Q- Solutions Prem Sattsangi Copyright 2008

#1 HYDROGEN-BONDS(5 kcal) Two components form H-bonds e.g. in NH 3 ; Donor (H 3 N:  … ), (donates U.S.electron pair) Acceptor (  H-NH 2 ) (attracts electron pair) [H 3 N:  …  H-NH 2 ] Show two ends, and H-bond in water.

#2 POLAR MOLECULES have: Polar bonds (  E.N.) Dipole moment [Polarities cancel in SYMMETRICAL molecules] Q. Which is polar? (a) NH 3 or (b) CO 2 Show polarities. (a) NH 3

#3 SOLUBILITY of SOLUTES in WATER PRINCIPLE: “Like dissolves like” Water, the solvent, is polar. Water can form Hydrogen-bonds. Q. Which solute is more soluble in water? (a):NH 3 (polar, has Donor and Acceptor ends, forms H-bonds with water). [CO 2 is non-polar, only Donor end is available, H-bonds with water, very limited]

#4 Table 4.1: Solubility Guidelines for Common Ionic Compounds in Water SOLUBLE: Acids Gr.IA, and NH 4 + salts Acetates, and Nitrates Halides (Except: Ag +, Pb 2+, Hg 2 2+ ) Sulfates (Except: Pb 2+, Hg 2 2+ ) Ba 2+, Sr 2+ ) Ag+ Borderline ? INSOLUBLE: Carbonates Phosphates Except: Gr.IA, NH 4 + Hydroxides Except: Gr.IA, NH 4 + Ca 2+, Sr 2+, Ba 2+ Mg(OH) 2 (Insoluble)

#5 INTERMOLECULAR FORCES (CH-11, p. 445) Give examples of compounds or systems, which have the following as the major intermolecular force. Use kerosene, water, ethanol, ammonia, carbon dioxide, salt water, and CH 3 Cl. Dipole CH 3 Cl Hydrogen bonding ammonia London Dispersion Kerosene, Carbon dioxide

#6 INTERMOLECULAR FORCES (CH-11, p. 446) Give examples of compounds or systems, which have the following as the major intermolecular force. Use kerosene, water, ethanol, ammonia, carbon dioxide, salt water, and CH 3 Cl. Hydrogen bondingWater, ammonia, ethanol Ion-dipole Salt water

# 7 SOLUTE(2 dr.)-SOLVENT(5 drs.) interactions A. Solute-Solvent (water) interactions are strong. ethanol (2 drops) + Water (5 drops) Solute (H-bonds) Solvent (H-bonds) Solute forms a homogenous solution. B. Solute-Solute interactions are strong. Water (2 drops) + Kerosene (5 drops) Solute (H-bonds) Solvent (London Dispersion) Solute will not dissolve. C. Solvent-solvent interactions are strong. Kerosene (2 drops) + Water (5 drops) Solute (London Dispersion) Solvent (H-bonds) Solvent will squeeze the solute out of solution.

#8 Solute/Solvent Interactions True or False: ___ Solute/Solute interactions are strong. ___ Solute/Solvent interactions are strong. ___ Solvent/Solvent interactions are strong. ___ Solute is squeezed out of solution. ___ Solute comes out of solution. For a mixture of Water (2 drops) + Kerosene (5 drops)

#9 WRITING EQUATIONS-1 Eq.-1 Ionization in water. Na 3 PO 4 (s) + H 2 O(l)  3Na + (aq) + PO 4 3- (aq) Show that Na 3 PO 4 (aq) is basic. Eq.-2 Reaction of PO 4 3- (aq) with water. PO 4 3- (aq) + H 2 O(l)  HPO 4 2- (aq) + OH - (aq) (Anion + Anion) CT(P.7-7) small size high charge anions split the water molecule.

#10 WRITING EQUATIONS-2 Show that AlCl 3 (aq) is acidic. Eq.-1 Ionization in water. AlCl 3 (aq) + H 2 O(l)  Al 3+ (aq) + 3Cl - (aq) Eq.-2 Reaction with water. Al 3+ (aq) + H 2 O(aq)  AlOH 2+ (aq) + H + (aq) (Cation + Cation) CT(P.7-6) small size high charge cations split the water molecule.

#11 WRITING EQUATIONS-3 For the following REDOX equation, write: 2CuSO 4 (aq) +2KI(aq)  I 2 (aq)+Cu 2 SO 4 (aq)+K 2 SO 4 (aq) A. 2Cu 2+ (aq) + 2SO 4 2- (aq) + 2K + (aq) + 2I - (aq)  I 2 (aq) 2Cu + (aq) + 2SO 4 2- (aq) + 2K + (aq) B. 2Cu 2+ (aq) + 2I - (aq)  I 2 (aq) + 2Cu + (aq) C. 2Cu 2+ (aq)  2Cu + (aq) (O.N. +2 to +1) D. 2 I - (aq)  I 2 (aq) (O.N. –1 to 0) A. Complete ionic equation B. Net Ionic equation C. What is getting reduced? D. What is getting oxidized?

#12 WRITING EQUATIONS-4 For the following PRECIPITATION equation, write: CuSO 4 (aq) +2NaOH (aq)  Cu(OH) 2 (s) +Na 2 SO 4 (aq) A. Cu 2+ (aq) + SO 4 2- (aq) + 2Na + (aq) + 2OH - (aq)  Cu(OH) 2 (s) + 2Na + (aq) + SO 4 2- (aq) B. Cu 2+ (aq) + 2OH - (aq)  Cu(OH) 2 (s) B. Net Ionic equation C. Spectator ions: 2Na + (aq) + SO 4 2- (aq) A. Complete ionic equation

#13 Redox reaction Explain a Redox reaction: Electrons are transferred from species getting OXIDIZED to species getting REDUCED. Give an Equation: 2Na + 2H 2 O  2NaOH + H 2 (g) “0” “1+” 1+ “0” [Ox.Nos.] REVIEW: Ox.No. is (a) the charge on atom = 0 (b) Charge on ion = as present in each case.