SIFAT SIFAT LARUTAN PROF SBW. LARUTAN CAMPURAN ZAT TERLARUT: GULA, GARAM, ASAM, BASA, GARAM-2 ALKALI DLL ZAT PELARUT: AIR, ETANOL, METANOL, HEKSAN, ETER.

Presentasi berjudul: "SIFAT SIFAT LARUTAN PROF SBW. LARUTAN CAMPURAN ZAT TERLARUT: GULA, GARAM, ASAM, BASA, GARAM-2 ALKALI DLL ZAT PELARUT: AIR, ETANOL, METANOL, HEKSAN, ETER."— Transcript presentasi:

1 SIFAT SIFAT LARUTAN PROF SBW

2 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

3 Solutions Solutions are homogeneous mixtures of two or more pure substances. In a solution, the solute is dispersed uniformly throughout the solvent.

4 Solutions How does a solid dissolve into a liquid? What ‘drives’ the dissolution process? What are the energetics of dissolution?

5 Solutions How Does a Solution Form? 1.Solvent molecules attracted to surface ions. 2.Each ion is surrounded by solvent molecules. 3.Enthalpy (  H) changes with each interaction broken or formed. Ionic solid dissolving in water

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

7 Solutions 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.

8 Solutions Energy Changes in Solution To 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.

9 Solutions Enthalpy Changes in Solution The enthalpy change of the overall process depends on  H for each of these steps. Start End Start

10 Solutions Enthalpy changes during dissolution The enthalpy of solution,  H soln, can be either positive or negative.  H soln =  H 1 +  H 2 +  H 3  H soln (MgSO 4 )= -91.2 kJ/mol --> exothermic  H soln (NH 4 NO 3 )= 26.4 kJ/mol --> endothermic

11 Solutions Why do endothermic processes sometimes occur spontaneously? Some processes, like the dissolution of NH 4 NO 3 in water, are spontaneous at room temperature even though heat is absorbed, not released.

12 Solutions 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)

13 Solutions 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.

14 Solutions 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.

15 Solutions Factors Affecting Solubility The stronger the intermolecular attractions between solute and solvent, the more likely the solute will dissolve. Example: ethanol in water Ethanol = CH 3 CH 2 OH Intermolecular forces = H-bonds; dipole-dipole; dispersion Ions in water also have ion-dipole forces.

16 Solutions Factors Affecting Solubility Glucose (which has hydrogen bonding) is very soluble in water. Cyclohexane (which only has dispersion forces) is not water- soluble.

17 Solutions Factors Affecting Solubility Vitamin A is soluble in nonpolar compounds (like fats). Vitamin C is soluble in water.

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

19 Solutions 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.

20 Solutions PENURUNAN TEKANAN UAP UNTUK LARUTAN NON-ELEKTROLIT Hukum Raoult: Dimana: P = tekanan uap jenuh larutan, = tekanan uap jenuh pelarut murni Xp = fraksi mol pelarut, XA atau Xzt = fraksi molekul zat terlarut, ΔP = penurunan tekanan uap pelarut. (14)

21 Solutions ΔP = P 0 (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 @20 0 C adalah 18 mmHg. Jawab: P larutan : P = P 0 (1 – XA); P = 18 (1 – 0,25) = 13,5 mmHg. MOL GLUKOSA = berat glukosa/BM = 45/180 = 0,25.

22 Solutions Larutan non-elektrolit X1 = fraksi mol pelarut; X2 = fraksi mol zat terlarut, maka X1 + X2 = 1 X 1 = 1-X 2 Persamaan HK. RAOULT : P pelarut = P 0 pelarut X pelarut Ppelarut = P 1 o (1-X 2 ) atau P 1 o -P =P 1 o X 2 (P 1 o -P)/ P 1 o = Δp/ P 1 o =X 2 = n 2 /n 1 +n 2

23 Solutions 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/ P 1 o = 0,5/55,5 + 0,5 = 0,5/60 = 0,0089 = 0,89% karena ada 0,5 mol sukrosa.

24 Solutions Kenaikan titik didih 16 T larutan = T pelarut + Kx, bp X2 17

25 Solutions T bp, soln = Tbp,1 + Kbp,m R = 8,314 X 10 -3 kJK -1 mol -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

26 Solutions Penurunan titik beku T fp, soln = Tfp,1 + Kfp,m 18

27 Solutions 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 = 0,037 0 C.

28 Solutions Tekanan osmosis

29 Solutions 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

30 Solutions 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

31 Solutions 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 -1 K -1 ) Jawab : Ba(OH) 2 merupakan elektrolit Ba(OH) 2 → Ba 2+ + 2 OH¯, n = 3 mol Ba(OH)2 = 1,71 gram / 171 gram/mol = 0,01 mol M = 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 -1 K - 1 × 310 K × (1 + (3 – 1)1)= 7,626 atm

32 Solutions LATIHAN SOAL