Practice Final Exam
1.
A chemist needs a buffer with pH 4.35. How many milliliters of pure acetic acid (density = 1.049 g/mL) must
be added to 465 mL of 0.0941 M NaOH solution to obtain such a buffer?
HC2H3O2 + OH-  H2O + C2H3O2-
Neutralization (limiting reactant) first. We know that the sodium hydroxide is the limiting reactant because we
are left with a buffer. If the NaOH were in excess or in a stoichiometric ration with the acetic acid we would not
have a buffer
nOH − =
465 mL ×
0.0941 mol
=
43.76 mmol
L
HPy
I
-43.76
I-43.76
Initial moles
Loss of moles
Final moles
OH43.76
-43.76
0
C2H3O2− 
43.76
4.74 + log
4.35
pH =
pK a + log 
=
=
I − 43.76
[HC2H3O2 ]
43.76
−0.39
0.4073802
= 10
=
I − 43.76
43.76
= ( 0.4073802 )( I − 43.76
=
) 0.4073802I − 17.82696
43.76 + 17.82696
0.4073802
I = 151.1780 mmol HC2H3O2
I=
151.1780 mmol HC2H3O2 ×
10 − 3 mol 60.0520 g
1 mL
×
×
=
8.7 mL HC2H3O2
1 mmol
mol
1.049 g
Py0
+43.76
43.76
2.
A study of the gas-phase oxidation of nitrogen monoxide at 25C and 1.00 atm pressure gave the following
results:
2 NO + O2  NO2
Conc. NO, mol/L
Conc. O2,mol/L
Initial Rate
4.5 × 10
4.5 × 10-2
9.0 × 10-2
3.8 × 10-1
2.2 × 10
4.5 × 10-2
9.0 × 10-2
4.6 × 10-3
0.80 × 10-2 mol/(Ls)
1.60 × 10-2 mol/(Ls)
1.28 × 10-1 mol/(Ls)
?
Exp. 1
Exp. 2
Exp. 3
Exp. 4
-2
-2
a. What is the experimental rate law for the reaction above?
Between experiments 1 and 2 the concentration of O2 is doubled and the rate is also doubled so the order with
respect to O2 is 1.
To find the order with respect to NO we need to do a little math…
k [NO ]3 [O2 ]3
R3 1.28 × 10 −1 mol/L s
8.00
=
=
=
=
1
n
R2 1.60 × 10 −2 mol/L s
k [NO ]2 [O2 ]2
n
1
( 9.0 × 10 ) ( 9.0 × 10 )=
( 4.5 × 10 ) ( 4.5 × 10 )
−2 n
−2 1
−2 n
−2 1
2n = 4
n log2 = log 4
n
=
log 4 0.6020599
=
= 2
log2 0.3010299
So the rate law is : Rate = k [NO ] [O2 ] where k is:
2
Rate = k [NO ] [O2 ]
2
=
k
0.80 × 10 −2 mol/L s
Rate
= 1.8 × 102 L2 / mol2 s
( 4.5 × 10−2 mol/L ) (2.2 × 10−2 mol/L )
=
2
[NO] [O2 ]
2
b. What is the initial rate of the reaction in Experiment 4?
=
Rate
(1.8 × 10
2
L2 / mol2 s ) [NO ] [O2 ]
2
=
(1.8 × 102 L2 / mol2 s )( 3.8 × 10−1 mol/L ) ( 4.6 × 10−3 mol/L )
2
= 1.2 × 10 −2 mol/L s
2n ⋅ 2
3.
Consider the following cell reaction at 25C.
2Cr + 3Fe2+  2Cr3+ + 3Fe
Calculate the standard cell potential of this cell from the standard electrode potentials, and from this obtain
G for the cell reaction. Use data below to calculate H. Use these values of H and G to obtain S for
the cell reaction.
Substance
Hf (kJ mol-1)
-89.1
Fe2+
3+
-143.5
Cr 
The cathode is Fe2+ to Fe and the anode is Cr to Cr3+. The number of moles of electrons transferred is 6.
E(Fe2+|Fe) = -0.447 V E(Cr3+|Cr) = -0.744 V



