1. finance
2. bank
3. atms

This is a sample Midterm #2 (longer then one midterm) that I (Rolf Unterleitner) made up
for Dr. T’s 2B sections (it’s good for others prof’s too). I will go over the solutions on Wed.
13th in 26 Wellman 5:40 PM-??? (likely over 2 hours). The answers are given at the end.
Please do not ask the Prof. or the TA's for the solutions they do not have them thank you.
1) Consider the following reaction allowed to come to equilibrium:
P4(s) + 4 NaOH(aq)+ 2 H2O(l) ⇐⇒ 2 PH3(g) + 2Na2HPO3(aq)
ΔH°<0
If equilibrium is perturbed by the following changes, what will be the effect on
the indicated quantity when equilibrium is reestablished.
1. If PH3(g) is added then K will?
a) increase b) decrease c) not change
2. If water is added to the solution the mass of P4(s) will?
a) increase b) decrease c) not change
3. If HCl(l) is added to the solution the reaction will shift to
a) left
b) right
c) will not shift
4. If P4(s) is added the reaction will shift to
.
a) left
b) right
c) will not shift
5. If NaOH(s) is added the reaction will shift to
a) left
b) right
c) will not shift
6. If the temperature is increased the rxn will shift to
a) left
b) right
c) will not shift
7. If the temperature is increased K will
.
a) increase b) decrease c) not change
8. If P4 (s) is added the reaction will shift to
.
.
a) left
b) right
c) will not shift
9. If the volume of the container is increased the rxn will shift to
a) left
b) right
.
.
.
c) will not shift
2) Given: 4NH3(g) + 7O2(g) ⇐⇒ 2N2O4(g) + 6H2O(g).
If, initially, P(NH3)=P(O2)=1.00 atm and at equilibrium P(H2O)=.125. Calculate the
partial pressure of oxygen at equilibrium.
a) .979 atm
b) .854 atm
c) .125 atm
d) 1.00 atm
1
e) need Kp to calculate
3) Which of the following 1.0 M solutions is acidic.
a)
K2HPO4 b) KHCO3
c) CoCl2
d) NaF
e) Ca(NO2)2
4)
Which of the following is correct if the initial pressures are as follows CH4=2 atm,
H2O=2 atm, CO=2 atm and H2=4 atm.
CH4(g) + H2O(g) ⇐⇒ CO(g) +3 H2(g) Kp=25.6
a)
b)
c)
d)
e)
The
The
The
The
The
reaction
reaction
reaction
reaction
reaction
favors
favors
favors
favors
favors
the
the
the
the
the
products and shifts to the products to get to equilibrium.
reactants and shifts to the products to get to equilibrium.
reactants and shifts to the reactants to get to equilibrium.
products and is at equilibrium.
products and shifts to the reactants to get to equilibrium.
5) Calculate Kc for NH3(aq) + NH3(aq) ⇐⇒ NH2-(aq) + NH4+ (aq) @ 25°C
given Kb= 1.75x10-5 for NH3(aq), Ka= 1x10-35 for NH3(aq) @ 25°C
a) 5.71x1025
b) 1.75x10-26
c) 1.75x10-40
d) 1.75x10-5
c) 1.75x10-54
6)
Calculate the pH of a solution that is made by mixing 100 mL of 0.50M HCl with
100 mL of 0.25 M in KNO2. ( Ka=4.47x10-4M for HNO2)
a) 0.903
b) 7.00
c) 5.47
d) 10.6
2
e) 3.35
7)
Which of the following would not always be true?
a) Kw=[H3O+][OH-]
b) pH=-log[H3O+]
c) A solution is acidic when [H3O+]>[OH-]
d) pH+POH=14
e) A solution is basic when [H3O+]<[OH-]
8)
Calculate the concentration of cyanate (CNO-) in a solution that is made by mixing
50.0 mL if 0.25M in HCNO(aq) and 20 mL of .25 M in NaOH(aq). (Ka=2.19x10-4 for HCNO)
a) 0.107
b) 0.0714
c) 2.19x10-4
d) 3.18
e) 2.88x10-4
9) Calculate the Ka for ClCH2COOH(aq) given that a 0.100 M ClCH2COOH(aq) has a %
dissociation of 10.96.
a) 1.35x10-5
b) 1.35x10-3 c) 9.12x10-2 d) 1.35x10-7 e) 1.35x10-1
10) The conjugate acid of HCrO4-(aq) is:
a) HCrO411)
b) H3O+
c) CrO4-2
d) H2CrO4
e) H3CrO4+
The conjugate base of the weak acid C2H5NH3+ is
a) H3O+
b)
OH-
c)
NH4+ d)
C2H5NH- e) C2H5NH2
12)
Calculate the pH of a solution made by mixing equal volumes of 0.10M HCNO,
0.10M HCN and 0.050M HI. Given Ka=2.19x10-4 for HCNO, Ka=4.79x10-10 for HCN
a) 2.33
b) 1.78
c) 1.30
d) 2.56
3
e) need more info
13) The pH of a 3.0x10-8 M Ca(OH)2(aq) at 25°C is.
a) 7.00
b) 7.52
c) 7.22
d) 6.77
e) 7.12
14)
If you had a 0.10 M solution of each one of the following acids which one would be
the most acidic? H3PO4 pka1=2.12 ,pka2=7.21 pka3=12.68, HCOOH pka=3.75,
HClO pka=7.52, HClO2 pka=2.00
a) H3PO4
b) HCOOH
c) HClO
d) HClO2
e) need more information to tell.
