1
Final 2000 [3 questions short due to change of coverage] solutions
1. Which of the following solutions will have a pH of 1?
(A) 0.1 M CH COOH (B) 0.1 M HF (C) 1 M CH COOH (D) 0.1 M NH
3
3
3
(E) 0.1 M HNO
3
solution
For a strong acid, pH = - log[H+] = - log 0.1 = 1
2. Which of the following will NOT produce a buffer solution?
(A) NaCl and CH COOH (B) NH Cl and NH (C) KCN and HCN
3
4
3
(D) NaHCO and H CO (E) CH COOH and CH COONa
3
2
3
3
3
solution
A buffer solution contains a weak acid and its conjugate base or a weak base and its conjugate
acid in a ratio of 1:15 -15:1.
NaCl is a salt, Cl- is not a conjugate base for the weak acid CH3COOH
NH4+/NH3 This is a conjugate acid/weak base pair, will form a buffer
CN-/HCN conjugate base/weak acid pair, will form a buffer
HCO3-/H2CO3 conjugate base/weak acid pait will form a buffer
CH3COOH/CH3COO- weak acid conjugate base pair, will form a buffer
2+
3. The ground state electron configuration of Zn is:
2
10
1
9
2
8
8
10
(A) [Ar]4s 3d (B) [Ar]4s 3d (C) [Ar]4s 3d (D) [Ar]3d (E) [Ar]3d
solution
Atomic number of Zn = 30
Electron configuration for Zn is [Ar]4s23d10 and Zn2+ = [Ar]3d10
4. 0.10 g of an unknown substance is dissolved in 100 mL of water, and the osmotic pressure is
measured to be 0.015 atm at 25 ºC. What is the molar mass of the unknown substance?
solution
П = molarity x RT ={ moles/Volume(L)} x RT} = (mass/MMV) x RT
MM = (mass x RT)/(V x П)
MM= (0.1 g x 0.0821atm/K mol x 298 K)/(0.1 L x 0.015 atm) = 16,00 g/mol
MM = molar mass
5. What is the equation for what occurs at the anode during the electrolysis of molten
magnesium chloride?
-
-
-
(A) 2 Cl (l) → Cl (g) + 2 e
-
2
-
(D) 2 Cl (aq) → Cl (g) + 2 e
2
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-
+
-
(B) Cl + 2e → 2 Cl (C) 2 H O → O (g) + 4 H + 2 e
2
2+
-
(E) Mg (l) + 2 e → Mg(s)
2
2
2
solution
Oxidation occurs at the anode and Cl- (oxidation number -1) is oxidized to Cl2(g) ( oxidation
number 0)
6. Calculate the equilibrium constant at 298 K for the decomposition reaction below given that
∆Gº (HF) = -275 kJ/mol
f
solution
2 HF(g)
H (g) + F (g)
2
2
∆G° = ∑∆G° f(products) - ∑∆G°f(reactans) = 0 – (2 x -275) = 550 kJ/mol
∆G° = -(RT)lnK
lnK = ∆G°/-RT = 550/ (- 8.3145 x 10-3 x 298) = -221.98
K = e-222 = 3.95 x 10-97
∆G°f is always zero(0) for elements or molecules in their natural state. That is in nature
hydrogen, fluorine, oxygen etc occurs in nature as H2, F2, O2 etc
7. Which of the following will be diamagnetic?
3+
2+
4+
2+
(A) Co (B) Mn (C) Ti (D) Fe (E) Ni
solution
Co2+ = [Ar]3d6 Mn2+ = [Ar]3d5
Ti4+ = [Ar]
Fe = [Ar]4s23d6 Ni2+ = [Ar]3d6
Diamagnetic means all the electrons are paired as opposed to paramagnetic, where one or more
electrons are unpaired.
8. Given the equilibrium 2H2O(l) <==> OH-(aq) + H3O+(aq) and that Ka = 4.93 x 10-10 for
hydrocyanic acid [HCN], determine the equilibrium constant of the reaction.
