1. page 1
EXAM 2-FORM A
Chem201
Fall 2016
What is the pH of a 0.045 M solution ofK[CH3CH2C02]?
(a)~/'
See. (byrf (~~
rorll\-. ~ P- \ ~
7.1 r
~
XL.
~
l<. \
S'c*-~
0- \'0
D ,0"<1
L0\+-:) x. ~
f o\{
'C
2. Vs .t1'lG.'1
-
)~ 7..2f ~
-"b
~lu ~
S' ,'2."i.___
8_......---,
S- ,'1O("f
rt.-7'S -=-LK.1'7.r
What is the pH of a buffer solution containing 0.35 M CH)CH2C02H and 0.10 M K[CH 3CH2C02]? (a) 8.99 &4.34
(c) 9.65
l"J
(d) 5.01
(e) 5.43
l:::)' Lum,
\<c.- ~
...{ ")J
3. -,
J(O,04.i)('"7.'11: \0 ... ~~) "::
:; - L.:!Oj
C. c; ~ q o "IT .... l 0'6 ) ~
r'\--\-:'
(ab) 10.88
(e) 9.69
@.78
(ab) 8.57
\"J(g:~~) 14J,!1
-I-
-\oJ (t.:b~lO-') -::.
'1 . S'lf6
Lets say you want to make a pH 5.00 butler with the mixture in the previous question. If
the concentration ofCH3CH2 CO,-H is 0.20 M, what concentration ofK[CH3CH2C02 ] will
be needed to get the desired pH?
(a) 15M
(b)[email protected]
::. 'f \<.~
f' Yt
\.,
.-Ir
\
\
<-'j ( ;;.. ~~) -J2.?
C(''''1
( \.00 - 'i. ti (;) 10 (d)0.65M
(e)0.15M
\0
~H'" ~\c~
(U' t
(u{1
L:> -:
J
[ C1-11:. ( \-4 l. (d H.
..::
l.
-'
-:
EXAM 2 ~ FORM A
Chem201
Fa112016
4. page 2
Dilution of an acid-base buffer does not change the pH. Why?
(a) The pH ofa buffer solution depends on the pKa ofthe acid component, which changes
upon dilution so to keep the pH ofthe buffer the same.
(b) A buffer solution contains a weak acid and a weak base, and the pH's of solutions of
weak acids or weak bases don't change upon dilution.
(c) A buffer solution contains a conjugate acidlbase pair and the pH of all conjugate acid­
base pair containing solutions won't change upon dilution.
!{d.i1he pH of a solution of a buffer depends on the ratio of weak base to weak acid,
~~h won't change upon dilution.
5. You want to make a pH 9.24 boric acid, H3BO]. buffer. Mixing together equal volumes
of which two solutions will give you the desired result? MARK TWO ANSWERS on
your scantron.
~.lOMH'lB03
(b) 0.10 M NaOH
@.050 M NaOH
(d) 0.10 M HeI
(e) 0.050 M Hel
.~ l
\ h
\f\\ ')I.
\ (...-
oU RO-tl-
r
J. b 0 ~
L
~e.
CCL\-(.
\:-\
"-.) 0\ 0
-1-
I,A
~d
-::J
"'-Oi.O~O ~
o
~
be......­
,r­
l!\J e- ~ J-. (1 0 1
o lOlO ~ort.
0 tl 0 "",ve.. -0 .om~
:'>.>
() I-OS-O ~
V")cV'\
\-\~ ~0""S 0
6. page 3
EXAM 2 - FORM A
Chem201
Fall 2016
Fill in the blanks.
/2N1 *\.0 fY
is a measure of disorder. A decrease in this is
W'F"V'
The answers in order are
(a) Entropy, favorable
(d) Enthalpy, w1favorable
7. /&)\ntropy, unfavorable
~
(c) Enthalpy, favorable
The three laws of thennodynamics are as follows:
(1) The total E.,.. H::.'(\O"'t of the universe is constant.
(2) The~kl"1_~(J(J'-t of the universe increases for a spontaneous process.
(3) The _f£a•.r 1 ~ot'"1 of a perfectly ordered system at 0 K is zero.