=
− Eanode
=
−0.297 V
Ecell
Ecathode
( −0.744 V ) − ( −0.447 V ) =

∆G =
−nFEcell
 6 mol e−   96485 C 
=
−
−0.297 V ) =
+1.719362 × 105 J mol−1

− (
 mol   mol e 
∆H 
 2 mol Cr 3+   −143.5 kJ   3 mol Fe   0.00 kJ    2 mol Cr  0.00 kJ   3 mol Fe2+   −89.1 kJ  
+



  − 

+

2+  
3+  
mol
  mol Fe  
 mol   mol Cr   mol   mol Fe    mol  mol Cr  
=
−19.7 kJ mol−1 =
−1.97 × 104 J mol−1
∆G  =∆H  − T ∆S 
=
∆S 
∆H  − ∆G  −1.97 × 104 J mol−1 − 1.72 × 105 J mol−1
= −577 J mol−1 K −1
=
T
298.15 K
4.
Crystals of AgBr can be removed from black-and-white photographic film by reacting the AgBr with
sodium thiosulfate.
AgBr + 2 S2O32-  [Ag(S2O3)2]3- + Br-
a. What is the equilibrium constant for this dissolving process?
The overall equation is a combination of the solubility of silver bromide and the complex ion formation of
the dithiosulfatoargentate(I) ion.
AgBr  Ag+ + Br-
Ksp = 5.410-13
Ag+ + 2 S2O32-  [Ag(S2O3)2]3-
Kf = 2.91013
AgBr + 2 S2O32-  [Ag(S2O3)2]3- + Br-
K = Ksp  Kf
K = K sp × K f = ( 5.4 × 10 −13 ) × ( 2.9 × 1013 ) = 15.66
b. In order to dissolve 2.5 g of AgBr in 1.0 L of solution how many moles of Na2S2O3 must be added?
Dissolving 2.5 g of AgBr in 1.0 L of solution tells us that the equilibrium concentration of the Br- must be:
2.5 g AgBr
1 mol AgBr
1 mol Br −
0.013314 M Br −
×
×
=
1.0 L
187.772 g AgBr 1 mol AgBr
S2O32I
-2x
1-2x
I
C
E
 Ag ( S2O3 )3−  Br − 
2 

K 
=
=
2
 S2O32− 
3.957271 =
( x=
)
2
( I − 2x )
2
Ag(S2O3)230
+x
x
Br0
+x
x
( 0.013314 )
2
( I − 2 ( 0.013314 ) )
2
15.66
=
( 0.013314 )
( I − 2 ( 0.013314 ) )
3.957271I − 0.026628 =
0.013314
0.013314 + 0.026628
I = 0.010 M S2O23−
3.957271
0.10 mol S2O23−
1.0 L ×
0.010 mol S2O23−
nS O2− =
=
2 3
L
5.
ln
Tritium, or hydrogen-3, is formed in the upper atmosphere by cosmic rays, similar to the formation of
carbon-14. Tritium has been used to determine the age of wines. A certain wine that has been aged in a
bottle has a tritium content only 70% of that in a similar wine of the same mass that has just been bottled.
How long has the aged wine been in the bottle? The half-life of tritium is 12.3 y.
Nt
ln2
ln2
=
−kt
k=
=
=
0.05635342 y −1
No
t 12 12.3 y
N
ln t
ln ( 0.70 )
N0
=
t =
= 6.3 y
−k
− ( 0.05635342 y −1 )
6.
25.00 mL of a 0.1765 M solution of formic acid is titrated with a 0.2144 M solution of potassium hydroxide.
(a) What is the pH of the solution when 13.68 mL of the sodium hydroxide solution have been added?
Neutralization first!
0.2144 mol KOH 1 mol OH−
2.932992 mmol OH−
×
=
L
1 mol KOH
0.1765 mol HCHO2
25.00 mL
4.41250 mmol HCHO2
=×
=
L
13.68 mL ×
nOH − =
nHCHO2
HCHO2 + OH-  H2O + CHO24.413 2.933
0
-2.933 -2.933
+2.933
___________________________
1.480
0
2.933
We have some of the weak acid left over and have created it’s conjugate base so we have a buffer.
= pK a + log
pH
=
3.74 + log
[B]
[ A]
2.933
=
4.04
1.480
(b) What is the pH of the solution at the equivalence point?
At the equivalence point the moles of OH- is equal to the moles of HCHO2. The total volume can be found from:
1 mol KOH
L
×
=
20.58 mL KOH used
−
1 mol OH 0.2144 mol KOH
total volume = 20.58 mL + 25.00 mL = 45.58 mL
4.41250 mmol OH− ×
All of the moles of HCHO2 are converted to the conjugate base, CHO2-. We can find the equivalence point pH by
doing the weak base calculate with the conjugate base.
CHO2− =

