Chemistry 2300 (Loader)
Winter 2004
Page 1 of 3
Chemistry 2300
Y
TOTAL 50 MARKS
NAME:___________________________
March 24th 2004
MUN STUDENT#:____________________
Answer ALL of the questions in the spaces provided. Show your calculations or
explanations and give final numerical answers to the correct number of significant digits. The
mark that you obtain for this test will be used in calculating your final grade for the course.
1.
(a)
A cell is constructed according to the following diagram using equal volumes
of solution in the electrodes
Ni|Ni2+(aq, 1.0 mol L–1)||Ag+(aq, 1.0 mol L–1)|Ag
For
[3]
Ni2+ + 2 e- → Ni(s); E° = -0.26 V
Ag+ + e- → Ag(s); E° = 0.800 V at 298 K
Write the cell reaction and calculate E o298 K for the cell.
(i)
The cell reaction:
2 Ag+(aq , 1.0 mol L–1) + Ni(s) → 2 Ag(s) + Ni2+ (aq, 1.0 mol L–1)
Ni(s) → Ni2+ + 2 e-;
E° = 0.26 V
2 Ag + 2 e → 2 Ag(s); E° = 0.800 V
+
-
adding gives 2 Ag+ + Ni(s) → 2 Ag(s) + Ni2+
E°cell = 1.060 V
E o298 K =
[6]
Ans:
1.06 V
(ii) When the cell is operated for a while the concentration of the Ag+(aq)
solution drops to 0.80 mol L–1. Calculate the cell potential at this time.
First calculate the final concentrations of the Ni2+ in the other electrode. 0.20 mol L–1
of the Ag+ has been used so from the equation 0.10 mol L–1 of additional Ni2+ must
have formed and [Ni2+] = 1.1 mol L–1.
Now, E = E° -
RT
F
[Ni 2+ ]
ln( [Ag + ] 2 ) = 1.060 −
8.314 % 298
2 % 96485
ln
1.1
(0.80 ) 2
V
E = 1.053 V
Ans: 1.05 V
(b)
Fe3+ + 3 e- → Fe(s);
Fe2+ + 2 e- → Fe(s);
Ni2+ + 2 e- → Ni(s);
E° = -0.036 V
E° = -0.440 V
E° = -0.26 V at 298 K,
calculate the cell potential for the standard cell described by the cell diagram
Pt(s)|Fe2+(aq, 1.0 mol L –1),Fe3+(aq, 1.0 mol L-1)||Ni2+(aq, 1.0 mol L –1)|Ni(s)
at 298 K as written.
The equation for the overall reaction can be obtained by reversing the first equation,
addint to the second, doubling the sum and adding it to the third. The reduction
potentials cannot be added as the electron transfer is not balanced.
But it we convert to Gibbs energies they can be added
2 Fe(s) → 2 Fe3+ + 6 e-; G o = - 6FE° = -6F(0.036 V)
2 Fe2+ + 4 e- → 2 Fe(s); G o = - 4FE° = 4F(0.440 V)
Ni2+ + 2 e- → Ni(s);
[6]
overall reaction
E ocell = −
G oreact
2F =
Ans: 1.032 V
G o = -2FE° = 2F(0.26 V)
Ni2+ + 2 Fe2+ → Ni(s) + 2 Fe3+
2F(0.26 V) + 4F(0.440 V) - 6F(0.036 V)/2F = 1.032 V
Chemistry 2300 (Loader)
2.
[4]
Winter 2004
Silver carbonate, Ag2CO3(s), is stable at room temperature but decomposes to
give silver(I) oxide, Ag2O(s), and carbon dioxide when heated. For the equilibrium
Ag2CO3(s) î Ag2O(s) + CO2(g)
at 298 K, the equilibrium constant, Kp = 3.16 x 10-6 and at 350 K the equilibrium
constant is 3.98 x 10–4. Calculate
(a) G )298 ,
G )298 = − RT ln K p, 298K =
[4]
Page 2 of 3
−8.314 J K −1 mol −1 % 298 K %ln(3.16%10 −6 )
1000 J k −1
= 31.3 8 kJ mol −1
Ans: 31.3 8 kJ mol −1
(b) H )298 ,
ln
K2
K1
ln
3.98 x10 –4
3.16 x10 −6
=
−H o
R
=
1
T2
−
1
T1
−H o
8.314 % 10 −3 kJ K −1 mol −1
o(
) −1
( 3501 K −
1
298 K
)
4.836 = −H −0.0599 kJ mol
H )298 = 80.6 4 kJ mol −1
[4]
=3]
Ans: 80.6 kJ mol −1
(c) S )298 , for the reaction.
G = H − TS
31.38 = 80.64 − 298S o298 K
S )298 = 165 J K −1 mol −1
Ans : = 165 J K −1 mol −1
(d)
Calculate G298 for the reaction if carried out open to the atmosphere where
p CO2 = 30.4 Pa?
G 298 = G )298 + RT ln Q p, 298K = 31.38 kJ mol −1 +
Ans:
s
11.3 kJ mol
−1
−3
x 10
−8.314 J K −1 mol −1 % 298 K %ln( 30.4
101.3 kPa )
1000 J k −1
= 11.3 kJ mol −1
(e) A similar equilibrium reaction is possible for calcium carbonate:
CaCO3(s) î CaO(s) + CO2(g)
for which Kp,298 K = 1.6 x 10-23. What reaction (if any) would you expect if some
CaO(s) was mixed with solid Ag2CO3, write the equation for it and calculate the
equilibrium constant at 298 K.