HCN(aq) + OH-(aq) <==> H2O(aq) + CN- (aq)
Solution
2H2O(l) <==> OH-(aq) + H3O+(aq) Kw = 1 x 10-14
HCN(aq) + H2O (aq) <==> H3O+(aq) + CNka = 4.93 x10-10 = [H3O+][CN- ]/[ HCN]
[CN- ]/[ HCN] = 4.93 x10-10/ 1 x 10-7 = 4.93 x 10-3 ; since kw of water = [H3O+][ OH-] = 1 x 10-14
HCN(aq) + OH-(aq) <==> H2O(aq) + CN- (aq)
Ka = [CN-]/[ HCN][ OH-] = 4.93 x 10-3/ 1 x 10-7 = 4.93 x 104
This is a good one; it took me time to figure it out. Very simple though, you just need some
algebra and play around with it. Come to ACS help desk for help if you cannot do this, I’ll be
glad to explain.
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3
9. Identify the unknown particle X in each of the following:
16
20
I. O + X → Ne
31
31
II. P → S + X
243
Am → α + X
III.
241
(A) β, α,
239
Np (B) α, β,
241
Np (C) α, β,
239
Np (D) α, β,
239
Pa (E) β, γ,
Pa
solution
16
31
O + 4 2α →
20
Ne ( Ne has 10 protons (8 from oxygen + 2 from alpha)
31
P → S + β ( beta has -1 and sulfur 16, this adds up to 15 protons for phosphorus)
243
Am → α + 239Np ( neptunium has 93 and 2 protons from alpha particle will add up to give
americium 95 protons)
10. What is the pH of a 1 L solution containing1.04 g of Ca(OH) [assume it all dissolves]
2
solution
Molarity = moles/Volume = mass/MMV = 1.04 g/74 g/mol x 1L = 0.01405 M
Ca(OH)2
Ca2+ + 2OH[OH] = 2 x 0.0135 = 0.0281 M
pOH = - log [OH-] = - log 0.027 = 1.55
pOH + pH = 14
pH = 14 - pOH
pH = 14 – 1.55 = 12.45
11. What is the pH of a 0.100 M aqueous solution of sodium azide (NaN ) at 25 ºC if the K for
-5
3
a
HN is 1.9 x 10 ?
3
solution
N3- + H2O
HN3 + OHS 0.1
0
0
C -x
x
x
E 0.1 – x
x
x
Kb = [HN3][OH-]/[N3-]
Since Ka x Kb = 1 x 10-14; K b = 1 x 10-14/Ka
1 x 10-14/ 1.9 x 10-5 = x2/(0.1-x); since Kb is small, the product and hence x is assumed to also be
small
5.26 x 10-10 = x2/0.1
x = 7.25 x 10-6
pOH = - log [OH-] = - log 7.25 x 10-6 = 5.14
pH = 14 – 5.14 = 8.86
12. State whether each of the following will be more soluble in water or hexane [C H ]
6
I. NaCl
II. CH OH
3
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14
4
III. octane (C H )
8
18
(A) hexane, water, water (B) water, water, hexane (C) hexane, water, hexane
(D) water, water, water (E) water, hexane, water
Ans. B
NaCl and CH3OH are polar and will be more soluble in water, a polar molecule whereas, octane
is non polar and will be more soluble in hexane a non-polar molecule.
Remember, like dissolves like
13. Which of the following statements is/are true?
I. The rate at which a reaction proceeds decreases with time
II. The rate law for any reaction can be determined from the overall balanced equation of the
reaction.
III. The rate constant for a reaction depends only on the energy difference between the reactants
and products and the temperature
IV. An increase in temperature causes an increase in the rate constant for a reaction
(A) All are true (B) I only (C) II and IV (D) I and IV (E) I, III and IV
14. A flask contains the following system at equilibrium:
2+
-
Mg(OH) (s) <==> Mg (aq) + 2 OH (aq)
2
Which of the following reagents could be added to increase the solubility of Mg(OH) ?
2
(A) MgCl (B) NH (C) NaOH (D) H O (E) HCl
2
3
2
solution
When HCl is added, the H+ will formed water with the OH-, and more Mg(OH)2 will dissolve to
replace the hydroxide ions used up. Le Chatelier’s principle
15. What is the balanced equation for the following cell notation?
2+
2+
Mn(s) | Mn (aq) || Cd (aq) | Cd(s)
2+
2+
(A) none (B) Mn(s) + Cd (aq) → Cd(s) + Mn (aq)
2+
2+
2+
-
(C) Cd(s) + Mn (aq) → Mn(s) + Cd (aq) (D) Mn (aq) + 2 e → Mn(s)
2+
-
(E) Cd (aq) + 2 e → Cd(s)
Just take the first of each half cell and add together to form the reactants and the second of each
half cell and add to form the products.
