SIFAT SIFAT LARUTAN

PROF SBW

LARUTAN

• CAMPURAN ZAT TERLARUT: GULA, GARAM, ASAM, BASA, GARAM-2 ALKALI DLL

• ZAT PELARUT: AIR, ETANOL, METANOL, HEKSAN, ETER DLL

• JENIS-JENIS LARUTAN:• ELEKTROLIT KUAT• ELEKTROLIT LEMAH• LARUTAN NON-ELEKTROLIT

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• Solutions are homogeneous mixtures of two or more pure substances.

• In a solution, the solute is dispersed uniformly throughout the solvent.

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How does a solid dissolve into a liquid?

What ‘drives’ the dissolution process?

What are the energetics of dissolution?

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How Does a Solution Form?1. Solvent molecules attracted to surface ions.2. Each ion is surrounded by solvent molecules.3. Enthalpy (DH) changes with each interaction broken or

formed.

Ionic solid dissolving in water

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How Does a Solution Form?1. Solvent molecules attracted to surface ions.2. Each ion is surrounded by solvent molecules.3. Enthalpy (DH) changes with each interaction broken or

formed.

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How Does a Solution Form

The ions are solvated (surrounded by solvent).

If the solvent is water, the ions are hydrated.

The intermolecular force here is ion-dipole.

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Energy Changes in SolutionTo determine the enthalpy

change, we divide the process into 3 steps. 1. Separation of solute

particles.2. Separation of solvent

particles to make ‘holes’.

3. Formation of new interactions between solute and solvent.

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Enthalpy Changes in Solution

The enthalpy change of the overall process depends on DH for each of these steps.

Start

End

EndStart

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Enthalpy changes during dissolution

The enthalpy of solution, DHsoln, can be either positive or negative.

DHsoln = DH1 + DH2 + DH3

DHsoln (MgSO4)= -91.2 kJ/mol --> exothermicDHsoln (NH4NO3)= 26.4 kJ/mol --> endothermic

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Why do endothermic processes sometimes occur spontaneously?

Some processes, like the dissolution of NH4NO3 in water, are spontaneous at room temperature even though heat is absorbed, not released.

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Enthalpy Is Only Part of the Picture

Entropy is a measure of: • Dispersal of energy in

the system.• Number of microstates

(arrangements) in the system.

b. has greater entropy, is the favored state

(more on this in chap 19)

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Entropy changes during dissolution

Each step also involves a change in entropy. 1. Separation of solute

particles.2. Separation of solvent

particles to make ‘holes’.

3. Formation of new interactions between solute and solvent.

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Factors Affecting Solubility

• Chemists use the axiom “like dissolves like”:Polar substances tend to

dissolve in polar solvents.Nonpolar substances tend

to dissolve in nonpolar solvents.

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Factors Affecting Solubility

The stronger the intermolecular attractions between solute and solvent, the more likely the solute will dissolve.Example: ethanol in water

Ethanol = CH3CH2OH

Intermolecular forces = H-bonds; dipole-dipole; dispersion

Ions in water also have ion-dipole forces.

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Factors Affecting Solubility

Glucose (which has hydrogen bonding) is very soluble in water.

Cyclohexane (which only has dispersion forces) is not water-soluble.

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Factors Affecting Solubility• Vitamin A is soluble in nonpolar compounds

(like fats).• Vitamin C is soluble in water.

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Which vitamin is water-soluble and which is fat-soluble?

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SIFAT KOLIGATIF

• LARUTAN NON-ELEKTROLIT: LARUTAN DIMANA ZAT TERLARUT TIDAK MENGION BILA DILARUTKAN DALAM AIR, SEHINGGA TIDAK MEMBAWA ALIRAN LISTRIK MELALUI LARUTAN TERSEBUT. LARUTAN NON-ELEKTROLIT: SUKROSA, GLUKOSA, GLISERIN, NAFTALENA DAN UREA.

