1. 1 Chapter 8A Solutions

2. 2 CHAPTER OUTLINE  Type of Solutions Type of Solutions  Electrolytes & Non-electrolytes Electrolytes & Non-electrolytes  Equivalents of Electrolytes Equivalents of Electrolytes  Solubility & Saturation Solubility & Saturation  Soluble & Insoluble Salts Soluble & Insoluble Salts  Formation of a Solid  Precipitation Reactions Precipitation Reactions

3. 3 TYPE OF SOLUTIONS  A solution is a homogeneous mixture of two substances: Solute: Solvent: substance being dissolved present in smaller amount substance doing the dissolving present in larger amount Solutes and solvents may be of any form of matter: solid, liquid or gas.

4. 4 TYPE OF SOLUTIONS TypeExampleSoluteSolvent Gas in gasAirOxygenNitrogen Soda waterCO 2 WaterGas in liq. VinegarAcetic acidWaterLiq. in liq. SeawaterSalt Water Solid in liq.

5. 5 TYPE OF SOLUTIONS TypeExampleSoluteSolvent Dental amalgam MercurySilver BrassZincCopper Liquid in solid Solid in solid

6. 6 SOLUBILITY  Solutions form between solute and solvent molecules because of similarities between them. Like dissolves Like Ionic solids dissolve in water because the charged ions (polar) are attracted to the polar water molecules. Non-polar molecules such as oil and grease dissolve in non-polar solvents such as kerosene.

7. 7 SOLUBILITY Water (polar) CH 2 Cl 2 (non-polar) I 2 (non-polar) Ni(NO 3 ) 2 (polar)

8. 8 ELECTROLYTES & NON-ELECTROLYTES  Solutions can be characterized by their ability to conduct an electric current.  Solutions containing ions are conductors of electricity and those that contain molecules are non-conductors.  Substances that dissolve in water to form ions are called electrolytes.  The ions formed from these substances conduct electric current in solution, and can be tested using a conductivity apparatus.

9. 9 STRONG ELECTROLYTES  Electrolytes are further classified as strong electrolytes and weak electrolytes.  In water, a strong electrolyte exists only as ions.  Strong electrolytes make the light bulb on the conductivity apparatus glow brightly.  Ionic substances such as NaCl are strong electrolytes. NaCl (s) Na + (aq) + Cl  (aq) Only ions present after solution

10. 10 WEAK ELECTROLYTES  Solutions containing weak electrolytes contain only a few ions.  These solutions make the light bulb on the conductivity apparatus glow dimly.  Weak acids and bases that dissolve in water and produce few ions are weak electrolytes. HF (aq) H + (aq) + F  (aq) Few ions present after solution

11. 11 NON- ELECTROLYTES  Substances that do not form any ions in solution are called non-electrolytes.  With these solutions the bulb on the conductivity apparatus does not glow.  Covalent molecules that dissolve in water but do not form ions, such as sugar, are non-electrolytes. C 12 H 22 O 11 (s) C 12 H 22 O 11 (aq) No ions present after solution

12. 12 ELECTROLYTES & NON-ELECTROLYTES

13. 13 Example 1: Identify the predominant particles in each of the following solutions and write the equation for the formation of the solution: NH 4 Br Strong electrolyte(only ions) NH 4 Br (s) NH 4 + (aq) + Br  (aq)

14. 14 Example 1: Identify the predominant particles in each of the following solutions and write the equation for the formation of the solution: CH 4 N 2 O Non-electrolyte(only molecules) CH 4 N 2 O (s) CH 4 N 2 O (aq)

15. 15 Example 1: Identify the predominant particles in each of the following solutions and write the equation for the formation of the solution: HClO Weak electrolyte(few ions) HClO (aq) H + (aq) + ClO  (aq)

16. 16 EQUIVALENTS OF ELECTROLYTES  Body fluids typically contain a mixture of several electrolytes, such as Na +, Cl –, K + and Ca 2+.  Each individual ion is measured in terms of an equivalent (Eq), which is the amount of that ion equal to 1 mole of positive or negative electrical charge.  For example, 1 mole of Na + ions and 1 mole of Cl – ions are each 1 equivalent (or 1000 mEq) because they each contain 1 mole of charge.

17. 17 EQUIVALENTS OF ELECTROLYTES  An ion with a charge of 2+ or 2– contains 2 equivalents per mole.  Some examples of ions and their equivalents are shown below: Ion Electrical Charge No. of Equivalents in 1 Mole Na + 1+1 Eq Ca 2+ 2+2 Eq Fe 3+ 3+3 Eq Cl – 1–1 Eq SO 4 2– 2–2 Eq

18. 18 EQUIVALENTS OF ELECTROLYTES  In body, the charge of the positive ion is always balanced by the charge of the negative ion.  For example, a solution containing 25 mEq/L of Na + and 4 mEq/L of K + must have 29 mEq/L of Cl – to balance.  Shown next are examples of some common intravenous solutions and their ion concentrations.