Filling in the blanks, the answers in order are
!(a~ergy, entropy, entropy
(b) entropy, energy, entropy (d) entropy, energy, energy ~nergy, entropy, energy
8. What is ~S 0 for the following reaction?
SO of different species in JIK: HN03(g) = 266.9
HN03(aq) = 146 NO(g) 210.8
N02(g) 240.1 ~164
JIK
&288 JIK
(c) 49 JIK
(b) -49 Jik
(ab) ~ 164 JIK
6~ 0 =-
2. '.S 'fAJc-t
.""";L S(). (-t') +
0
--- z. (\ '16) -t
~..5()vr'AA
-
l-'\lU,:)1
l' 0
(d) 288 JIk
'S,<3
'~~)
•.1' -
~ - In ,} 'J"/I~
'"
-
(3. S"N"
L
('1 ('--<0. \)
1- '2 i'f ;;/~J
()
-+
'5
~"\I,."("'~
-r7 O. 0 )
) )
Chem201
Fall 2016
page 4
EXAM 2 - FORM A 9.
~ (\-5+ (<:lI{)) \-
10. PO .... '3 '- (~G)
'\J
p,,rr t-JOl"
What happens if an aqueous solution of silver nitrate is added to a saturated aqueous
solution of silver phosphate?
(' C).,u.}f~ [As 1'":) l' -' o...Ujd v\'V'-'\ V).;:.~ S h\.'-b
.
U FI to reVV\ GW't e'lC<:.1'"~.J [~-.!~::J j M.<.l ve <. -:>\ ~
~Nothing happens.
~o s ol ......... dv~~ cA.e.t-"'\"'<.ii"',!"
The silver phosphate solubility equilibrium shitts to the LEFT and the solubility of
sIlver phosphate DECREASES.
(c) The silver phosphate solubility equilibrium shifts to the RIGHT and the solubility of
silver phosphate DECREASES.
(d) The silver phosphate solubility equilibrium shifts to the LEFT and the solubility of
silver phosphate INCREASES.
(e) The silver phosphate solubility equilibrium shifts to the RIGHT and the solubility of
silver phosphate INCREASES.
C5:l
11. What will be the solubility of silver phosphate in 0.10 M silver nitrate? K,p of silver
phosphate = 8.89xl017
~3.2XlO-6M
cY8.9xlO-14 M (b) 8.9xlO- 16 M EXAM2
Chem201
Fall 2016
12. FORMA pageS
Ph ~'l-
Knowing only that lead(I1) iodide is a poorly soluble ionic compound with Ksp
9.8 x lO- 9, what would you calculate to be the maximum concentration of iodide that could
be present in a 0.025 M lead(Il) nitrate solution without causing lead(II) iodide to
precipitate?
/(a))lst under 6.3xlO- 4 M
~~ under 3.9xl0- 7 M
(e) just under 8.7xl03 M
\<)~ -:
(b) just under 1.3 x 10- 8 M (d) just under 1.1 xl0- 4 M f\;,:J-t.. ()) <==-- P\ol.A-(~)
[(J'<J1,t
2'2~(~,
1­
J [I-,} 1- ~ [;J.-..J ~ ( \(' .Jl. \"t.=. fetE
l[r.,'<))
=. 13. ,'(L
)£(01)
,,10 ,Cfls}
~,2 & I~O~-""f_---,
1"'
"lG'lo-~'h \
As it turns out the assumptions going into the calculation in the precedmg question are
faulty because lead(II) forms a complex ion with 4 iodides. What is the Kr expression for
formation of this complex ion?
(a) ~
(b) [Pb2+][rJ
[pbI/-]
1
[Pb +]W]4
2
(c) , bI 2 ­
[Pb 2+]Wt
(e) [Pb2+][r]4 (d) [Pb2+][rt
[PbI/-] ~,
14. it\
Although we discusscd Lc Chatelier's Principle in the contcxt of solubility equilibria,
applies to all types of equilibria. Consider the reaction to form the complex ion discussed
in the previous question. Assume you have a solution ofthe complex ion at equilibrium.
How will the addition of iodide affect the equilibrium position?
//~(a) It will cause the reaction to shift LEFT and the concentration of complex ion will
INCREASE.
