SIFAT SIFAT LARUTAN

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. How Does a Solution Form? • Solvent molecules attracted to surface ions. • Each ion is surrounded by solvent molecules. • Enthalpy (DH) changes with each interaction broken or formed. Ionic solid dissolving in water

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

7. 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. Energy Changes in Solution To determine the enthalpy change, we divide the process into 3 steps. • Separation of solute particles. • Separation of solvent particles to make ‘holes’. • Formation of new interactions between solute and solvent.

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

10. Enthalpy changes during dissolution DHsoln = DH1 + DH2 + DH3 The enthalpy of solution, DHsoln, can be either positive or negative. DHsoln (MgSO4)= -91.2 kJ/mol --> exothermic DHsoln (NH4NO3)= 26.4 kJ/mol --> endothermic

11. 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.

12. 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. Entropy changes during dissolution Each step also involves a change in entropy. • Separation of solute particles. • Separation of solvent particles to make ‘holes’. • Formation of new interactions between solute and solvent.

14. 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. 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.

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

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

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

19. 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. PENURUNAN TEKANAN UAP UNTUK LARUTAN NON-ELEKTROLIT (14) HukumRaoult: Dimana: P = tekananuapjenuhlarutan, = tekananuapjenuhpelarutmurni Xp = fraksimolpelarut, XA atauXzt = fraksimolekulzatterlarut, ΔP = penurunantekananuappelarut.

21. ΔP = P0 (1 – XA) (15) • Contoh: Hitunglahpenurunantekananuapjenuh air, bila 45 g glukosa (BM=180) dilarutkandalam 90 g air. Diketahuitekananuapjenuh air murni @200C adalah 18 mmHg. • Jawab: P larutan : P = P0 (1 – XA); • P = 18 (1 – 0,25) = 13,5 mmHg. • MOL GLUKOSA = beratglukosa/BM = 45/180 = 0,25.

22. Larutan non-elektrolit • X1 = fraksimolpelarut; X2 = fraksimolzatterlarut, maka X1 + X2 = 1 • X1 = 1-X2 • Persamaan HK. RAOULT : • P pelarut = P0pelarutX pelarut • Ppelarut = P1o(1-X2) atau P1o-P =P1oX2 • (P1o-P)/ P1o=Δp/P1o=X2=n2/n1+n2

23. 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: molsukrosa= 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% karenaada 0,5 molsukrosa.

24. Kenaikantitikdidih 16 T larutan = T pelarut + Kx, bp X2 17

25. T bp, soln = Tbp,1 + Kbp,m • R = 8,314 X 10-3kJK-1mol-1 • HITUNG: PeningkatanTbpuntuklarutan air yang mengndung urea = 0,1, jikaperubahantekananuaplarutan = 40,656 kJ/molpada 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. Penurunantitikbeku 18 • T fp, soln = Tfp,1 + Kfp,m

27. Berapatitikbekularutan yang mengandung 3,42 g dalam 500 g air, bilaKfp = 1,86, BM sukrosa = 342 • Solution: • ΔTfp = Kf m = Kf 1000 w2/w1m2 • ΔTfp= 1,86 X 1000x3,42/500x342 = 0,0370C.

28. Tekanan osmosis

29. SIFAT KOLIGATIF LARUTAN ELEKTROLIT KUAT • ELEKTROLIT KUAT: SPT: GARAM NaClatauHCldsbakanmengionmenjadi ion Na danCldalam air, makakonsentrasi solute ataujumlah solute meningkatdalamlarutan air, makarumussifatkoligatifberbedadenganlarutan non-elektrolit

30. Rumus-2 sifatkoligatiflarutanelektrolitkuat • ΔP = XA ×P ×i • ΔTb = Kb ×m× i (19) • ΔTf = Kf ×m× i • π = M× R×T × i

31. Contoh: Padasuhu 37 °C kedalam air dilarutkan 1,71 gram Ba(OH)2hingga volume 100 mL (Mr Ba(OH)2 = 171). Hitungbesartekananosmotiknya! (R = 0,082 L atm mol-1K-1) • Jawab : • Ba(OH)2merupakanelektrolitBa(OH)2 → Ba2+ + 2 OH¯, n = 3 molBa(OH)2 = 1,71 gram / 171 gram/mol  = 0,01 molM = n / V = 0,01 mol / 0,1 L = 0,1 mol ⋅ L-1π = 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

32. LATIHAN SOAL