base:
electron pair donor
8) List the following molecules in order of decreasing acid strength
NaOH, H2O, CH3CO2H, HCl
_____ > _____ > _____ > _____
HCl > CH3CO2H > H2O > NaOH
9) Circle the one(s) of the following that are effective buffer systems:
a) NaCl (aq)/ HCl (aq)
b) NaF (aq) / HF (aq)
c) NaH2PO4 (aq) / Na2HPO4 (aq)
10) Which of the following statements about buffer solutions is false?
a) A buffer resists changes in pH
b) A buffer consists of a weak acid and a soluble conjugate salt of the weak acid.
c) A buffer solution can react only with H3O+ ions.
d) Most body fluids contain natural buffer systems.
11) Which of the following is correct? Assume 25 C.
a) pH + pOH = 7.00
b) [OH-] = 10-pOH
c) pKa = log(Ka)
d) [OH-][H3O+] = 1.0x10-7
12) Which of the following combinations are buffer solutions?
a) HCN and NaCN
b) NH3 and NH4Cl
c) HNO3 and NH4NO3
d) H2SO4 and Na2SO4
13) Consider a solution prepared from 20 mL of 0.15 M HF and 10 mL of 0.20 M KOH
(dissociation constant of HF is Ka = 7x10-4) What main species are in the solution after
mixing
a) K+, OHb) H+, Fc) K+, F-
d) K+, H+, F-, OHe) K+, F-, HF
Buffer solutions: Calculate the pH of a buffer solution composed 1.0 L of 0.50 M formic acid
(HCO2H) and 0.70 M sodium formate (NaHCO2) before and after adding 0.01 mol NaOH
(solid). (Ka = 1.8x10-4)
pH = pKa + log[base]/[acid]
pKa = -log(1.8x10-4) = 3.74
before: pH = pKa + log[0.70]/[0.50] = 3.74 + 0.15 = 3.89
after addition NaOH:
Before
After
HA
0.50
0.49
A0.70
0.71
OH0.01
0.00
HA + OH- A- + H2O
0.50 0.01
pH = pKa + log[base]/[acid]
pH = pKa + log[0.71]/[0.49] = 3.74 + 0.16 = 3.90
Knowing that the Ka value of magnesium fluoride is 5.2x10. calculate the solubility of the salt in
a) moles per liter and b) grams per liter.
MgF2 (s) Mg2+ (aq) + 2F- (aq)
K = [Mg2+][F-]2; [F-] = 2[Mg2+]
X = [Mg2+]
Wanted: [Mg2+], b.c MgF2 solubility = [Mg2+]
K = x(2x)2 = 4x3 = 5.2x10
b)(2.4x10-4 mol/L)(62.5 g/mol) = 0.015 g/L MgF2
Practice Test
Predict standard molar entropy values for the following molecules at 298 K. Circle the one that
has the highest, and underline the one that has the lowest value F2(g), Cl2(g), Br2(l), I2(s).
In the reaction 2NO (g) + O2 2NO2(g) the entropy increases / decreases / stays the same / is
zero (circle one!)
2 + 1 2 entropy decreases
Sort the following molecules in the order of increasing standard molar entropy values.
H2O(l), CH3OH(l), H2O(g), CH3OH(g)
H2O(l) < CH3OH(l) < H2O(g) < CH3OH(g)
Determine wheather the reactions listed below are entropy favored or disfavored under standard
conditions.
a) N2(g) + 2O2(g) 2NO2(g)
disfavored
1+ 22
b) 2C(s) + O2(g) 2CO(g)
favored
12
c) CaO(s) + CO2(g) CaCO3(s)
disfavored
10
d) 2NaCl(s) 2Na(s) + Cl2(g)
favored
01
Practice Test
1) Larger and more complex molecules have greater / smaller / more negative molar entropy
value than simpler molecules (circle one!)
2) Predict standard molar entropy values for the following molecules at 298K. List them in
the order of increasing values: F2(g), H2(g), N2(g)
H2 < N2 < F2 Molecular weight
3) Entropy on the molecular scale (L. Boltzmann): name three types of motion that
molecules can exhibit:
rotation
vibration
translational
4) In the reaction 2NH3(g) N2(g) + 2H2(g) the entropy increases / decreases / stays the
same / is zero (circle one!).
2 moles 1 + 3 = 4 moles
5) Sort the following molecules in the order of increasing standard molar entropy values.
HF(g), CO2(g), H2O(g)
HF < H2O < CO2
6) Sort the following molecules in the order of increasing standard molar entropy values.
MgO(s), C(diamond)(s), CaCO3(s)
C(diamond)(s) < MgO < CaCO3
7) If G > 0, the reaction will be spontaneous / nonspontaneous / at equilibrium /
reversible / rechargeable / irreversible (circle one!)