(b) It win cause the reaction to shift LEFT and the concentration of complex ion will
DECREASE. IfC})It will cause the reaction to shift RIGHT and the concentration of complex ion will \..H("CREASE. (d) It will cause the reaction to shift RIGHT and the concentration of complex ion will
DECREASE.
C~~~
~ \-\ "k b
["I -l
~ \ L H""T
~ ['P~ :L"t~j-t-
\-0
.1-1>
'f(! """" 0 V't!.
~;X:( e J 1
::I..-
CAV"\Ot
.BJ v- ; 'D
L.
Chem201
Fall 2016
15. page 6
EXAM 2 - FORM A
Continuing the discussion ofthe complex ion from the previous page, if excess iodide is
added to a saturated aqueous solution oflead(ll) iodide, the following reaction will take
place:
In the above case, the molar solubility ofPbIis) would egual the concentration of
PbI/-(ag) at equilibrium. Assume 0.10 M NaI(aq) is added to solid PbI2 Which of the
following equations could be used to solve for the molar solubility ofPbI2 under these
conditions? x the molar solubility of PbIz, K f = formation constant of PbI/- (aq), ~ =
solubility constant for PbI2(s).
(HINT: Detennine how the equilibrium constant for the above reaction is related to Kr
and K sp, then set up a reaction table for the above reaction.)
(a) ~ / Kr= x / (0.10 ~2x)
(b)
~·Kr=x/(O.lO
-2x)
(c) Ksp / K f = x I (0.10 _2X)2
@<-sp.~. = x / (0.10 - 2xf
(e) ~p / Kr= x / (0.10 _X)2
(ab) Ksp'Kr= x / (0.10 _X)2
EXAM2-FORMA
Chem201
Fall 2016
page 7
Consider the following acid-base titration for the next 5 questions.
30.0 mL of 0.054 M LiOIl(aq) (in flask) is titrated with 0.090 M IIN03 (aq) (in buret)
16. What is the overall reaction occurring in the titrution?
17. What volume ofHN03(aq) needs to be added to reach the equivalence point of this
titration?
(c) 20. mL
(b) 30. mL
~e e roy
18. V'
\'::,
(e) 0.055 mL
(d) 15. mL
'+t- 1-­
What is in the flask after 20.0 mL of the HNOlaq) solution have been added?
...
Il..-
f'I~,eIt..
@~,J
~,
(a) LiOH(aq) and all products
~~<?H.(aq) and HN03 (aq)
o.~(aq) and all products
(c) all products (e) LiOH(aq), HN03(aq) and all products
®fiN
19. How is the acid-base nature oftbe solution in the flask best characterized after 20.0 mL of
the HN03(aq) solution have been added?
~ongbase
(a) buffer
(e) weak acid
20. (c) weak base
(d) neutral
®trongacid
What is the pH of the solution in the flask after 20.0 mL of the HNOlaq) solution have
been added?
(a) 2.22
~.44
[\..\ ! l)....
1
=-
(c) 11.55
(d) 10.46
t \--\0 u).J '(J"''' ~
-
("'~v
- c...
H ""':l
hT
V ....~
V
~ 1tb'" I D .... lV\
V'~ ~ . . l"S (~t6'"
1
f" "
(e) 11.77
M ()\ H t-.Jo~.
n,-1V
(ab) 3.54
Co II.
~ J ~c:..l \-
<"''t:t'''' ~
"'0""'''" ':- ( C!.. ,0"1 i) .... (1.. ~ ....
(lO. M.-y
\D-')
J
y
page 1
EXAM 2-FORMB
Chem201
Fal12016
Consider the following acid-base titration for the next 5 questions.
{k(l.,yu6 ~,f\"~
S'1 f\.() ~L At-hO
30.0 mL of 0.054 M LiOH(aq) (in flask) is titrated with 0.090 M HN03(aq) (in burct)
1. What is the overall reaction occurring in the titration?
2. What volume ofHN03(aq) needs to be added to reach the equivalence point ofthis
titration?
(b) IS. mL
(a) 20. mL
(c) 0.055 mL (d) 30. mL
e €tfA\vc;.AeV\.c:.c.
