Download - 005 Larutan Koloid(1)
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LARUTAN DAN KOLOID
Macam-Macam Campuran
LarutanPerbandingan larutan dispersi Koloid & Suspensi
Dispersi KoloidSuspensi1
LARUTANDISPERSI KOLOIDSUSPENSISemua bentuk partikel dari atom, ion atau molekul (0,1 1 nm)Parikel paling sedikit satu komponen atom, ion atau molekul kecil (1 1000 nm)Partikel paling sedikit satu komponen yang dapat dilihat di bawah mikroskopStabil terhadap gravitasiKurang StabilTidak stabilHomogenPerbatasan homogenTidak HomogenTembus CahayaBuramTidak tembusTidak ada efek TyndallEfek TyndallTidak transparanTidak ada gerak BrownGerak BrownPartikel terpisahTidak dapat dipisahkan dengan penyaringanTidak dapat dipisahkan dengan penyaringanDapat dipisahkan dengan penyaringan - SUSPENSI
Dapat dipisahkan dengan penyaringan atau dengan sentrifugasi
DISPERSI KOLOID2
JENISFASA TERDISPERSIMEDIA PENDISPERSICONTOHBuasGasCairBusa sabunBusa PadatGasPadatBatu apungAerosol CairCairGasKabut, halimun, awanEmulsi CairCairKrim, susu, saosEmulsi PadatCairPadatMentega, kejuAsapPadatGasDebu, partikulat dalam asapSolPadatCairPati dalam air, jeli, catSol PadatpadatpadatAloy, mutiara -
Muatan Elektron partikel Koloid
FIGURE Colloidalparticlesoftenbearelectricalchargesthatstabilize
the dispersion. On the left is a particle whose extremely large molecules
carry negatively charged groups. On the right the colloidalparticles has
attracted chloride ions to itself. In either case, these colloidal particles repel
each other and cannot join together.
Colloidal particle with
organic ionic groups
Colloidal particle with
adsorbed chloride ions
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Efek Tyndall
The Tyndall effect,
A pencil line thin red laser beam
passes through the liquid in three
Rest tubes. The first contain a
colloidal dispersion of starch, the
second a solution of sodium
chrornate, and the third a colloidal
dispersion of Fe2O3, All three
appear transparent, and in the ab
sence of the ryndall effect we
might think they are all solutions,
However, the Tyndilll effect reveals
that the fist and third are coilds,
Not true solutions
LARUTAN
CAMPURAN HOMOGEN
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Larutan Gas
Gas dalam gasCair dalam gasPadat dalam gasUdara
Sistem koloid
Sistem koloid
Larutan Cairan
Gas dalam cairCair dalam cairPadat dalam cairCoca-cola
Cuka, Gasolin
Gula dalam cair
Larutan Padat
Gas dalampadatCair dalam padat Padat dalam padatAloy hidrogen dim paladium
Benzen dalam karet
Karbon dalam besi
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Mengapa Terbentuk Larutan
Larutan Dalam Cairan-
+
Ion ion force of attraction asin sodium chploride
+ -
+ -
Polar molecule
Dipole-dipole force of attraction as in sugar of water
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- Larutan cair Dalam Cair
Figure Ethyl alcohol molecules, C2H5 O H, experience hydrogen bonding ( ) between themselves in pure alcohol. Hydrogen bonds also occur in pure water. When these two liquids for a solution, hydrogen bonds can easily form between molecules of water and those of alcohol thus, attractive forces between molecules in the pure liquids are replaced by similar forces in the solution, and the solution easily forms
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- Larutan padat dalam cair
Portion of surface and edge of NaCl cristal in contact with water
FIGURE
Hydration of ions
Hydration involves a complex
redirection of force of attrac-
tion and repulsion. Before this
solution forms, water mole-
cules are attracted only to
each other; and Na+ and CI-
ions have only each other in
crystal to be attracted to.
In the solution, the ions
have water molecules to
takes the places
of their oppositely
charged counterparts; and
water molecules
find ion more attactive than
even other water molecules.
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Panas Larutan
Terjadi pertukaran energi sistem dan sekelilingnya apabila 1 mol zat terlarut dilarutkan dalam bentuk ( pada tekanan konstan) untuk membuat larutan encer.