19. 19 EQUIVALENTS OF ELECTROLYTES

20. 20 Example 1: Indicate the number of equivalents in each of the following: 2 mol K + 2 mol x 1 Eq 1 mol = 2 Eq 0.5 mol Mg 2+ 0.5 mol x 2 Eq 1 mol = 1 Eq 3 mol CO 3 2  3 mol x 2 Eq 1 mol = 6 Eq

21. 21 Example 2: A typical concentration for Ca 2+ in blood is 8.8 mEq/L. How many moles of Ca 2+ are present in 0.50 L of blood? 0.50 L x Liter blood mEq moles Ca 2+ 8.8 mEq 1 L 1 mol 2 Eq x 1 Eq 10 3 mEq x = 0.0022 mol

22. 22 Example 3: An IV solution contains 155 mEq/L of Cl –. If a patient received 1250 mL of the IV solution, how many moles of chloride were given to him? 1250 mL x 1 L 10 3 mL 1 mol 1 Eq x 155 mEq 1 L x = 0.194 mol 1 Eq 10 3 mEq x

23. 23 Example 4: A sample of Ringer’s solution contains the following concentrations (mEq/L) of cations: Na + 147, K + 4, and Ca 2+ 4. If Cl – is the only anion in the solution, what is the concentration of Cl – in mEq/L? Total cation mEq/L =147 + 4 + 4= 155 mEq/L Total cation mEq/L =Total anion mEq/L Total Cl  mEq/L =155 mEq/L

24. 24 SOLUBILITY  Solubility refers to the maximum amount of solute that can be dissolved in a given amount of solvent.  Many factors affect the solubility of a solute in a solution. Type of solute Type of solvent Temperature Solubility is measured in grams of solute per 100 grams of solvent at a given temperature.

25. 25 SATURATION  A solution that does not contain the maximum amount of solute in it, at a given temperature, is called an unsaturated solution.  A solution that contains the maximum amount of solute in it, at a given temperature, is called a saturated solution. Un-dissolved solid in solution

26. 26 SOLUBILITY  Solubility of most solids in water increases as temperature increases.  Using a solubility chart, the solubility of a solute at a given temperature can be determined.  For example, KNO 3 has a solubility of 80 g/100 g H 2 O (80%) at 40  C.

27. 27 SOLUBILITY OF GASES  Solubility of gases in water decreases as temperature increases.  At higher temperatures more gas molecules have the energy to escape from solution.  Henry’s law states that the solubility of a gas is directly proportional to the pressure above the liquid.  For example, a can of soda is carbonated at high pressures in order to increase the solubility of CO 2. Once the can is opened, the pressure is reduced and the excess gas escapes from the solution.

28. 28 SOLUBLE & INSOLUBLE SALTS  Many ionic solids dissolve in water and are called soluble.  However, some ionic salts do not dissolve in water and do not form ions in solution. These salts are called insoluble salts and remain solid in solution.  Chemists use a set of solubility rules to predict whether a salt is soluble or insoluble.  These rules are summarized next.

29. 29 SOLUBILITY RULES SOLUBLESOLUBLE NO 3  No exceptions Na +, K + NH 4 + No exceptions Cl , Br , I  Except those containing Ag +, Pb 2+ SO 4 2  Except those containing Ba 2+, Pb 2+, Ca 2+

30. 30 SOLUBILITY RULES INSOLUBLEINSOLUBLE S 2 , CO 3 2  PO 4 3  Except those containing Na +, K +, NH 4 + OH  Except those containing Na +, K +, Ca 2+, NH 4 +

31. 31 Example 1: Use the solubility rules to determine if each of the following salts are soluble or insoluble: K 3 PO 4 soluble CaCO 3 insoluble(most carbonates are insoluble) (all salts of K are soluble)

32. 32 Example 2: Using the solubility chart, determine if each of the following solutions is saturated or unsaturated at 20  C: 25 g NaCl in 100 g water Solubility of NaCl at 20  C is 40% Solution is unsaturated

33. 33 Example 2: Using the solubility chart, determine if each of the following solutions is saturated or unsaturated at 20  C: 11 g NaNO 3 in 25 g water Solubility of NaNO 3 at 20  C is 85% Solution is unsaturated

34. 34 Example 2: Using the solubility chart, determine if each of the following solutions is saturated or unsaturated at 20  C: 400. g of glucose in 125 g water Solubility of glucose at 20  C is 80% Solution is saturated

35. 35 FORMATION OF A SOLID  Solubility rules can be used to predict whether a solid, called a precipitate, can be formed when two solutions of ionic compounds are mixed.  A solid is formed when two ions of an insoluble salt come in contact with one another.  For example, when a solution of K 2 CrO 4 is mixed with a solution of Ba(NO 3 ) 2 a yellow insoluble salt BaCrO 4 is produced.