V
fr.&\J..1;;.
g~~v '8
c""
C!'fVPr,CO<1L.... -l
-::
~
V~ Ef"c:rt\.cz
fL.......-
Cr?
Vp..t(.J"j-
~ G..~~-.-vrJ---"1
~a U \ \J
­
'(' \' .
(b) LiOH(aq) and HNOlaq) (d) HNOlaq) and all products How is the acid-base nature of the solution in the flask best characterized after 15.0 mL of
the HN03(aq) solution have been added?
(a) buffer
(e) weak acid
5. Vk~\ b.::.<-,< or~
CO ~O~-'-{ t'\A.) (.'10 .0 ~L.)
. (a) LiOH(aq) and all products :
all products
(e) LiOH(aq), HN03(aq) and all products
4. "
""
,
~
(0 l DC1.0 If'V'\ )
What is in the flask after 15.0 mL ofthe HN03(aq) solution have been added?
t
3. \;"'\.01 ~,. Od-4...A..
@18.mL
~ong base
(c) weak base
~)- strong acid
(d) neutral
What is the pH ofthe solution in the flask after 15.0 mL ofthe HNOJ(aq) solution have
been added?
(c) 11.55
(ab~
@11.77
-;',. ~~ L, 0\-\ W!~""''''-J
~
CM...s
~ L~o
+dtV
\-t
-
"""<>\
\-UJU,\
lot-v
-; (O.C>\--t ~ ) (~o~u ~L)
4 '".(.) "",L.
~
c~c.t~t:,( _~ CL\Ijv<V~l.J. . .- t -C\.tIl>6"lo V\-4N«!r
.+ok"V
(0. c.)~O Vh1( \ ).0",,'-) ::. D )':
- loS (b.O ~ \0 -1) -::
t\
"Z.. 1.2..1...
10 3 M
~\-\ ~ \~tG} 'ill 1- 2. .1.t.L
=D'L1I1
EXAM 2 - FORM B
Chem201
Fall 2016
6.
page 2
What is the Ksp expression for silver phosphate?
a[Ag+][P04-]
e Ag+P[pO/-]
@
I\~ ~ eO", (\) F
. ~ f'-,
"y (tM(.)
.~
"!:,­
r0~
(at)
k \~.~ [~\j~ 1"., [\?o . . .t 'j
7. 8. k.b!. po '"
What happens if an aqueous solution of sodium phosphate is added to a saturated aque s
solution of silver phosphate? CD..u...;..
PO ~~ -;3 ....,.. I c:L...s S.... l "'''''.......... V").;.<A SlA... ~ LfZF/
+0 ve""".,;.)~
e·,>"c.~...»
'Po'-t'!r.~.; a."'Cl Su\ ...."'.h"t") ~c....'V('c..se.\
(a) The silver phosphate solubility equilibrium shifts to the LEFT and the solubility of silver phosphate INCREASES. (b) The silver phosphate solubility equilibrium shifts to the RIGHT and the solubility of
Jll~er phosphate INCREASES.
(£9JThe silver phosphate solubility equilibrium shifts to the LEFT and the solubility of silver phosphate DECREASES. (d) The silver phosphate solubility equilibrium shifts to the RIGHT and the solubility of
silver phosphate DECREASES.
(e) Nothing happens.
r
What will be the solubility of silver phosphate in 0.10 M sodium phosphate?
phosphate = 8.89xlO- 17
~.2xIO-6 M
l-(e) 8.9xlO-
16
(b) 1.5xlO- 7 M
~p
of silver
(d) 4.3xlO- s M
(c) 8.9xlO- 14 M
M
::> Ps " ('c.C)
o
-\: ~~
)3(1" \O'M
-
1
9. r\.., ·11..
Knowing only that lead(II) iodide is a poorly soluble ionic compound with Ksp =
9.8x I 0- 9 , what would you calculate to be the maximum concentration of iodide that could
be present in a 0.75 M lead(II) nitrate solution without causing lead(Il) iodide to
precipitate?
~Lust under 1.6x I 0 M
(a) just under 1.3xlO- 8 M
(c) just under 6.3 x 10- 4 M
(e) just under 3.9xlO~7 M
10. page 3
EXAM2-FORMB Chem201
Fall 2016
3
~ustunder 1.lxlO- 4 M
As it turns out the assumptions going into the calculation in the preceding question are
faulty because lead(II) forms a complex ion with 4 iodides. What is the Kr expression for
formation of this complex ion?
