Jeopardy
Salt
1
solutions
substance
the
Pick
.
:
whose
aqeous
solution
is
acidic
a) NANO }
b) NHyN03
c)
Naf
←
HF weak acid
d) KCIO 4
2
Pick the
.
a) KNOZ
substance
←
whose
HNOZ is
weak
aqeous solution
is
basic
acid
b☒ NH4Cl
c)
Back
☒
KCI
Buffers
At
1.
point
which
on this curve is
a
buffered
solution
obtained?
¥
2
.
A is
endpoint
:
C. is equivalence
Which
of
highest
a)
F)
f)
d)
the
buffer
I
M
HNOZ
v16
M
M
.
-01
•
016M
following
buffered
solutions has
capacity ?
M
NANOZ
0.04
M
Nanoz
+
0.01
M
Nana
+
0.004 M
t
0.1
HNOZ
t
HNOZ
HNOZ
NANOZ
*
higher
higher
the
buffer
the
concentration
capacity
.
3.
A
microbiologist
to
culture
pH
fermentation
E.
solutions
of
to
135 mL
make
Kai
which
on
buffers the medium
She
.
acid
-
producing
volumes of eauimolar aqueous
KzHP04 and kHz Poy
7.2×10-3
=
medium
a
minimize the effect of
What
.
preparing
coli bacteria
7.00 to
at
is
of
Kaz
the
=
pH =p ka
pH
Kaz
=
4.2×10-13
[+11-042]
log
t
buffer?
7.00
6.3×10-8
she combine
must
Pkaz
ftp.i]
=
7.2
log
EEE.ae?-.z-log ¥
7.0=7-2
82.82
52.2
mL of
-
10--2=-631
HzP04
HPOLF
mL of
+
+82.82=135
×
634+-1--135
•
4=82.82
1. 634=135
✗
or
49.18
=
4.
pH -3.74
pka =3 -7459
-
3-74=3.745
+
log '¥=
HCOO
1-11001-1 NaOH
[1-100]=11 NaOH)(
→
+
-
+
'
-
log
.
vi. 7L
(based
←
Cacia,
98855
Natt H2O
)
Naoh
.⇐D -4111
.
)]
-
=
98855
=
✗=
.
4675
C. 7-
7¥ -2¥
-
>
.
'
Notum
•
zx
-
Final concentration of HCOOH
(1) ( ¥) C) 0¥ )
"
log
•
[1-11001-1]=4
base = acid
Close
=
-
.
.
NaOH
46 > 5)
34M
HCOOH
HCOOH
L
=
.
=
232 LNa
Titrations
1.
The
pH
the
graph
below
a) Weak
base
for
curve
.
a
The
certain
initial
titration is
sample
shown
in
contains
b) Strong acid
c) weak acid
d) strong base
*2
B.
B)
.
W/
pKa
What
1-13 As 04
acid
Arsenic
,
=
pKa 2=7.0
2.2
is
the
pH
is
triprotic
a
pka 3=11.5
valve
of
a
l
m
solution
HzAsO4
+
1-120
HzAsO4
_- HAS 04
11.52+7.02
r
of
NazHAS 04?
-
H3As04t Hz0
-
acid
"
=
t
t
9.25
1-130
1-130
OHI
HA
→
42A
-
-
+
→ AZ
OH
-
+
HA
-
+
+120
Hzo
3
a=
•
•
;¥÷
ran
b.
040.2=-008
04
.
.
I
004
=
•
moles
moles
ptkpka
MA
NaOH
f. 008 )
-
+
{÷
log
C.0043=-004
pH=pRa=
=
•
.
04
-
•
2=1.008 moles Ma
(
•
016
•
008)=
•
left
2.85
008
ptl-pkaz-logc.jo#ka=ygxl0¥4T
^
•
16
.
.
I
=
•
016
•008_
"
4
=
.
04
moles NaOH
k•=
[OH-1=489
=
←
•
000014142
14-4.89=9.15
NaOH
Acid
-
Base
Indicators
1
*
at
a
will see
pH you
change
what
color
2.
*B
be
because
close
point
to
it
need to
equivalence
Solubility
I
a)
10-20
Sp
Ksp
-
-
=
b) 10-10
82=1×10-10
C) 10-5
d) 10--2-5
2
5=1×10-5
.
Ksp
Ksp
3
[AgF[ 01-3=82
>
[ Fe"][ of
=
=
[ Agt ]
'
[ crzoi ]
KCI
5=3.7
4.
100mL
ksp=Ck+][ CT]
a)
b)
i
[ Kt
]=3?zM•#
[a)
Entropy
Qsp
=
' 85M
'
=
=
5
(3. 7)
'
=
Ksp 13.69
__
kspcnopercipitate forms)
Qsp€1.85] [1.85+6]--14.52
.
=
1.85m
3×2=6
+I-=4.85-
[1.85-4.85]=8.9725
Qsp
bz )
<
Qsp > Ksp precipitate forms
13.69=[1.85-5] [7.85-11]
✗
=
-0866
MOI
l
0086*21%5,5%01
.
c
1.29g
llkeivin
•all
2
f/
Krslnw
Boltzman
¥
microstates
constant
5- entropy
3
.
DS
-_9r¥
4.
* count
gas
molecules
positive negative positive
,
,