LARUTAN DAN KOLOID

Macam-Macam Campuran

Larutan

Perbandingan larutan dispersi Koloid & Suspensi

Dispersi KoloidSuspensi

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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 KOLOID

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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 gas

Udara

Sistem koloid

Sistem koloid

Larutan Cairan

Gas dalam cairCair dalam cairPadat dalam cair

Coca-cola

Cuka, Gasolin

Gula dalam cair

Larutan Padat

Gas dalampadatCair dalam padat Padat dalam padat

Aloy 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

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 % berat

Penurunan Tekanan Uap

Tek uap campuran turun dengan adanya komponen lainTek uap larutan (zat terlarut : non volatil) < tek uap pel murniHukum Raoult

Plarutan = Xpelarut . Popelarut

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Sifat Koligatof 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.

But the total pressure is the sum of the partial pressures, so

= 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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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 didih

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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 beku

Dialisis 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 cairan

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% 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

=