Ag 2 CO 3 î Ag 2 O(s) + CO 2 (g); K p = 3.16 % 10 −6
This equation can be combined with the one for CaO if it is reversed
Ag 2 CO 3 î Ag 2 O(s) + CO 2 (g); K p = 3.16 % 10 −6
CaO(s) + CO2(g) î CaCO3(s); Kp2 = 1/(1.6 x 10-23)
Adding gives
Ag2CO3(s) + CaO(s) î Ag2O(s) + CaCO3(s);
and Kp3 = (3.16 x 10-6) + 1/(1.6 x 10-23) = 2.0 x 1017 for the new reaction. Since Kp3
is very large the reaction is spontaneous
Ans: 2.0 x 1017 Since Kp3 is very large the reaction is spontaneous and almost
complete reaction would be observed at equilibrium.
Chemistry 2300 (Loader)
3.
[3]
(a)
Winter 2004
Page 3 of 3
A 0.010 m aqueous solution of the ionic compound Co(NH3)5Cl3 is a strong
electrolyte and has a freezing point depression of 0.0558 K. Calculate the
apparent concentration of the Co(NH3)5Cl3 solution and explain what this
indicates about the structure of the Co(NH3)5Cl3 in aqueous solution. The
molal freezing-point depression constant for water is Kf = 1.86 K mol –1 kg.
t = K f mwhere ∆t is the temperature depression and m is the molality of the
solution. So
K
−1
m = 1.860.0558
K mol −1 kg = 0.0300 mol kg .
Since the solution is three times the analytical concentration the compound must
give three ions for each Co(NH3)5Cl3 dissolved (possibly to give [Co(NH3)5]2+ and
2Cl-)
(b) 25.00 mL of a solution of propanoic acid, CH3CH2COOH, requires 22.00 mL
of 0.100 mol L –1 NaOH solution to reach the stoichiometric endpoint in the
reaction
CH3CH2COOH(aq) + NaOH(aq) → NaOOCCH2CH3(aq) + H2O(l)
(i) When the stoichiometric endpoint for the titration was reached an
additional 25.00 mL of the propanoic acid solution as added to the
reaction mixture and the measured pH was 4.87. Calculate the acidity
constant, Ka for propanoic acid.
Assuming that self-ionization of the water can be ignored then
[CH CH COO − ]
pH = pKa + lg [CH 33CH 22COOH] , and here [CH3 CH2 COO− ] and [CH3 CH2 COOH]
are equal so pH = pKa. Now pKa = - lg Ka so Ka = 10 −pKa = 10-4.87 = 1.34x 10-5
Ans: 1.3 x 10-5
(ii)
Calculate the pH of the titration solution at the stoichiometric endpoint.
[4]
At the stoichiometric endpoint the solution is a solution of the salt of the acid
NaOOCCH2CH3(aq) containing the anion –OOCCH2CH3. The anion is from a weak
acid and is hydrolyzed by water
CH3CH2COO–(aq) + H2O(l) → CH3CH2COOH(aq) + OH–(aq)
[CH CH COOH][OH− ]
[5]
(c)
K
−14
3
2
1.0 x 10
−10
Kb = [CH
.
= Kwa = 1.34
−
x 10 −5 = 7.41 x 10
3 CH 2 COO ]
Here, [CH3 CH2 COOH] = [OH−] so if [OH−] << [CH3 CH2 COOH− ] then
[OH − ] 2
[OH − ] 2
L −1 % 22.00 mL
Kb = [CH 3 CH 2 COO − ] = 7.41 x 10 −10 = C o . C o = 0.100 mol47.00
= 0.0468 mol L −1
mL
-6
-1
Substituting and solving for [OH− ] gives 5.89 x 10 mol L
so pOH = 5.230 and pH = 14 -5.230 = 8.77
Ans: 8.77
Calculate the concentration of all of the species in a 0.20 mol L –1 solution of
the oxalic acid, HOOCCOOH, (a diprotic acid) for which pKa1 = 1.19 and
pKa2 = 4.21.
There is a factor of 103 difference in pKa1 and pKa2 so this dibasic acid can be
treated in the same way as carbonic acid.
HOOCCOOH(aq) → HOOCCOO–(aq) + H+(aq)
[H + ] 2
K a1 = [HOOCCOOH] = 0.06457 using the usual method and solving the quadratic gives
[H+] = [HOOCCOO–] = 0.0858 mol L-1. Since this ionization is, by far the main
source of H+ , within experimental error, this is the concentration the second
ionization will see.
HOOCCOO–(aq) → –OOCCOO–(aq) + H+(aq)
[H + ][ − OOCCOO − ]
[HOOCCOO − ]
–
–
but [H+] = [HOOCCOO–]
so [ OOCCOO ] = 6.17 %
mol L-1 and our approximations are valid as the
[H+] from this source also = 6.17 % 10−5 mol L-1 and is much less than 0.0858 mol
L-1.. That leaves [HOOCCOOH] = (0.20 - 0.0858) mol L-1 = 0.114 mol L-1 and
w
[OH-] = [H+K(aq)]
= 1.17 x 10-13 mol L-1
K a2 =
= 6.17 % 10 −5
10 −5
Ans: [H+] = [HOOCCOO–] = 0.086 mol L-1., [–OOCCOO–] = 6.2 % 10−5 mol L-1,
[HOOCCOOH] = 0.11 mol L-1 and [OH-] = 1.2 x 10-–-13 mol L-1