I
[pb2+][r]4 (a) (d) [Pb2+][r]
[PbIl~]
Although we discussed Le Chatelier's Principle in the context of solubility equilibria, it
applies to all types of eqUilibria. Consider the reaction to form the complex ion discussed
in the previous question. Assume you have a solution ofthe complex ion at equilibrium.
How will the addition of iodide ,affect the equilibrium position?
~
~t will cause the reaction to shift RIGHT and the concentration of complex ion will
LfirCREASE.
(b) It will cause the reaction to shift RIGHT and the concentration of complex ion will
DECREASE.
(c) It will cause the reaction to shift LEFT and the concentration of complex ion will
INCREASE.
(d) It will cause the reaction to shift LEFT and the concentration of complex ion will
/fa
1I
\
\
\\
\
ECREASE.
~
".'
oo~
-r -~
Jev-.oV'(...
J
C:J..-1
&)< c.(lU
\v-o~~~~~"'j;~ ..
1"
•
,;
)
~'i\ ~- tU.:;!Uv'e
ht<.
C
"-}(.V'\
e
.{
\A\~~..(
Rl6j.-(.,
{bv ~I.'-"'-'
~
~
page 4
EXAM 2 - FORM B
Chem201
Fall 2016
12. Continuing the discussion of the complex ion from the previous page, if excess iodide is
added to a saturated aqueous solution of lead(II) iodide, the following reaction will take
place:
In the above case, the molar solubility ofPbI2(s) would equal the concentration of
Pbl q2- (ag) at equilibrium. Assume 0.10 M NaI(aq) is added to solid PbI2 Which of the
following equations could be used to solve for the molar solubility ofPbI2 under thesc
conditions? x ~ the molar solubility ofPbI2 , Kr= formation constant ofPbIl(aq), ~p =
solubility constant for PbIls).
(HINT: Determine how the equilibrium constant for the above reaction is related to ~
and ~P' then set up a reaction table for the above reaction.)
(b) ~p I Kr= x I (0.10 -xY~
(c) ~'Kr= x 1(0.10 -2x)
(d)
~p
I Kr
x I (0.10 -2x)
V
i
-}<
\<,
-
= [" \.;, :J '1t.-J
t:1~:)~
~
What is t~e pH of a 3.2 M ~lution ofK[CH3CH2C02]?
13. ne,-,,~ c;l."'t01o ........,J.. .~
(b~
(a) 8.78
S I "" l e
\-<. V\. ..::.
CU,... H~
~
\("
(,.w
\
\-..
,. \+ ~
bGo to.('
J
(d) 11.81
f..<...)-(
~I
C\.~!> C'"""2 (() 'L-
(a) 0.15 M @026M
r
~
C: ' - ~hC.t"...
M.....
\(' l--\ < 7
\.-t ~
r-
~
(c) ~
1.~ ')t' c.,...J( c:::.J.."'­
\.::, G. )(:
\:;,
Cc:..IV',) ~G4.""
0. (.U~ "'<::4".'C/
e C"
c::;u....)
~
:;
(c) 0.65 M
(e) ~
\., <;.."')C'-
~.69
( Cot ' "
~ \.. " ~ \ "'" c:-l:-~
'D-~
"),731< (O-to
(d) 1.5 M
\O,e::" '" 'OJ.( ~4~"\ ~
( t.( .00 - y.~ ... TS_)
~
'0 dt ~ C:.J }
;;:.,0.1'3> _- . (
(e) 0.26 M
V",,"--p"~-= ~~\.4I.
t0 Le\A1 cu.· L. fO il,..­
LCHl
c.(~1. (\Ai.. ('o~ 1::-
e.
u- CV"J C \l- LCO"'\-\-,1
Lets say you want to make a pH 4.00 buffer with the mixture in the previolls question, If
the concentration ofCH3CH2C02H is 0.20 M, what concentration ofK[CH3CH2 CO:J will
be needed to get the desired pH?