H : Fungsi keadaan yang tidak bergantung pada jalannya perubahan
FIGURE Enthalpy diagram for a solid dissolving in liquid. In the real word, the solution is formed directly as indicated by the red arrow. We can analyze the energy change by imagining the two separate steps, because entrhalpy changes are fuctions of state and are independent of path. The energy change along the direct path is the algebraic sum of step 1 and step 2.
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FIGURE The Formation of aqueous potassium iodide
Step 1:
Step 2:
Kl(s) K+(g) + l-(g)
K+(g) + l-(g) K+(g) + l-(g)
H = +632 kJ
H = -619 kJ
Net:
Kl(s) K+(g) + l-(g)
Hlarutan = +13 kJ
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- Larutan cairan dalam cairanLarutan gas dalam cairan Energi
solvasi eksoterm
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Pengaruh Suhu pada Kelarutan
Kelarutan : Massa zat terlarut yang membentuk larutan jenuh dengan massa pelarut pada suhu tertentu
Satuan : gram zat terlarut / 100 gram pelarut
Solut (tidak larut)Solut (larut)
Kelarutan naik jika mengabsorpsi panas
Solut (tidak larut) + Panas Solut (larut)
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FIGURE Solubilty in water versus temperature for several substances
Gas larut secara eksoterm dalam cairan pada semua konsentrasi
Gas (tidak larut)Gas (larut) + Panas
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Pengaruh Tekanan Pada Kelarutan dalam Gas
Kelarutan gas dalam cairan naik dengan niaknya tekanan
Gas + pelarutLarutan
FIGURE
Solubility in water versus pressure for two gases
FIGURE
How pressure inueases the solubility of a gas in a liquid. (a) At some specific pressure, equilibrium exists between the vapor phase and the solution. (b) An Increase in pressure puts stress on the equilibrium. (c) More gas dissolves and equilibrium is restored
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Hukum Henry : Konsentrasi gas dalam cairan pada suhu yang diberikan secara langsung sebanding dengan tekanan gas pada larutan.
Cg = Kg . Pg
Kelarutan gas yang terhidrasi kuat
SO2, NH3 & CO2 lebih mudah larut dibanding S2 & N2
NH3 Ikatan H, SO2 & CO2 bereaksi dengan air kesetimbangan
CO2(aq) + H2O H2CO3(aq) H+(aq) + HCO3-(aq)
SO2(aq) + H2O H+(aq) + HSO3-(aq)
NH3(aq) + H2O NH4+(aq) + OH-(aq)
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* % konst. % berat (% b/b) : jumlah gram zat terlarut / 100 g larutan
% volume (% v/v) : jumlah mL zat terlarut / 100 mL larutan
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*
Konsentrasi
* Fraksi mol dan % mol
Mol fraksi : Perbandingan jumlah mol suatu komponen terhadap
jumlah mol total komponen yang ada.
XA =
nA
nA + nB nC + dst
Hukum gas ideal : nA =
PA . V
R. T
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Perubahan di antara satuan dan konsentrasi
Merubah dari % berat ke molalMerubah dari % berat ke fraksi molMenghitung % berat dari fraksi molMerubah molal ke fraksi molMerubah % berat ke molarMerubah dari molar ke % beratPenurunan Tekanan Uap
Tek uap campuran turun dengan adanya komponen lainTek uap larutan (zat terlarut : non volatil) < tek uap pel murniHukum RaoultPlarutan = Xpelarut . Popelarut
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Sifat Koligatof Larutan
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CALCULATING THE MOLE FRACTION OF A GAS FROM PARTIAL PRESSURE
Problem : What are the mole percents of nitrogen and oxygen in air when the partial pressure are 160 torr for oxygen and 600 torr for nitrogen ? (Asumme no other gases are present)
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Solution : Let us use Equation to find the mole fraction of N2 fist.
But the total pressure is the sum of the partial pressures, so
= 0,789, or 78.9 mole percent of N2
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Now we do the same for oxygen
You can easlly see that the two mole percents add up to 100%
CALCULATING MOLAL CONCENTRATION
Problem : An experiment calls for an aqueous 0,150 m solution of sodium chloride. To prepare a solution with this concentration, how many grams of NaCl would have to be dissolved in 500 g of water ?