15) Which of the following is the strongest acid and which has the strongest conjugate
base HClO2, HCl , HBrO2, HBr , HIO2, HI.
a)
b)
c)
d)
e)
HI strongest acid HIO2 has strongest conjugate base
HCl strongest acid HClO2 has strongest conjugate base
HI strongest acid HClO2 has strongest conjugate base
HBr strongest acid HBrO2 has strongest conjugate base
HClO2 strongest acid HI has strongest conjugate base
16) Which of the following indicator(s) would most likely be suitable for the titration NH3 with
HCl.
a)
c)
e)
Bromothymol blue (pkin=7.1)
Bromocresol green (pkin=4.9)
Methyl Orange (pkin=3.4)
b) Thylmol blue (pkin=1.7)
d) Alizarin yellow (pkin=11.2)
4
End of the multiple choice
No Partial Credit Problems
1) For each of the following pairs CIRCLE the one that is more acidic and state why in
two words or less.
Why
1)
HClO2
HClO3
2)
HClO2
HBrO2
3)
H2SO3
H3PO4
4) ClCH2COOH
FCH2COOH
5) ClCF2OH
ClCH2COOH
2) For each of the following tell if the pH is > 7, < 7, =7 or not enough information given
(NEI), by putting the correct (>7, <7, =7 or NEI) sign in the space.
a) 5.0 M C6H5CO2H
b) 0.01M NaC6H5CO2
c) 0.2M (CH3CH2)2NH
d) 0.2M (CH3CH2)2NH2I
e) Ba(HSO4)2
f) Fe(ClO3)2
g) Fe(C2H3O2)2
Work out section. Partial Credit Problems
3) Given that the pH of a 0.010 M Ba(CN)2(aq) solution is 10.8, calculate the Ka for
HCN.
Ka=___________________________
5
4) Sketch a representative titration curve for the titration of 100 mL of 1M HF(aq)
with 2M NaOH(aq). Ka=6.76x10-4M HF(aq). Label the axes. Indicate the pH at the
start, mid point and the equivalence point and excess
pH
Volume (
) added
6
5) a) Assume that you have available the following compounds H2SO3, NaHSO3, Na2SO3.
Describe how you would prepare a buffer with a pH=7.60.
Ka1(H2SO3)=1.70x10-2, Ka2(H2SO3)=6.46x10-8
b) If the concentration of SO32- was 0.500 M how many grams of NaHSO3 would
you need to add to get to a pH=7.60 if there was 500.0 mL of solution?
Grams NaHSO3 =
7
6) Nitrogen and hydrogen at initial partial pressure of 0.1545 atm and 0.0502 atm
respectively, were allowed to react at 227° C. At equilibrium the total pressure is
found to be .2034 atm. Calculate the value of Kp and Kc for the following reaction at
227°C, and the % yield for the rxn.
N2(g) + 3H2(g) ⇐⇒ 2NH3(g)
Kp=
Kc=
%yield=
7) Calculate the equilibrium partial pressures and the total pressure if 10.0 grams of
HI(g) is place in a 5.0 L container at 500 K and allowed to go to equilibrium given the
reaction and K below.
H2(g) + I2(g) ⇐⇒ 2 HI(g) K= 160.
P(HI)=
P(I2)=
P(H2)=
8
Pt=
8) Calculate the pH of 0.35 M H2SO4(aq). Given and pKa2=1.92 (For H2SO4 the first
proton is strong H2SO4 is the exception and NOT the RULE).
pH=__________________
9) Calculate the pH and equilibrium concentrations of the species in a 0.500 M
solution of H3PO4(aq). pka1=2.15, pka2=7.21, pka3=12.36 for H3PO4(aq)
pH=
[H3PO4]=
[H2PO4-]=
[HPO42-]=
[PO43-]=
[H3O+]=
[OH-]=
9
Warning: There may be mistakes by the end of the review all the mistakes should have
been corrected, if you have to go early do not spend hours on one problem that could
simply just be the wrong answer typed down here.
Page 1 #1 1) c 2) a 3) a 4) c 5) b 6) a 7) b 8) c 9) b 2) b 3) c
4) e 5) b 6) a 7) d 8) b 9) b 10) d 11) e
12) b 13) e 14) d 15) a 16) c
Fill-in )
1) 1) HClO3 more resonance, 2) HClO2 inductive effect, 3) H2SO3 inductive effect (same #
of terminal O’s so same # of resonance forms)
4) FCH2COOH inductive effect 5) ClCH2COOH more resonance
2) a) < b) > c) > d) < e) < f) < g)NEI
Long answer/workout
3) Ka=4.89x10-10
4) pH at the start is 1.59 at the mid pt pH=3.17(25ml's added)
stoich pt pH=8.50 (50ml's added), excess could use any amount over 50 ml's I used 60 and
got a pH of 13.10
past equivalence point is excess
equivalence point
pH
mid. point
25
50
Volume in mL's
5) a) Would need NaHSO3 and Na2SO3 b/c pka2=7.19 is closest to the needed pH. To
get the pH=7.60 the ratio of [SO32-]/[HSO3-]=2.57 so could mix 2.57 moles of Na2SO3
with 1 mole of NaHSO3 and add water to reach a final volume of 1 L (although any
volume would be fine and any amount with the same ratio [SO32-]/[HSO3-]=2.57 would
work) b) 10.1 g of NaHSO3
6) Kp=9.78x10-2, Kc=1.65x102 , 3.88% yield (small as expected b/c of small K)
7) P(HI)=2x=0.554 atm, P(I2)=P(H2)=x=4.38x10-2 atm, Pt=0.642 atm (same as the start
because Δn gases=0)
8) 0.442
9) pH=1.25 [H3PO4]=0.444, [H2PO4-]=5.61x10-2, [H3O+]=5.61x10-2,
[OH-]=1.78x10-13, [HPO4-2]=6.16x10-8,[PO4-3]=4.80x10-19
O
10