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PENURUNAN TEKANAN UAP UNTUK LARUTAN NON-ELEKTROLIT

Hukum Raoult:Dimana: P = tekanan uap jenuh larutan, = tekanan uap jenuh pelarut murniXp = fraksi mol pelarut, XA atau Xzt = fraksi molekul zat terlarut, ΔP = penurunan tekanan uap pelarut.  

(14) 

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•ΔP = P0 (1 – XA) (15)• Contoh: Hitunglah penurunan tekanan uap

jenuh air, bila 45 g glukosa (BM=180) dilarutkan dalam 90 g air. Diketahui tekanan uap jenuh air murni @200C adalah 18 mmHg.

• Jawab: P larutan : P = P0 (1 – XA); • P = 18 (1 – 0,25) = 13,5 mmHg. • MOL GLUKOSA = berat glukosa/BM =

45/180 = 0,25.

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Larutan non-elektrolit

• X1 = fraksi mol pelarut; X2 = fraksi mol zat terlarut, maka X1 + X2 = 1

• X1 = 1-X2

• Persamaan HK. RAOULT :• P pelarut = P0 pelarut X pelarut • Ppelarut = P1

o(1-X2) atau P1o-P =P1

oX2

• (P1o-P)/ P1

o= Δp/ P1o=X2= n2/n1+n2

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LATIHAN• HITUNG PENURUNAN TEK. UAP

LARUTAN @20 0C, UNTUK LARUTAN BERISI SUKROSA 171,2 g BM SUKROSA 342,3. DALAM 1 LITER AIR, BM AIR 18,02.

• SOLUTIONS: mol sukrosa= 171,2/342,3 = 0,5; mol air = 1000/18,2 = 55,5;

• Δp/ P1o = 0,5/55,5 + 0,5 = 0,5/60 = 0,0089

= 0,89% karena ada 0,5 mol sukrosa.

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Kenaikan titik didih

16 T larutan = T pelarut + Kx, bp X2 17

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• T bp, soln = Tbp,1 + Kbp,m• R = 8,314 X 10-3kJK-1mol-1• HITUNG: Peningkatan Tbp untuk larutan

air yang mengndung urea = 0,1, jika perubahan tekanan uap larutan = 40,656 kJ/mol pada 373,15 K

• Kx,bp = 8,314 (373,15)/40,656 = 28,47 K• Tbp = 373,15 + (28,47) (0,1) = 376,00 K

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Penurunan titik beku

• T fp, soln = Tfp,1 + Kfp,m18

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• Berapa titik beku larutan yang mengandung 3,42 g dalam 500 g air, bila Kfp = 1,86, BM sukrosa = 342

• Solution:• ΔTfp = Kf m = Kf 1000 w2/w1m2• ΔTfp= 1,86 X 1000x3,42/500x342 =

0,0370C.

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Tekanan osmosis

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SIFAT KOLIGATIF LARUTAN ELEKTROLIT KUAT

• ELEKTROLIT KUAT: SPT: GARAM NaCl atau HCl dsb akan mengion menjadi ion Na dan Cl dalam air, maka konsentrasi solute atau jumlah solute meningkat dalam larutan air, maka rumus sifat koligatif berbeda dengan larutan non-elektrolit

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Rumus-2 sifat koligatif larutan elektrolit kuat

• ΔP = XA ×P ×i• ΔTb = Kb ×m× i

(19)• ΔTf = Kf ×m× i• π = M× R×T × i

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Contoh: Pada suhu 37 °C ke dalam air dilarutkan 1,71 gram Ba(OH)2 hingga volume 100 mL (Mr Ba(OH)2 = 171). Hitung besar tekanan osmotiknya! (R = 0,082 L atm mol-1K-1)

• Jawab : • Ba(OH)2 merupakan elektrolit  Ba(OH)2 → Ba2+

+ 2 OH¯, n = 3 mol Ba(OH)2 = 1,71 gram / 171 gram/mol  = 0,01 molM = n / V = 0,01 mol / 0,1 L = 0,1 mol ⋅ L-1

π = M × R × T × i= 0,1 mol L-1 × 0,082 L atm mol-1K-1 × 310 K × (1 + (3 – 1)1)= 7,626 atm

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LATIHAN SOAL