16. Which of the following solutions has the highest melting point? Which has the highest
boiling point?
I. 10.0 g of glucose (C H O ) in 100 mL of water
6
12
6
II. 10.0 g of glucose in 50.0 of water
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5
III. 10.0 g of ethylene glycol (C H O ) in 100. mL of water
2
6
2
(A) I, I (B) I, III (C) III, III (D) I, II (E) III, I
Solution
The least concentrated will have the highest melting point and the most concentrated will have
the highest boiling point.
I = 10 g/180 g/mol x 0.1L = 0.56 M
II =10 g/180 g/mol/0.05L = 1.1 M
III = 10 g/62 g/mol x 0.1L = 1.6 M
Look at it this way, melting goes down and boiling goes up. The least concentrated goes with the
highest melting point and the most concentrated goes with the highest boiling point.
17. If 0.35 g NaF is added to 150 mL of a 0.30 M HF solution, what is the resultant pH of the
-4
solution. For HF, K = 7.2 x 10
a
(A) 3.15 (B) 2.41 (C) 4.21 (D) 5.55 (E) 3.90
solution
Moles of NaF = mass/ MM = 0.35 g / 42 g/mol = 0.0083 moles
[F-] = moles / Volume(L) = 0.0083/0.15 = 0.056 M
[HF] = 0.3 M
HF
+ H2O <==>
H3O+ + FS
0.3
0
0.056
C
-x
+x
+x
E
0.3 - x
x
0.056+ x
+
Ka = [H3O ][F ]/[HF]
7.2 x 10-4 = x(x+ 0.056)/ (0.3 –x); since Ka is small, we can asummed that the product and hence
x is also small
7.2 x 10-4 = 0.056x/0.3
x = 3.86 x 10-3
pH = - log[H3O+] = - log 3.86 x 10-3 = 2.41
18. What would predict when 0.0045 g KCl is added to 300 mL of a 0.12 M Pb(NO ) solution?
3 2
-5
K for PbCl = 1.7 x 10
sp
2
(A) KCl precipitates (B) KNO precipitates
3
(C) solution not saturated, no PbCl2 will precipitate (D) cannot predict anything
(E) solution more than saturated, PbCl precipitates
2
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6
solution
2KCl + Pb(NO3)2 <==> PbCl2 + 2KNO3
[KCl] = 0.0045/7(4.55 x 0.3L) = 2 x 10-4 M = [K+][Cl-]
PbCl2 <==> Pb2+ + 2ClQ = (2 x10-4/2)2 x (0.12/2) = 6.0 x 10-6 < K(1.7 x 10-5) , solution not saturated, no PbCl2 will
precipitate.
A precipitate will not form when Q < Ksp
19. Name the strongest oxidizing agent among the following substances:
Eº (V)
-
-
I (s) + 2 e → 2 I (aq)
2
-
0.53
-
Br (l) + 2 e → 2 Br (aq)
2
-
1.07
-
F (g) + 2 e → 2 F (aq)
2
-
2.87
-
Cl (g) + 2 e → 2 Cl (aq)
2
2+
1.36
-
0.34
Cu (aq) + 2 e → Cu(s)
(A) Cu(s) (B) I (s) (C) Br (l) (D) F (g) (E) Cl (g)
2
2
2
2
Solution
The strongest oxidizing agent will be reduced with the most Eº (V)
20. Which of the following pairs has the stronger acid listed first?
I. H AsO , H AsO
3
3
3
4
II. HF, HBr
III. HOCl, HOBr
(A) III only (B) I and II (C) I and III (D) II only (E) I only
solution
In I the stronger acid is the one with more oxygen;
in II, HBr is a stronger acid since acidity increases down the period for binary acids;
and in III, HOCl since for oxyacid acidity increases up the group
21. Which substance has the greatest molar entropy?
(A) Rn(g) (B) Ne(g) (C) Ar(g) (D) Kr(g) (E) Xe(g)
solution
Radon, Rn with the biggest molecular mass has the highest entropy.
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22. Balance the redox equation in basic solution and select the correct smallest integer
-
coefficient for the ClO ion.