Solution : As with almost all problems involving concentrations; our first step is to prepare a conversion factor. Thus, 0,150 m NaCl gives us the following, two rations, where we substitute 1000 g for 1 kg.
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and
To calculate the moles of NaCl we need for 500 g of H2O, we use the first ratio, because then the units will cancel property
This gives us the moles of NaCl needed. We next convert 0.0750 mol of NaCl to grams of NaCl. (The formula weight of NaCl is 58,5 which means, of course, 58.5 g NaCl/mol NaCl)
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Thus, when 4,39 g of NaCl is dissolved in 500 g of H2O, the concentration is ), 150 m NaCl. With a little practice, you will be able to set up a string of conversion factors and do the calculation at the end. For example,
USING WEIGHT/WEIGHT PERCENT
Problem : How many grams of a 4.00% (w/w) solution of NaCl needed to obtain 0.500 g of NaCl ?
Solution : The given concentration gives us the following conversion factors
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and
We want 0.500 g of NaCl from this solution, so we use the second conversion factor
Thus, if we take 12.5 g to the 4.00% (w/w) NaCl solution, we will also be taking 0.500 g of
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FIGURE
The vapor pressure of an ideal, two-component solution of volatile compounds
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- Larutan ideal dan penyimpangan hukum Raoult
FIGURE
Typical deviations from ideal behavior, of the total vapor pressure of real, two-componen solutions of volatile substances.
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KenaikanTitik Didih
th = kb . m
FIGURE
Boiling point elevation: Shown here are plots of vapor pressures versus temperatures for a solvent (upper curve) and for a solution of a non volatile solute in the same solvent (lower curve).
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Tabel. molal boiling point elevation and freezing point depression constants
Menghitung kenaikan titik didih dari molar harga konstanta kenaikan titik dan molalMenghitung BM dari kenaikan titik didih28
SolventBp (oC)KbMp (oC)KfWater1000.1501.86Acetic acid118.33.0716.63.57Benzen8022.535.455.07Chloroform61.23.63--Camphor--178.437.7Cyclohexane80.72.696.520.0 -
Penurunan Titik Beku
th = kb . m
Menghitung penurunan titik beku dari molar harga konstanta penurunan titik dan molalMenghitung BM dari kenaikan titik bekuDialisis dan Osmosis
Dialisis : Jika 2 larutan dengan konstrasi berbeda dipisahkan oleh suatu membran, konst akan berubah hingga setimbang. Membran bersifat semipermiabel (hanya ion dan molekul kecil yang dapat lewat)
Osmosis : Jika hanya molekul pelarut yang dapat lewat pada membran
Tekanan Osmosis : Tekanan untuk menjaga aliran osmosis
= MRT
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Sifat sifat Koligatif pada Larutan Elektrolit
Memperkirakan sifat koligatif pada larutan elektolitIneraksi ion-ion dalam larutan cairan32
% ionisasi elektrolit elektrolit lemah
pelarut
kg
terlarut
zat
mol
Molal
Konst.
=
=
m
tot
A
A
P
P
X
=
(
)
(
)
elektrolit
non
an
penghitung
f
pengukuran
f
t
t
:
Hoff
t
Van'
faktor
=
i
ada
yang
asam
mol
i
terionisas
asam
mol
%ionisasi
K
t
m
f
=
=
=
total
total
N
N
P
torr
600
P
P
X
2
2
=
=
torr
160
torr
600
torr
600
X
2
N
+
=
2
O
O
to
up
add
percents
mole
21.1
or
0.211,
torr
160
torr
600
torr
160
X
2
=
+
=
O
H
g
1000
NaCl
mol
0,150
2
NaCl
mol
0,150
O
H
g
1000
2
NaCl
mol
0,0750
O
H
g
1000
NaCl
mol
0,150
x
O
H
500g
2
2
=
NaCl
g
4,39
NaCl
mol
1
NaCl
g
58,5
x
NaCl
mol
0,0750
=
NaCl
g
4,39
NaCl
mol
1
NaCl
g
58,5
O
H
g
1000
NaCl
mol
0,150
x
O
H
g
500
2
2
=
=
solution
g
100
NaCl
g
4.00
NaCl
g
4.00
solution
g
100
solution
(w/w)
4.00%
of
g
12.5
NaCl
g
4.00
solution
g
100
x
NaCl
g
0.500
=