Spontaneous only if G < 0 (-), G = 0 : at equilibrium
The normal melting point of benzene, C3H8, is 5.5 C. For the process of melting, what is the sign
of each of the following?
G = H - T S
G < 0 : spontaneous
a) H
Change in enthalpy
endothermic: H > 0
b) S
Change in entropy
S > 0
c) G at 5.5 C
G = 0
d) G at 0 C
G > 0 nonspontaneous
e) G at 25 C
G < 0 spontaneous
The effect of molecular structure can be seen as atoms or molecules of similar molar mass, e.g.
CO2
M = 44 g/mol
Ar
M = 40 g/mol
C3H8
M = 44 g/mol
Predict which substance has the highest entropy. Which one has the lowest?
Highest C3H8
Lowest Ar
Which substance has the higher entropy?
a) 1 mole NaCl(s)
1 mol HCl(g) at 25 C
b) 1 mole H2O(s)
1 mol H2O(l) at 0 C
c) 2 mole NaCl(g)
1 mol HCl(g) at 25 C
d) 1 mole CH4(g)
1 mol Ar(g)
e) 1 mole CO(g)
1 mol CO2(g) at 25 C
at 25 C
The standard free energy change, G, for the formation of NO(g) from its elements is +86.58
kJ/mol at 25 C. Calculate the equilibrium constant at this temperature for the equilibrium
N2(g) + O2(g) 2NO(g)
Comment on the sign of G and the magnitude of K. (R = 8.3145 J/K mol)
K = [NO]2/[N2][O2]
G > 0 : nonspontaneous
G = -RTlnK; 25C 298.15 K
G = 86.58 kJ/mol = 86580 J/mol
-86580/(8.3145*298.15) = lnK
e-86580/(8.3145*298.15) = K
K << 1 : reactant favored
The standard free energy change, G , for the formation of 1.00 mol NH3(g) from nitrogen and
hydrogen is -16.37 kJ/mol. Use this value to calculate the equilibrium constant for this reaction
at 25 C. Is the reaction product/reactant favored? (R = 8.3145 J/K mol)
N2(g) +3H2(g) 2NH3
25 C 298.15 K
-16.37 kJ/mol = -16370 j/mol
G = -RTlnK
K >> 1 product favored
Assign an oxidation number to the atom in each ion, molecule, or compound.
a) O2
0
b) PO43+5
c) NO2+3
d) NO3+5
e) Na2SO4
+6
f) MgCl2
+2
g) NH3
-3
Balance the following net ionic equations
Ag+(aq) + Cu(s) Ag(s) + Cu2+(aq)
Balance the following net ionic equation.
Al(s) + Cu2+(aq) Al2+(aq) + Cu(s)
Aluminum reacts with acid to give Al(aq) H2(g). The unbalanced equation is
Al(s) + H+(aq) Al3+(aq) + H2(g)
Write balanced equations for the half reactions and the balanced net ionic equation. Identify the
oxidizing agent and the reducing agent.
Use the half-reaction method to write the balanced net ionic equation for the reaction in acid
solution.
MnO4-(aq) + Fe3+(aq) Mn2+(aq) + Fe3+(aq)
Use the half reaction method to write the balanced net ionic equation for the reaction in acid
solution.
Cr2O72-(aq) + I-(aq) Cr3+(aq) + I2(s)
Use the half reaction method to write the balanced net ionic equation for the following redox
reaction (in acid).
Cu(s) + NO3(aq) Cu2+ + NO2(g)
Balance the following redox equations. All occur in acid solution.
Ag(s) + NO3(aq) NO2(g) + Ag+
MnO2(s) + Cl(aq) Mn2+ + Cl2(g)
Zn(s) + NO2(aq) Zn2+(aq) + N2O(g)
MnO4(aq) + HSO3-(aq) Mn2+ + SO42-(aq)
Write balanced equations for the following half reactions. Specify whether each is an oxidation
or reductions.
a) Cr(s) Cr3+(aq)
(in acid)
Oxidation
b) VO3(aq) V2+
Reduction
(in acid)
c) Ag(s) Ag2O(s)
(in base)
Oxidation
d) H2O2(aq) O2(g)
(in acid)
Oxidation
e) SO42-(aq) SO2(g)
(in acid)
Reduction
A voltaic cell is constructed using the reaction
Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)
a) Rite equations for the oxidation and reduction half-reactions
Ox: Mg Mg2+ + 2eRed: 2H+ +2e- H2
b) Which reaction occurs in the anode compartment, and which occurs in the cathode
department?
anode compartment ox
cathode department - red
c) Complete the following sentences: Electrons in the external circuit flow from the _____
to the _____. Negative ions move in the salt bridge from the _____ half-cell to the _____
half-cell.anode
cathode
cathode
anode
A voltaic cell is set up using the following half-reactions
Reduction: Ag+(aq) + e- Ag(s)
Oxidation: Ni(s) Ni2+(aq) + 2ea) Which is the anode and which is the cathode?
b) What is the direction of flow in an external wire connecting the two electrodes?
c) What is the overall cell reaction?
