Page 40 - E-MODUL LARUTAN ELEKTROLIT DAN NONELEKTROLIT(2)

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E - M o dul L a r ut a n E l ekt r o l i t d a n N o n E l e kt r o l i
a
n
E-Modul Larutan Elektrolit dan Non Elektrolittt

d
E
l

L
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l
o
dul
i
E
M
o
n
r
ut
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N
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E
e
kt
n
Dis
y
co
ver
ed
B
Berbasis Guided Discovery Learningningning
B erba s is Guid ed Dis co ver y L ea r
erba
Guid
is
s
ea
r
L
ap
d
d
ajat
ag
u
n
io
is
an
at
ilai
n
eb
er
lk
d
i
s
p
is
Dari nilai derajat ionisasi diatas dapat disimpulkan sebagaiaiai
m
i
d
Dar i n ilai d er ajat io n is as i d iatas d ap at d is i m p u lk an s eb ag
as
i
Dar
iatas
ik
er
u
b
b er ik u
berikut:t:t:
n
aja
ak
a
s
in
in
b
k
an
es
m
ter
y
an
d
g
io
t
er
r
y
ai
ar
e
b
zat
ak
u
a.
as
a. Semakin banyak zat yang terurai semakin besar derajat ionisasi,i,i,
m
is
Se
a. Se m ak in b an y ak zat y an g ter u r ai s e m a k in b es ar d er aja t io n is as
in
s
elek
artinya sifat elektrolit semakin kuat.at.at.
at
if
a
m
e
s
lit
o
ar
tin
k
tr
y
ar tin y a s if at elek tr o lit s e m ak in k u
u
ak
Se
s
k
u
in
k
m
an
ai
io
s
ecil
y
r
ed
er
d
t
a
ter
m
z
b
ik
n
g
aja
it
e
in
ak
b . . Se m ak in s ed ik it z at y an g ter u r ai s e m a k in k ecil d er aja t io n is as
b. Semakin sedikit zat yang terurai semakin kecil derajat ionisasi,i,i,
is
as
at
tr
lit
tin
o
elek
le
m
at
s
a
if
e
s
ar
m
artinya sifat elektrolit semakin lemahahah
ak
ar tin y a s if at elek tr o lit s e m ak in le m
in
y
Co
l
o
nt
h
Contoh soal :::
Co nt o h s o a l
o
s
a
p
a
l
Amatilah penguraian mol zat terlarut dalam larutanya padaaa
ma
A ma tila h p en g u r a i a n mo l z a t ter la r u t t d a la m la r u ta n ya p a d
r
z
r
A
n
i
ter
a
u
mo
a
m
la
t
a
p
d
en
g
u
ya
h
n
d
r
a
la
tila
la
ta
u
er
in
a
g
b
t
r
mb
a
gambar berikut iniii
g a mb a r b er iku t in
iku
O
OO
C
C
OO
C
C
C
CH COONaNaNa C H 333 C OO CO(NH )))
O
NH
(
H
(
H
NH
C
C
H
CH COOHHH
OO
C
2 222
2
2
333
r
L
Larutannn
u
a
a
Larutannn
r
L
u
t
t
L a r u t a L a r u t a
a
a
O
O
O
C
O
C
O
H
N
N
(
O
H
(
CH
C
CH COONaNaNa
C
H
C
O
C
O
C
H
CH 333 C O O CH COOHHH CO(NH )))
2 222
333
2
2
u
u
a
=
a
o
u
a
a
l
=
m
m
l
-
u
-
l
l
m
o
a
l
M
m
M o l l m u l l a - m u l l a = Mol mula-mula = 444 M o l l m u l l a - m u l l a =
Mol mula-mula = 444
l
M
o
a
o
a
u
m
M
u
M
m
u
Mol mula-mula = 444
a
=
m
-
m
-
=
u
o
o
o
m
molll
molll
m
m o molll m o
m
o
m
2
o
r
o
o
m
m
=
n
o
o
o
m
2
m
o
=
4
n
r
Mol terion = 0 molll
M
o
o
l
o
o
l
e
r
r
e
e
t
i
0
t
i
M
=
=
Mol terion = 4 molll
M
M
n
n
M o l l t t e r i i o n = 4 m o Mol terion = 2 molll M o l l t t e r i i o n = 0 m o
e
o
–
n
l
Dar i d ata d iatas d i p er o leh m o l l m u l l a – m u l l a , , m o l t e r i o n , , d an m o l y a n
o
r
Dari data diatas diperoleh mol mula–mula, mol terion, dan mol yanggg
an
m
e
d
m
o
i
Dar
t
o
u
d
leh
p
i
o
er
m
m
o
ata
m
a
u
a
y
a
iatas
n
l
d
d
i
t
o
i
e
id
n
n
e
t
r
r
b
tu
t id a k t t e r i i o n y ai tu b e r ik u t i i n i
u
y
tidak terion yaitu berikut ini :::
ik
ai
a
k
u
u
L a r u t a Jumlah Molll
Larutannn
L
J
M
J
h
h
m
m
u
r
a
t
a
a
a
l
l
o
M
o
–
o
r
k
d
r
a
d
o
a
o
T
a
k
Terionnn
r
m
u
e
t
u
t
e
i
i
e
T
T
i
M
M u l l a – m u l l T e r i i o Tidak terionnn
i
Mula–mulaaa
4 molll
OO
C
OO
C
o
CH COONaNaNa 4 molll 4 m o ---
m
o
C
C
m
m
4
4

H
4
o
H
333
o
2
C
m
m
C H 333 C O O 4 molll 2 m o 2 molll
4
4
2
CH COOHHH
C
o
m
O

o
O
2
H

o

o

m
2 molll
m
CO(NH ))) 4 molll --- 4 molll
C
C

4
4
(
NH
NH
O
O
(

m

o
o
m
o
m
m
4
4
o
2 222
2
2
3
3
3444
KELAS X SMA/MAAA
M
KE L A S X S M A / / M
KE
S
S
X
A
A
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