15. I'J
page 5
EXAM2-FORMB
Chem201
Fall 2016
1
­
C ~1..(c:..)L\+
A. c.t.&..
1
0 \ \, [<:~:) C\..-('l..(O 1...\"\.;;;.1_--_
-.: 0 \ \3. (0, 'l <.> """')
~
to.
-
0
1.{'V"\ \
16. page 6
EXAM2-FORMB
Chem201
Fall 2016
Dilution of an acid~base buffer does not change the pH. Why?
(a) A buffer solution contains a conjugate acidlbase pair and the pH of any conjugate
acid-base pair containing solution won't change upon dilution.
~e pH ofa solution ofa buffer depends on the ratio ofweak base to weak acid,
~h won't change upon dilution.
(c) A buffer solution contains a weak acid and a weak base, and the pH's of solutions of
weak acids or weak bases don't change upon dilution.
(d) The pH ofa buffer solution depends on the pKa ofthe acid component, which
changes upon dilution so to keep the pH of the buffer the same.
17. You want to make a pH 9.24 boric acid, H3B03, buffer. Mixing together equal volumes
of which two solutions will give you the desired result? MARK TWO ANSWERS on
your scantron.
~.(a.10 MNaH B0
2
~.050MHCl
'L~ '1J
f\<'&-\ ~
3
(b) 0.10 MNaOH
(c) 0.050 M NaOH
\4 ~ ~ 0'3 -:: S \ '8 .,... \0-\0
-l c:eJ (S-, 1''' \0·- .....·) :::­
(d) 0.10 M Hel
18. page 7
EXAM2-FORMB
Chem201
Fal12016
Fill in the blanks.t), ~0-\
6.:Vr..v-~
f<l=- is a measure of heat energy.
A decrease in this is
The answers in order are
(a) Entropy, favorable
(d) Enthalpy, unfavorable
19. (b) Entropy, unfavorable
@nthalPY, favorable
The three laws of thermodynamics are as follows:
(1) The total tlJEM of the universe is constant.
"
.'.
(2) The(;'!JIV'--O'P'1 of the universe increases for a$1?,pntaneous
process.
(3) Thel:a,..l"lf\.O'(?'-I of a perfectly ordered system at 0 K is zero. Filling in the blanks, the answers in order are &ergy,
(a) entropy, energy, energy
(c) entropy, energy, entropy
20. entropy, energy
) e ergy, entropy, entropy
What is AS 0 for the following reaction?
SO of different species in J/K: HNOig) = 266.9
HNOlaq) = 146 NO(g) 210.8
NOlg) = 240.1 (b) -288 JIK
(ab) -49 JIk
(a) 164 JIK
(e) 49 JIK
.6 ~ ~ ~
t
~"f ~ -
2
-: 2.s. "1<"'''' C.o;) -t
- L (2btLq) ~
- - I b'i · f"
(d) 288 JIk
j
0
~c;."~
.
S':...",) - ( '3
2\0.8' -
J""/I<.
'
t
oS:"'..,,) ... ~~....<> <.,,)
("1 (2.'1
0 .. \')
I 6'( T I ~
+- \f? .. r)
ADDITIONAL INFORMATION and EQUATIONS
Some K.. Values at 25°C
Kal
CH~CO,H
1.8
xl0~5
CH3CH2C02H
1.3
xlO~5
C"II.,CH,CO,H
4.9
xlO~5
H 3B03
5.8 xlO- lo
H,C,04
5.6 xlO- 2
HC02 H
1.8 X 10- 4
H~P04
7.5 xlO- 3
6.2 X 10- 8
H2 SO4
very large
1.2 X 10-2
HF
3.5 xlW 4
Some K.. values at 25°C
NH3
1.8 X 10- 5
HOCH,CH,NH1
3.2 X 10- 5
(CH3)2CHNH2
4.7 xlO- 4
C SH 5N
1.7
C(\HSNH2
4.0 xlO- lO
xlO~9
K..K., = ~ = 1.008x 10- 14
[base]
pH = pK + log..=..---=:.
[acid]
a
AHO ""
~
L..
AHO
f
products
~
L..
6.Hf 0 reactants
5.4
xl0~5
4.8 xlO- 13