PRACTICE KEY EXAM 3 (Chemistry 102)
Aqueous Solid/Solution Equilibria(5 pts each)1) What pH is needed to precipitate Ni(OH)2 so completely that the Nickel concentration is less than
1.0 x 10-5 M (Ksp = 1.6 x 10-14) Is it ever possible to precipitate ALL the Nickel?
OH=SQRT(1.6E-14/1.0e-5)= / 4.0E-05 / pOH= / 4.39 / pH=9.602) The solubility of PbCl2 is 0.016M at 25C
a) What is its Ksp(25oC) (3pts)
=4x3= 4*0.016^3 = 1.64E-05
b) Would solubility change with temperature and what ADDED information would you need to determine trend/direction? (2pts)
YES, Ksp is a Equilibrium Constant that changes with temperature. We would need to know DH to determine the direction of the reaction. (If they say DG, give them one point)
3) What is the solubility in Molarity and (g/ml) of Ag2CO3 at 25oC (Ksp=8.1x10-12)
Molar Mass=276.4x3=Ksp / x= / 1.27E-04M / 2.12E-02g/L / 3.5e-5
4) Excess Calcium Phosphate Ca3(PO4)2 is placed into a 0.1 L volumetric. ( Ksp= 2.0x10-29).
How many gram will dissolve. Would you consider calcium phosphate to be very soluble?
x = 7.1x10-7moles/L = 7.1e-8moles in 0.1 L x 310.182= 2.20229E-05 g Very Insoluble
5) The Solubility of the solid salt Ag2SO3 in pure water is 1.3x10-4 M Hint(First determine Ksp=?);
WHAT is the solubility in 0.01AgNO3
Ksp=4x3= / 8.788E-12 / x*(0.01)^2=Ksp / 8.788E-08 M6) If Strong Acid where added to the following salts what would happen to their solubility
Compound / Solubility Change(increase/decrease/minimal change) and WHY?Mg3(PO4)2 / PO4+H+==HPO4 Increase Solubility (Remove PO4 from solution)
Al(OH)3 / OH +H+ == H2O Increase Solubility (Remove OH from solution)
7) 100 ml of 0.00040 M BaCl2(soluble) and 300ml of 0.00080M K2SO4(soluble) are mixed.
(Ksp for BaSO4 is 1.1x10-10 ) Determine Q and assess if precipitation will occur (2pts)
0.1X0.0004 moles of Ba / 0.00004 / moles in 0.4L / 0.0001 / M0.3x 0.0008 moles of SO4 / 0.00024 / moles in 0.4L / 0.0006 / M
Ba++ / SO4=
Q= / 0.0001 / 0.0006 / 6.00E-08 / > / 1.10E-10 / Precipitate
What would be the FINAL concentration of Barium in solution; SETUP ONLY(3pts)
SETUP ONLY: (0.0001-x)*(0.0006-x)= 1.1e-10
Extra Credit: Determine the actual final concentrations. (5pts)
x= / 9.95E-058) What is the Ksp expression for magnesium phosphate, Mg3(PO4)2?
Ksp=[Mg+2]3 [PO4-3]2
9) Consider a solution containing 0.181 M lead ions and 0.174 M mercury(II) ions. The Ksp for lead sulfide is 3.4 x 10-28 and that for mercury(II) sulfide is 4.0 x 10-53. Calculate the maximum concentration of sulfide ions that can be in solution without precipitating any lead ions.
[S=]=3.4E-28/0.181 / 1.87845E-2710) Will Mn(OH)2 precipitate from solution if the pH of a 0.050 M solution of MnCl2 is adjusted to 8.0 (Ksp = 1.8 x 10-11)? AND At what pH(approximately) will a precipitate form?
noQ = 5 x 10-14 < Ksp.
0.05*x2 = 1.8e-11 / 1.89737E-05
4.72184875 / pH= / 9.28
11) A solution is 10 millimolar in each of the metal ions in the following table (Fe+2,Ni+2,Pb+2,Cu+2)
H2S gas is bubbled through the solution to attain a [S=] concentration of 0.10 M.
