1. Chapter 4

2. H 2 O, The Universal Solvent Much of chemistry that affects each of us occurs among substances dissolved in water. Virtually all chemistry that makes life possible occurs in an aqueous environment. Aqueous Solution: a solution in which water is the dissolving medium or solvent. Solvent: the dissolving medium in a solution. Solute: a substance dissolved in a liquid to form a solution.

3. H 2 O, The Universal Solvent http://nrich.maths.org/7273 http://www.chem1.com/acad/webtext/chembond/cb05.html Waters polar characteristics: Electrons shared spend more of their time with the oxygen atom

4. H 2 O, The Universal Solvent http://www.biology.arizona.edu/biochemistry/tutorials/chemistry/graphics/n acl2.gif Hydration: hydration of ions tends to cause salts to “fall apart” or dissolve.

5. H 2 O, The Universal Solvent It is very important to recognize that when an ionic substance dissolves in water they break up into the individual cations and anions. NH 4 NO 3 (s) NH 4 + (aq) + NO 3 - (aq) H 2 O (l) Example: Ammonium Nitrate in water When cations and anions have been dissolved in solution they move around independently

6. H 2 O, The Universal Solvent Solubility: the amount of substance that dissolves in a given volume of solvent at a given temperature The solubility of substances in water varies greatly and for ionic compounds it depends on the relative attraction between the ions. “Like dissolves Like”, in general polar or ionic compounds are expected to be soluble in water than non polar substances

7. Strong and Weak Electrical Conductivity: the ability to conduct electricity Strong Electrolytes: solution that conducts electricity efficiently Weak Electrolytes: solution that conducts electricity inefficiently Non electrolytes: solution that prevents the flow of electricity

8. Strong and Weak Strong Electrolytes completely ionize when dissolved in water HCl H + (aq) + Cl - (aq) H 2 O (l) HNO 3 H + (aq) + NO 3 - (aq) H 2 O (l) H 2 SO 4 H + (aq) + HSO 4 - (aq) H 2 O (l) Strong Acids: a substance that produces H + ions (protons) when dissolved in water (virtually every molecule ionizes)

9. Strong and Weak Strong Electrolytes completely ionize when dissolved in water NaOH (s) Na + (aq) + OH - (aq) H 2 O (l) KOH K + (aq) + OH - (aq) H 2 O (l) Strong Bases: a substance that produces OH - ions (hydroxide) when dissolved in water (virtually every molecule ionizes)

10. Strong and Weak Weak Electrolytes do not completely ionize when dissolved in water Acetic Acid is very different from strong acids, for every 100 molecules of acetic acid only 1 will ionize (approximately 1%) HC 2 H 3 O 2 H + (aq) + C 2 H 3 O 2 - (aq) H 2 O (l) Weak Acid: any acid that dissociates only to a slight extent in aqueous solutions.

11. Strong and Weak Weak Electrolytes do not completely ionize when dissolved in water NH 3 + H 2 O NH 4 + (aq) + OH - (aq) Ammonia is the most common weak base. It also is very different from strong bases, for every 100 molecules of ammonia only 1 will ionize (approximately 1%) Weak Base: any base that dissociates only to a slight extent in aqueous solutions.

12. Molarity Whenever chemical reactions take place when two solutions mix and we want to perform Stoichiometry calculations we must know two things: 1. The nature of the reaction (the chemicals involved) 2. The amounts of chemical involved M = Molarity = Moles of Solute Liters of solution

13. Molarity Calculate the molarity of a solution prepared by dissolving 11.5 grams of Sodium Hydroxide in enough water to make 1.50 Liters of solution. First calculate the number of moles of solute: 11.5 grams NaOH 1 x 1 mol NaOH 40.00 g NaOH = 0.288 mol NaOH Then divide the number of moles by the volume of solution: 0.288 mol NaOH 1.50 L = 0.192 M NaOH

14. Molarity Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl in enough water to make 26.8 ml of solution. First calculate the number of moles of solute: 1.56 grams HCl 1 x 1 mol HCl 36.46 g HCl = 4.28 x 10 -2 mol HCl Then divide the number of moles by the volume of solution: 4.28 x 10 -2 mol HCl 2.68 x 10 -2 L = 1.60 M HCl 1L. 1000 ml 26.8 ml 1 x = 2.68 x 10 -2 L

15. Molarity Calculate the number of each type of Ion in a 0.5 M solution of Fe(ClO 4 ) 3. First determine the balanced reaction: Then multiply the molarity by the coefficients of the products: = 1.5 mol ClO 4 - ions Fe(ClO 4 ) 3 (s) Fe 3+ (aq) + 3ClO 4 - (aq) H 2 O (l) 3 ClO 3 - ions x 0.5 mol 1 Fe 3+ ions x 0.5 mol= 0.5 mol Fe 3+ ions

16. Molarity Typical blood serum is 0.14 M NaCl. What volume of blood would contain 1.0 mg of NaCl? First determine the number of moles represented by 1.0 mg NaCl: = 1.71109 x 10 -5 mol NaCl Next determine what volume of 0.14 M contains 1.71109 x 10 -5 mol NaCl: 1.0 mg NaCl 1 x 1 mol NaCl 58.44207g 1g NaCl 1000 mg x V x 0.14 M NaCl =1.7 x 10 -5 mol NaCl V = 1.7 x 10 -5 mol NaCl 0.14 M NaCl/ = 1.2 x 10 -4 L

17. Dilution & Molarity To save time and space solutions are often stored in a concentrated form. To achieve the molarity desired you can dilute a known concentration to produce a new solution. The key is to understand that we ARE NOT adding any solute only solvent. M 1 V 1 = M 2 V 2

18. Dilution & Molarity What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M solution? (M 1 V 1 = M 2 V 2 ) First, you know the volume of the final solution and the two molarity values. Plug in the values and solve for the unknown (16 M) V 1 = (0.10 M) (1.5 L) V 1 = (0.10 M) (1.5 L) (16 M) V 1 = 0.0094 L