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LARUTAN DAN KOLOIDMacam-Macam Campuran• LarutanPerbandingan larutan dispersi Koloid & Suspensi
LARUTAN DISPERSI KOLOID SUSPENSISemua 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 mikroskop
Stabil terhadap gravitasi Kurang Stabil Tidak stabil
Homogen Perbatasan homogen Tidak Homogen
Tembus Cahaya Buram Tidak tembus
Tidak ada efek Tyndall Efek Tyndall Tidak transparan
Tidak ada gerak Brown Gerak Brown Partikel terpisah
Tidak dapat dipisahkan dengan penyaringan
Tidak dapat dipisahkan dengan penyaringan
Dapat dipisahkan dengan penyaringan
• Dispersi Koloid • Suspensi
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• SUSPENSIDapat dipisahkan dengan penyaringan atau dengan sentrifugasi
• DISPERSI KOLOIDJENIS FASA
TERDISPERSIMEDIA
PENDISPERSICONTOH
Busa Gas Cair Busa sabun
Busa Padat Gas Padat Batu apung
Aerosol Cair Cair Gas Kabut, halimun, awan
Emulsi Cair Cair Krim, susu, saos
Emulsi Padat Cair Padat Mentega, keju
Asap Padat Gas Debu, partikulat dalam asap
Sol Padat Cair Pati dalam air, jeli, cat
Sol Padat padat padat Aloy, mutiara
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Muatan Elektron partikel Koloid
FIGURE Colloidal particles often bear electrical charges that stabilizethe dispersion. On the left is a particle whose extremely large moleculescarry negatively charged groups. On the right the colloidalparticles hasattracted chloride ions to itself. In either case, these colloidal particles repeleach other and cannot join together.
Colloidal particle withorganic ionic groups
Colloidal particle withadsorbed chloride ions
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Efek TyndallThe Tyndall effect,
A pencil line thin red laser beampasses through the liquid in three
Rest tubes. The first contain acolloidal dispersion of starch, the
second a solution of sodiumchrornate, and the third a colloidal
dispersion of Fe2O3, All threeappear transparent, and in the ab
sence of the ryndall effect wemight think they are all solutions,
However, the Tyndilll effect revealsthat the fist and third are coilds,
Not true solutions
LARUTAN CAMPURAN HOMOGEN
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Larutan Gas Gas dalam gas Cair dalam gas Padat dalam gas
UdaraSistem koloidSistem koloid
Larutan Cairan Gas dalam cair Cair dalam cair Padat dalam cair
Coca-colaCuka, GasolinGula dalam cair
Larutan Padat
Gas dalampadat Cair dalam padat Padat dalam padat
Aloy hidrogen dim paladiumBenzen dalam karetKarbon 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 9
• Larutan padat dalam cair
Portion of surface and edge of NaCl cristal in contact with water
FIGUREHydration of ions
Hydration involves a complexredirection of force of attrac-
tion and repulsion. Before thissolution forms, water mole-cules are attracted only to
each other; and Na+ and CI-
ions have only each other incrystal 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 LarutanTerjadi 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 kJH = -619 kJ
Net : Kl(s) K+(g) + l-(g) Hlarutan = +13 kJ
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• Larutan cairan dalam cairan
• Larutan 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 pelarutSolut (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 konsentrasiGas (tidak larut) Gas (larut) + Panas
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Pengaruh Tekanan Pada Kelarutan dalam Gas
Kelarutan gas dalam cairan naik dengan niaknya tekananGas + pelarut Larutan
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 kuatSO2, 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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Konsentrasi* Fraksi mol dan % mol
Mol fraksi : Perbandingan jumlah mol suatu komponen terhadapjumlah mol total komponen yang ada.
XA = nA
nA + nB nC + … dst
Hukum gas ideal : nA = PA . VR. T
tot
AA P
PX
pelarut kgterlarutzat mol
Molal Konst. m
* % 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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*
Perubahan di antara satuan dan konsentrasi • Merubah dari % berat ke molal• Merubah dari % berat ke fraksi mol• Menghitung % berat dari fraksi mol• Merubah molal ke fraksi mol• Merubah % berat ke molar• Merubah dari molar ke % berat
Penurunan Tekanan Uap• Tek uap campuran turun dengan adanya komponen lain• Tek uap larutan (zat terlarut : non volatil) < tek uap pel murni• Hukum Raoult
Plarutan = Xpelarut . Popelarut
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Sifat Koligatif Larutan
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.
totaltotal
NN P
torr600PP
X 2
2
But the total pressure is the sum of the partial pressures, so
torr160 torr600 torr600X
2N
= 0,789, or 78.9 mole percent of N2
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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2
O
O toup add percents mole 21.1 or0.211, torr160 torr600
torr160X2
OH g 1000NaCl mol 0,150
2 NaCl mol 0,150OH g 1000 2and
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
NaCl mol 0,0750OH g 1000
NaCl mol 0,150 xOH 500g
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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)
NaCl g 4,39NaCl mol 1NaCl g 58,5
xNaCl mol 0,0750
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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,
NaCl g 4,39NaCl mol 1NaCl g 58,5
OH g 1000NaCl mol 0,150
xOH g 5002
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USING WEIGHT/WEIGHT PERCENTProblem : 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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solution g 100NaCl g 4.00
NaCl g 4.00solution g 100
and
We want 0.500 g of NaCl from this solution, so we use the second conversion factor
solution (w/w) 4.00% of g 12.5NaCl g 4.00
solution g 100 xNaCl g 0.500
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
FIGURETypical deviations from ideal behavior, of the total vapor pressure of real, two-componen solutions of volatile substances.
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KenaikanTitik Didihth = 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
Solvent Bp (oC) Kb Mp (oC) Kf
Water 100 0.15 0 1.86Acetic acid 118.3 3.07 16.6 3.57Benzen 802 2.53 5.45 5.07Chloroform 61.2 3.63 - -Camphor - - 178.4 37.7Cyclohexane 80.7 2.69 6.5 20.0
• Menghitung kenaikan titik didih dari molar harga konstanta kenaikan titik dan molal
• Menghitung BM dari kenaikan titik didih
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Penurunan Titik Bekuth = kb . m
• Menghitung penurunan titik beku dari molar harga konstanta penurunan titik dan molal
• Menghitung BM dari kenaikan titik beku
Dialisis dan OsmosisDialisis : 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 29
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Sifat – sifat Koligatif pada Larutan Elektrolit• Memperkirakan sifat koligatif pada larutan elektolit• Ineraksi ion-ion dalam larutan cairan
elektrolit non anpenghitungf
pengukuranf
Δt
Δt : Hofft Van' faktor i
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% ionisasi elektrolit – elektrolit lemah
ada yang asam moliterionisas asam mol
%ionisasiKΔt
mf