36. 36 AQUEOUS REACTIONS K 2 CrO 4 (aq) + Ba(NO 3 ) 2 (aq) BaCrO 4 (s) + 2 KNO 3 (aq) Solid BaCrO 4 forms precipitate

37. 37 PRECIPITATION REACTIONS  Double replacement reactions in which a precipitate is formed are called precipitation reactions.  To predict a precipitate, follow the steps outlined next.

38. 38 PRECIPITATION REACTIONS 1. Write the reactant ions that form after dissolution. 2.Write the product combinations possible when reactant ions combine. 3.Use solubility rules to determine if any of the products are insoluble. 4.If a precipitate forms, write the formula for the solid. Write other ions that form soluble salts as ions. If no precipitate forms, write “NO REACTION” after the arrow. 5.Cancel ions that appear the same on both sides of the equation (spectator ions), to form Net Ionic Equation.

39. 39 The reaction of K 2 CrO 4 and Ba(NO 3 ) 2 can be predicted as shown below 2 K + +CrO 4 2  Step 1: + Ba 2+ + 2 NO 3   ??? Step 2: 2 K + +CrO 4 2  +Ba 2+ + 2 NO 3   BaCrO 4 (?) + 2K + NO 3  (?) Step 3: Step 4: spectator ions Step 5:Ba 2+ + CrO 4 2   BaCrO 4 (s) precipitate PRECIPITATION REACTIONS 2 K + +CrO 4 2  +Ba 2+ + 2 NO 3   BaCrO 4 (s) + 2K + NO 3  (aq) 2 K + +CrO 4 2  +Ba 2+ + 2 NO 3   BaCrO 4 (s) + 2K + +2NO 3  Net Ionic Equation

40. 40 Predict the products, if any, for the reaction of AgNO 3 and NaCl and write the net ionic equation. Example 1: Ag + +NO 3  Step 1:+ Na + + Cl   ??? Step 2:Ag + +NO 3  +Na + +Cl   Ag + Cl  (?) + Na + NO 3  (?) Step 3:Ag + +NO 3  +Na + +Cl   Ag + Cl  (s) + Na + NO 3  (aq) Step 4:Ag + +NO 3  +Na + +Cl   AgCl (s) + Na + + NO 3  (aq) spectator ions Step 5:Ag + +Cl   AgCl (s) Net Ionic Equation precipitate spectator ions

41. 41 Predict the products, if any, for the reaction of Na 2 SO 4 and Pb(NO 3 ) 2 and write the net ionic equation. Example 2: 2 Na + +SO 4 2  Step 1: + Pb 2+ + 2 NO 3   ??? Step 2: 2Na + +SO 4 2  +Pb 2+ +2NO 3   2Na + NO 3  (?) + Pb 2+ SO 4 2  (?) Step 3: Step 4: Step 5:Pb 2+ +SO 4 2   PbSO 4 (s) precipitate 2Na + +SO 4 2  +Pb 2+ +2NO 3   2Na + NO 3  (aq) + Pb 2+ SO 4 2  (s) 2Na + +SO 4 2  +Pb 2+ +2NO 3   2Na + + 2NO 3  (aq) + PbSO 4 (s) (NIE)

42. 42 Predict the products, if any, for the reaction of PbCl 2 and KI and write the net ionic equation. Example 3: Pb 2+ + 2 Cl  Step 1: + K + + I   ??? Step 2: Pb 2+ + 2 Cl  + K + + I   Pb 2+ I 2  (?) + K + Cl  (?) Step 3: Step 4: Step 5:Pb 2+ + 2 I   PbI 2 (s) precipitate Pb 2+ + 2 Cl  + 2 K + +2 I   Pb 2+ I 2  (s) + 2 K + Cl  (aq) Pb 2+ + 2 Cl  + 2 K + +2 I   PbI 2 (s) + 2 K + + 2 Cl  (NIE)

43. 43 Predict the products, if any, for the reaction of NH 4 Cl and KNO 3 and write the net ionic equation. Example 4: NH 4 + + Cl  Step 1: + K + + NO 3   ??? Step 2: NH 4 + + Cl  + K + + NO 3   NH 4 + NO 3  (?) + K + Cl  (?) Step 3: Step 4: NH 4 + + Cl  + K + + NO 3   NH 4 + NO 3  (aq) + K + Cl  (aq) NH 4 + + Cl  + K + + NO 3   NH 4 + + NO 3  + K + + Cl   No Reaction

44. 44 THE END