Predict which of the sulfides precipitate under the given conditions.
FeS / 6.2 x 102 / 1.00E-03 / Q<Ksp / UndersaturatedNiS / 8.0 x 10-1
1.00E-03 / Q<Ksp / Undersaturated
PbS / 3.1 x 10-7
1.00E-03 / Q>Ksp / Precipitation
CuS / 6.1 x 10-16
1.00E-03 / Q>Ksp / Precipitation
Basic Thermodynamics (5pts each)
1) Draw a Diagram illustration The Influence of Temperature on Free Energy (DG); Note the FOUR possible
Scenarios! (Assume that H and S do not change with temperature.)
Draw G vs S diagram showing how four scenarios for G
Endothermic H+ è G+ at low temperature
If S+ then G will move from non-spontaneous to spontaneous at higher T
If S- then G will remain non-spontaneous at all T
Exothermic H- è G- at low temperature
If S+ the G will further decrease and always remain spontaneous
If S- the G will begin spontaneous but become non-spontaneous at higer
2) Under what conditions would a reaction that is spontaneous at HIGH temperatures and becomes non-spontaneous at LOWER temperatures? (Assume that H and S do not change with temperature.)
H > 0 Endothermic but S positive (increased entropy) which wins out at higher temperature
3) What are the appropriate conditions NECESSARY for each of the following thermodynamic factors
DSuniv / DG / E(EMF) / DSsystemEquilibrium / 0 / 0 / 0 / DSsystem + DSsurr = 0
Spontaneous / + / - / + / DSsystem + DSsurr > 0
Non-Spontaneous / - / + / - / DSsystem + DSsurr < 0
4) Use the following standard-state free energy of formation data to calculate the acid-dissociation equilibrium constant (pKa) for Carbonic acid
CompoundH2CO2(aq / -372.3
H+(aq) / 0.0
HCO2-(aq) / -351.0
DGrxn(kJ/mol) / 21.30
K / 0.000185
pK / 3.73
5) Calculate DH and DS for the following reaction. Is this reaction Endothermic or Exothermic
Then using these values; determine the DGo for this reaction at 25C AND illustrate why
NH4NO3 spontaneously dissolves is water at room temperature
NH4NO3(s) + H2O(l) < == > NH4+ (aq) + NO3- (aq)
Compound / Hfo(kJ/mol) / So(J/mol-K)NH4NO3 (s) / -365.56 / 151.08
NH4+ (aq) / -132.51 / 113.40
NO3- (aq) / -205.0 / 146.40
Compound / Hfo(kJ/mol) / So(J/mol-K) / G(compound)
NH4NO3(s) / -365.56 / 151.08 / -410.5
NH4+ (aq) / -132.51 / 113.40 / -166.3
NO3- (aq) / -205.0 / 146.40 / -248.6
For Reaction / 28.1 / 108.72 / -4.35
G0(rxm)= -4.35
K=5.78
Equilibrium(K) is large and far to product side thus high solubility!
6) Using a standard-state enthalpy of formation and absolute entropy data table for the reaction
2NO2 == N2O4
Hfo(kJ/mol) / So(J/mol-K)NO2(g) / 33.18 / 240.0 / -38.4
N2O4 (g) / 9.16 / 304.2 / -81.5
RXN / -57.2 / -175.8 / -4.8
DG0 = -4.8 kJ/mole
What is the Equilibrium Constant at a temperature of 25oC AND at 100oC
K= EXP(DS/R)*EXP(-DH/RT) / Keq / G0(T) J / Keq=exp(-G0/RT)K298)= / =EXP(-175.8/8.314)*EXP(57200/(8.314*298))= / 6.97 / -4811.6 / 6.97
K(273) / =EXP(-175.8/8.314)*EXP(57200/(8.314*273))= / 57.76 / -9206.6 / 57.76
K(373) / =EXP(-175.8/8.314)*EXP(57200/(8.314*373))= / 0.07 / 8373.4 / 0.07
Extra Credit: Note the direction of the change and explain in terms of DH and DS (5 pts)
With Exothermic reactions the equilibrium constant decreases with Temperature
This effect is independent of sign/direction of S
Conversely with Endothermic reactions the equilibrium constant increases with Temperature
7) For the Oxidation of Iron to Rust 4Fe + 3O2 è 2Fe2O3
DH rxn0 / -1648.4 kJ/molDS rxn0 / -543.7 J/(mol K)
What is the standard Gibbs free energy(DG rxn0 ) for this reaction?(3pts)
=-1648.4-0.298*(-543.7) / -1486.4 kJAt what Temperature do the forward and reverse rusting reactions occur in equilibrium (assuming that H0 and S0 are temperature independent?) (2pts)
=-1648.4/0.5437 / 3032K8) Based on the following thermodynamic data; calculate the boiling point of ethanol in degrees Celsius.