-
3
-
-
-
BH (aq) + ClO (aq) → H BO (aq) + Cl (aq)
4
3
2
3
(A) 3 (B) 5 (C) 6 (D) 4 (E) 2
Solution
-5
+5
-
+3
-
-1
-
-
BH (aq) + ClO (aq) → H BO (aq) + Cl (aq)
4
3
-
-
2
-
BH → H BO + 8e
4
-
2
-
3
-
ClO + 6e → Cl
3
3
x3
x4
-
-
-
-
3BH (aq) + 4ClO (aq) → 3H BO (aq) +4 Cl (aq) + 3H2O
4
3
2
3
23. What is the temperature at which the following process reaches equilibrium at 1 atm?
S(rhombic) → S(monoclinic), ∆Hº = 0.3 kJ, ∆Sº = 0.7 J / K
∆G° = ∆H° - T∆S°
For the process to react equilibrium, ∆G° must be zero
0 = ∆H° - T∆S°
T = ∆H° / ∆S° = 0.3 kJ/ 0.7 x10-3 kJ/K = 428 K
24. The mass of carbon-14 is 14.003241 g/mol. Calculate the binding energy of a carbon-14
nucleus if the mass of a proton is 1.00783 g/mol, the mass of a neutron = 1.00867 g/mol, the
8
2
2
speed of light is 3.00 x 10 m/s and 1 J = 1 kg m /s
C-14 has 6 protons and 8 neutron
Mass = (6 x 1.00783 + 8 x 1.00867) = 14.11634 g/mol
Mass deficit = 14.11634 – 14 .003241 = 0.113099 g/mol = 1.13099 x10-4 kg/mol
E = mc2 = 1.13099 x( 3.00 x108)2 = 1.02 x 1013 J/mol
25. Under which of the following conditions is a reaction spontaneous under standard-state
conditions.
(A) ∆Gº > 0, K > 1, Eº < 0 (B) ∆Gº < 0, K > 1, Eº > 0 (C) ∆Gº < 0, K = 1, Eº = 0
cell
(D) ∆Gº = 0, K = 1, Eº
cell
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cell
= 0 (E) ∆Gº = 0, K = 1, Eº
cell
cell
>0
8
26. Consider the following reaction, I (aq) + Cl (aq) <==> 2 ICl(g), ∆H = -27 kJ, for which K =
2
5
2
p
1.6 x 10 at 25 ºC. If the temperature is increased to 100 ºC, what changes if any will be
observed.
(A) [Cl ] will remain the same (B) no change because K = K (C) [ICl] will increase
2
c
p
(D) [I ] will increase (E) the partial pressure of ICl will increase
2
Solution
Increasing the temperature, the reaction will go to the positive direction. Since the forward
direction is negative, the backward direction is positive, more I2 or Cl2 will be formed.
27. What mass of copper will be deposited when 18.2 A are passed through a CuSO solution for
4
45 minutes?
Cu2+(aq)+ + 2e-
Cu(s)
Charge = ampere x time = 18.2 x 45 x 60 = 49140 amps
Charge = neF
n = charge/eF = (49140/ 2 x 96500) = 0.255 moles, e = number of electrons and F= faraday
constant.
Mass = moles x MM = 0.255 moles x 63.55 g/ mol = 16.2g
28. The most common oxidation state for ions of the inner transition elements is
(A) +5 (B) +4 (C) +3 (D) +6 (E) +2
29. What is the rate law for the reaction 2 NO(g) + Cl (g) → 2 NOCl(g) given the following
2
mechanism:
1. NO(g) + Cl (g) → NOCl (g) fast
2
2
2. NOCl (g) + NO(g) → 2 NOCl(g) slow
2
2
(A) rate = k[NO] [Cl2] / [NOCl]
(B) rate = k[NO]2[Cl2]
2
(D) rate = k[NOCl]
(E) rate = k[NO]2[Cl2]/[NOCl]2
(C) rate = K[NO][Cl2]
Solution
The slow step determines the rate law, and NOCl2 is an intermediate product formed in step 1
and used up in step 2.
We will have
NO(g) + Cl(g) + NO(g) ===> 2NOCl(g). I brought down reactants in step 1 and added it to step
2, then ignore NOCl2 since it will cancel out.
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9
30. Which of the following acids has the strongest conjugate base?
(A) chloroacetic acid, pKa = 2.9 (B) ascorbic acid, pKa = 4.1 (C) benzoic acis pKa = 4.2
(D) 3-cholobenzoic acid, pKa = 3.8 (E) 2-hydroxybenzoic acid, pKa = 3
solution
The weakest acid( highest pKa) has the strongest conjugate base.