i
4.413 mmol CHO2−
= 0.09682 M
45.58 mL
I
C
E
CHO20.09682
-x
0.09682-x
K=
b
[HCHO2 ] OH −  ( x ) (10−7 + x ) ( x )( x )
5.6 × 10 −11
Kb = −
=
≈
=
0.09682
0.09682
x
−
CHO2 
)
(
) (
x=
( 5.6 × 10 ) ( 0.09682) =
−11
− log OH −  =
5.6347
pOH =
2.319 × 10 −6 ≈ OH − 
14.00 − pOH =
8.37
pH =
Kw 1.00 × 10 −14
=
= 5.6 × 10 −11
−4
Ka
1.8 × 10
OH10-7
+x
10-7+x
HCHO2
0
+x
x
approximation is valid here (4.3%)
7.
Phosphorus(V) chloride, PCl5, dissociates on heating to give phosphorus(III) chloride, PCl3, and chlorine.
PCl5  PCl3 + Cl2
A closed 2.00-L vessel initially contains 0.0100 mol PCl5. What is the total pressure at 250C when
equilibrium is achieved? The value of Kc at 250C is 4.15 × 10-2.
First find Kp because we want the pressure of the system.
(
Kp =
K c ( RT ) =
( 4.15 × 10−2 ) ( 0.083144 L bar mol−1 K−1 ) ( 523 K )
∆n
)
(2−1)
=
1.804598
Initial pressure of PCl5:
nRT
P=
=
PCl5
V
( 0.0100 mol PCl5 ) ( 0.083144 L bar mol−1 K−1 ) ( 523 K )
=
(2.00 L )
Set up ICE table:
Calculate:
PPCl3 PCl2
Kp =
=
PPCl5
PCl5
0.2174
-x
0.2174-x
I
C
E
0.2174 bar
PCl3
0
+x
x
Cl2
0
+x
x
x2
= 1.80
( 0.2174 − x )
0
x 2 + 1.80 x − 0.391 =
−b ± b2 − 4ac − (1.80 ) ±
x=
=
2a
PT= PPCl5 + PPCl3 + PCl2=
8.
(1.80 ) − 4 (1)( −0.391)
= −1.9959
2 (1 )
( 0.2174 − x ) + x + x= 0.2174 + x= 0.431 bar
2
or 0.19590
Answer the following questions about transition metal chemistry.
a. [Fe(HOC6H4COO)3] is purple. The salicylate ion, HOC6H4COO-, is a bidentate ligand. What hybridization is
the iron ion using in the complex ion? Explain your reasoning.
The oxidation state of the iron is 3+ so it has 5 electrons in the d orbitals. Because the ligand is bidentate the
coordination number is 6 so the geometry is octahedral. Therefore the hybridization is either d2sp3 or sp3d2.
The color of complex is purple so it absorbs orange which means that the  is small which means that the
electrons pair up before moving to higher d levels which means that the two higher d orbitals are open so
the hybridization is d2sp3.
b. Cd(H2O)62+ is colorless. Explain why this would be expected.
The Cd2+ ion has 10 electrons in the d orbitals so there can be no transitions of electrons between d levels to
give rise to a color.
c. The fluoride ion is a weakly bonding ligand. What color would you expect to see in the FeF64- ion? Explain.
Because the ligand is weakly bonding the  should be small so the color absorbed should be in the
red/yellow/orange part of the spectrum so we would see the green/blue/purple color.
d. Name K2[Cr(CO3)2(en)2]. What is the coordination number of the chromium ion?
Potassium dicarbonatobis(ethylenediamine)chromate(IV). The coordination number is 6 because the
carbonate is monodentate (21) and the ethylenediamine is bidentate (22).