Substance / DHf0 (kJ) / S0 (J/K mol)C2H5OH(l) / -277.7 / 160.6
C2H5OH(g) / -235.1 / 282.6
Substance / DHf0 (kJ) / S0 (J/K mol)
C2H5OH(l) / -277.7 / 160.6
C2H5OH(g) / -235.1 / 282.6
Rxn / 42.6 / 122
H/S=T
=42600/122=349K 349-273 = 76C
9) The Reaction of Carbon Disulfide and Hydrogen Gas to form Carbon dioxide and H2S gas has a
DH of -32.1kJ and DS of -7.2 J/K. When the reactants are mixed, no observable change occurs.
The reason(s) for this include(s):
a) The reaction is not spontaneous.
b) The reaction is not favored thermodynamically.
c) It is a slow reaction kinetically.
d) The activation energy of the reaction is too small.
e) The Entropy of the Reaction is Negative which is not allowed
f) The activation energy of the reaction is large.
10) Explain how a non-spontaneous reaction/process {such as many found associated with life (biochemical)} can still take place. What do the laws of thermodynamics tells us. (Be concise here, irrelevant information will detract from grade)
A reaction can ONLY take place if DS Universe is + thus if a reaction DGsystem is not – then energy must be introduced into the surrounding to increase the entropy of the surroundings and make the total entropy of the universe positive
11) Using the equations on hand, show how the Electrochemical Potential E(EMF) is related to the Entropy of the Universe.
E = –n F DG = –n F DSuniverse/T
11 ) Electrochemical Cells
Half-Cell Reduction Potentials / VoltsI2+ 2e == 2I / 0.53
Br2 + 2e == 2Br- / 1.07
Using the above data construct a voltaic cell and determine the standard potential and the cell potential at 25oC if the
concentations of [I-] and [Br-] are 0.2M and 0.02M respectively (3pts)
Br2 + 2I- =2e = 2Br- + I2Standard Cell Potential / 0.54
If the concentations of I- and Br- are 0.2M and 0.02M respectively
What would be the voltage of this Voltaic Cell at 298K
E =0.54-( 0.0592/2)*LOG(0.02^2/(0.2)^2) / 0.599 / volts
What will be the voltage if we raise the temperature to 100oC? (2pts)
=0.54-(8.314*2.303*373/(2*96500))*LOG(0.02^2/(0.2)^2) / 0.614 / volts12)Balance the following reaction in a BASIC environment(3pts). Show number of electrons transferred(2pts)
Al(s) + NO2¯ (aq) ® AlO2¯ (aq) + NH3 (g)
H2O + 2Al(s) + NO2¯ (aq) + OH¯ ® 2AlO2¯ (aq) + NH3 (g)
6ELECTRONS!
13) The following initial rate data are for the oxidation of arsenate ion by ferric ion in solution (5pts total): +
AsO33- + 2 Fe3+ + 2OH- è AsO43- + 2 Fe2+ + H2O
If 2 micromoles of arsenate were used up in this reaction over a 2 second period, what current in amperes would be generated by this reaction? (3pts)
2.0E-6 moles of Arsenate4.00E-06 / moles of electrons
=0.000004*96500= / 0.386 / Coulombs
0.386C/2sec = / 0.193 / A