31. Which of the following elements can be isolated by electrolysis of the aqueous salt shown?
(A) hydrogen from CuSO (aq) (B) sodium from Na PO (aq) (C) sulfur from K SO (aq)
4
3
4
2
4
(D) silver from AgNO (aq) (E) potassium from KCl(aq)
3
solution
silver metal will be deposited at the cathode and can be isolated.
32. An experiment designed to determine the rate law for the reaction, H (g) + 2 ICl(g) → 2
2
HCl(g) + I (g) yields the following data:
2
Trial
Initial [H ] mol/L
Initial [ICl] mol/L
1
0.400
0.400
3.2 x 10
2
0.800
0.400
6.4 x 10
3
0.800
0.800
1.28 x 10
2
Initial rate mol / L s
-6
-6
-5
2
2
(A) rate = k[H ] (B) rate = k[H ][ICl] (C) rate = k [ICl]
2
2
1/2
(D) rate = k[H ][ICl] (E) rate = k[H ] [ICl]
2
2
Solution
Rate law of H2
0.4/0.4 x 0.8/0.4 = 6.4 x 10-6/3.2 x 10-6
2 = 2 so rate law for H2 is 1 I chosed where ICI has the same concentration in order to get rate
law for H2
Rate law for ICI
0.8/0.8 x 0.8/0.4 = 1.28 x 10-5/6.4 x 10-6
2 = 2, the rate law is also 1. I also chose where H2 has the same concentration in order to get the
rate law for ICI
Rate = k[H2][ICI]
33. The reaction, A → 2 B is first order in A. If it takes 2.00 minutes for the concentration of A
to decrease from 3.0 M to 0.75 M, what is the rate constant?
1n[A]t/[A]0 = -kt
ln (0.75/3) = - 2k
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10
k = 0.69 min-1
34. Arsenic acid, H AsO , is a polyprotic acid used to manufacture pesticides. Given that K =
3
-4
4
-8
a1
-13
2.5 x 10 , K = 5.6 x 10 , and K = 3 x 10 , what is the pH of a 0.500 M solution of the acid?
a2
a3
Solution
S
C
E
H2AsO4- + H3O+
0
0
x
x
x
x
H3AsO4 + H2O
0.5
-x
0.5 – x
Ka = [H2AsO4-][H3O+]/[H3AsO4]
2.5 x 10-4 = x2/0.5-x , let us assume x is small, then,
2.5 x 10-4 = x2/0.5
x = 1.12 x 10-2
pH = - log [H3O+] = - log 1.12 x 10-2 = 1.96
35. A 2.00 liter flask is filled with 1.5 moles SO , 2.5 moles SO and 0.5 moles of O , and
3
allowed to reach equilibrium, 2 SO (g)
2
2
2 SO (g) + O (g). At this temperature, K = 1.0. What
3
2
2
c
is the effect on the concentration of O as equilibrium is being achieved?
2
(A) [O ] will decrease as Q < K (B) [O ] will increase as Q > K (C) [O ] will decrease as Q > K
2
2
2
(D) [O ] will remain the same as Q + K (E) [O ] will increase as Q < K
2
2
Solution
2SO2 + O2
2SO3
[ ] 1.5/2= 0.75 2.5/2= 1.25 0.5/2= 0.25
0.75
1.25
0.25
Q= (1.25 )2(0.25) = 0.69
(0.75 )2
0.69<1, equilibrium will shift to the right and more [O2] and [SO2] will be produced
238
36. how many half-lives will take for a sample of 5.00 moles
238
U?
(A) 4 (B) 3 (C) 5 (D) not enough info given (E) 2.34
Solution
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U to decay to 0.3125 moles of
11
Half-life is the time taken for a sample to decay to half its original value
5 moles
2.5 moles
2.5 moles
1.25 mole
1.25 moles
0.625 moles
0.625 moles
0.3123 moles
It took 4( steps) half lives
37. A certain transition element has the stable oxidation states of +2, +4, +5 and +6. In which
state will be the element be most likely to form a covalent bond with chlorine?
(A) +4 (B) +5 (C) +2 (D) it will always form a covalent bond with chlorine (E) +6
Solution
Since its oxidation state is stable up to +6, chlorine being and oxidizing agent will definitely
oxidized the transition metal to its highest oxidizing state.
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