Which of the following elements is the best reducing agent, which is the best oxidizing agent?
a) Cu (E = +0.34V)
b) Ag (E = +0.80V)
c) Zn (E = -0.76V)
d) Al (E = -1.66V)
D is the best reducing agent because it is the most negative.
B is the best oxidizing agent because it is the most positive.
Which of the following metals will react with H+(aq) to produce H2 in a product favored reaction:
CU, Al, Ag, Fe, Zn?
Cu2+(aq) + 3 e- Cu(s)
E = +0.34V
Al3+(aq) + 3 e- Al(s)
E = -1.66V
Ag+(aq) + 1 e- Ag(s)
E = +0.80V
Fe2+(aq) + 2 e- Fe(s)
E = -0.44V
Zn2+(aq) + 2e- Zn(s)
E = -0.76V
Which metal from the list is easiest to oxidize?
Al
Which metal is the most difficult to oxidize?
Ag
Calculating Standard Cell Potentials:
The net reaction that occurs in a voltaic cell is
Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)
Identify the half-reactions that occur at the anode and the cathode, and calculate at potential for
the cell assuming standard conditions.
Zn2+(aq) + 2e- Zn(s)
E = -0.76V
anode
Ag+(aq) + 1e- Ag(s)
E = +0.80V
cathode
Anode: Zn Zn2+ + 2e- (Ox)
Cathode: Ag+ + e- Ag (Red)
E cell = E red(cathode) - E red(anode)
0.80 (-.76) = 1.56 V
Calculating Standard Cell Potentials: The net reaction that occurs in a voltaic cell is
Zn(s) + 2Ag+(aq) Zn2+(aq) + 2Ag(s)
Identify the half reactions that occur at theanode and the cathode, and calculate a potential for the
cell assuming standard conditions.
Zn2+(aq) + 2e- Zn(s)
E = -0.76V
anode
Ag+(aq) + 1e- Ag(s)
E = +0.80V
cathode
E cell = E red(cathode) - E red(anode)
E cell = 0.80 (-0.76) = 1.56V
Anode: Ox.
Zn Zn2+ + 2e-
Cathode: Red.
Ag+ + e- Ag
The following reaction has an E value of -0.76V.
H2(g) + Zn2+(aq) Zn(s) + 2H+(aq)
Calculate G for this reaction. Is the reaction product or reactant favored?
What is the standard cell potential, E cell, for the reaction of silver ions with copper metal at
25C? Clculate G for this reaction. (F = 96500 C/mol e-)
Cu2+ + 2e- Cu(s)
E = +0.34V
Anode, Oxidized
Ag+(aq) + 1e- Ag(s)
E = +0.80V
Cathode, Reduced
Anode: Ox.
Cu Cu2+ + 2e-
Cathode: Red.
Ag+ + e- Ag
Net ionic: Cu +2Ag+ Cu2+ +2Ag
E cell = E red(cathode) - E red(anode)
E cell = 0.80 0.34 = 0.46V
G = -nFE
G = - (2)(96500)(0.46) = 89200 CV
Calculate the equilibrium constant for the reaction at 25 C.
Fe(s) + Cd2+(aq) Fe2+(aq) + Cd(s)
Cd2+ + 2e- Cd
E = -0.40V
Fe2+ + 2e- Fe
E = -0.44V
A coltaic cell is set up at 25 C with the following half cells.
Al (0.0010 M Al3+)
Ni (0.50 M Ni2+)
Write an equation for the reaction that occurs when the cell generates an electric current and
determine the cell potential.
Al3+(aq) + 3e- Al(s) E = -1.66V
Ni2+(aq) + 2e- Ni(s)E = -0.25V
Al: Anode
Ni: Cathode
Ox:
Al Al3+ +3e-
Red:
Ni2+ + 2e- Ni
Net:
2Al + 3Ni2+ 2Al3+ + 3Ni
E cell = E red(cathode) - E red(anode) = -0.25V (-1.66V) = 1.41V
Nerst equation:
E = E - (0.0592V/n)logQ
= 1.41V (0.0592V/6)logQ
Q = -ln{[Ni2+]2/[Al3+]3} = -ln{[0.50]2/[0.0010]3}
E = 1.41V (0.0592V/6)log(Q)