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LARUTAN DAN KOLOID Macam-Macam Campuran • Larutan andingan larutan dispersi Koloid & Suspensi LARUTAN DISPERSI KOLOID SUSPENSI Semua 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 • Dispersi Koloid • Suspensi 1

005_Larutan Koloid

May 02, 2017



Raden Hendra
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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




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



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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

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• 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


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




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


Sifat Koligatif Larutan

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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)


Solution : Let us use Equation to find the mole fraction of N2 fist.




X 2


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

torr160 torr600 torr600X


= 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%


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.




O toup add percents mole 21.1 or0.211, torr160 torr600


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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


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


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


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



Δt : Hofft Van' faktor i


% ionisasi elektrolit – elektrolit lemah

ada yang asam moliterionisas asam mol