1. Solubility

2. John A. Schreifels Chemistry 212 Chapter 12-2 Types of Solution n Solution – homogeneous mixture of two or more substances of ions or molecules. E.g. NaCl (aq) 1) Solvent = component which is the component in greater amount. 2) Solute = component which is present in the smaller amount. –Gaseous = gases are completely miscible in each other. –Liquid = gas, liquid or solid solute dissolved in solute. –Solid = mixture of two solids that are miscible in each other to form a single phase. n Colloid – appears to be a homogeneous mixture, but particles are much bigger, but not filterable. E.g. Fog, smoke, whipped cream, mayonnaise, etc. n Suspension: larger particle sizes, filterable. E.g. mud, freshly squeezed orange juice.

3. Dissolving a salt... n A salt is an ionic compound - usually a metal cation bonded to a non-metal anion. n The dissolving of a salt is an example of equilibrium. n The cations and anions are attracted to each other in the salt. n They are also attracted to the water molecules. n The water molecules will start to pull out some of the ions from the salt crystal.

4. n At first, the only process occurring is the dissolving of the salt - the dissociation of the salt into its ions. n However, soon the ions floating in the water begin to collide with the salt crystal and are “pulled back in” to the salt. (precipitation) n Eventually the rate of dissociation is equal to the rate of precipitation. n The solution is now “saturated”. It has reached equilibrium.

5. Solubility Equilibrium: Dissociation = Precipitation In a saturated solution, there is no change in amount of solid precipitate at the bottom of the beaker. Concentration of the solution is constant. Dissolving NaCl in water Na + and Cl - ions surrounded by water molecules NaCl Crystal

7. An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. nonelectrolyteweak electrolytestrong electrolyte 4.1

8. Precipitation Reactions n Mix two aqueous solutions made by dissolving ionic compounds in water. n If a reaction happens, a precipitate (solid) is formed.

10. Predicting Products of Precipitation Reactions 1) Ionic Compounds are Strong Electrolytes –Determine charge on all ions of reactants 2) Using Ion Charges; Predict formula of products. ( + ion of one reactant forms compound with – ion of other reactant) 3) Balance Equation 4) Determine is product is solid or aqueous solution

11. Solubility Rules for Common Ionic Compounds In water at 25 0 C Soluble CompoundsExceptions Compounds containing alkali metal ions and NH 4 + THESE ARE INSOLUABLE NO 3 -, HCO 3 -, ClO 3 - Cl -, Br -, I - Halides of Ag +, Hg 2 2+, Pb 2+ SO 4 2- Sulfates of Ag +, Ca 2+, Sr 2+, Ba 2+, Hg 2+, Pb 2+ Insoluble CompoundsExceptions CO 3 2-, PO 4 3-, CrO 4 2-, S 2- Compounds containing alkali metal ions and NH 4 + OH - Compounds containing alkali metal ions and Ba 2+ 4.2

12. Predicting Products of Precipitation Reactions (Cont) 5) Determine spectator ions (Ions that are still dissolved in water in the product) 6) Write net ionic equation (Only shows ions involved in forming solid)

13. Precipitation Reactions Precipitate – insoluble solid that separates from solution molecular equation ionic equation net ionic equation Pb 2+ + 2NO 3 - + 2Na + + 2I - PbI 2 (s) + 2Na + + 2NO 3 - Na + and NO 3 - are spectator ions PbI 2 Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) precipitate Pb 2+ + 2I - PbI 2 (s) 4.2

14. AgNO 3 (aq) + NaCl (aq) AgCl (s) + NaNO 3 (aq) Ag + + NO 3 - + Na + + Cl - AgCl (s) + Na + + NO 3 - Ag + + Cl - AgCl (s) 4.2 Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

15. Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases 4.3

16. Acids  Produce H + (proton) or (H 3 O) + when dissolved in water  Proton donor HNO 3 (aq) + H 2 O (l)  H 3 O + (aq) + (NO 3 ) - (aq) HNO 3 (aq)  H + (aq) + (NO 3 ) - (aq) H2OH2O

17. Monoprotic acids; Produce one H + when dissolved in water HNO 3 H + + NO 3 - Strong electrolyte, strong acid Diprotic acids; Produce two H + when dissolved in water H 2 SO 4 2 H + + SO 4 -2 Strong electrolyte, strong acid Triprotic acids; Produce three H + when dissolved in water H 3 PO 4 3 H + + PO 4 -3 Weak electrolyte, weak acid 4.3

18. Bases Produce (OH) - when dissolved in water Proton (H + ) acceptor F - (aq) + H 2 O (l) HF (aq) + (OH) - (aq) Na(OH) (s) -----> Na + (aq) + (OH) - (aq) H2OH2O

19. Neutralization Reaction Acid + Base -> Salt + H 2 O

20. Acid + Carbonate -> Salt + CO 2 (g) + H 2 O (l) Carbonate; Contains (CO 3 ) -2 or (HCO 3 ) - Chalk; Ca(CO 3 )

21. Displacement Reactions – Metal Displaces H from acid or water  Metal + Acid -> Salt + H 2 (g)  Metal + Water -> Base + H 2 (g)  Use Activity Series to Know if a Reaction Will Happen

22. Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles of solute liters of solution What mass of KI is required to make 500. mL of a 2.80 M KI solution? volume KImoles KIgrams KI M KI 500. mL= 232 g KI 166 g KI 1 mol KI x 2.80 mol KI 1 L soln x 1 L 1000 mL x 4.5

24. Acid/Base Titrations n Experimental technique that determines the concentration (in Molarity) of an acid (or base) n This is based upon an acid/base neutralization reaction. –ACID +BASE -> SALT + H 2 O n Base (or acid) is added until there is the same amount (same # moles) of base and acid.

25. Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color 4.7

26. Fig. 4.17a,b

27. Acid-Base Titrations Introductory Chemistry 2/e by N Tro, Prentice Hall, 2006, pg 480 Base; (OH) - Acid; H + Acid + Base -> Salt + H 2 O

28. At the endpoint of an acid/base titration…. n Moles acid = Moles base n (MV) acid = (MV) base n Note –If solid; moles = mass/ MM –If aqueous solution; moles = MV

29. What volume of a 1.420 M NaOH solution is Required to titrate 25.00 mL of a 4.50 M H 2 SO 4 solution? 4.7 WRITE THE CHEMICAL EQUATION! volume acidmoles acidmoles basevolume base H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 M acid rx coef. M base 4.50 mol H 2 SO 4 1000 mL soln x 2 mol NaOH 1 mol H 2 SO 4 x 1000 ml soln 1.420 mol NaOH x 25.00 mL = 158 mL

30. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = MiViMiVi MfVfMfVf = 4.5

31. How would you prepare 60.0 mL of 0.2 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f 4.5 V i = MfVfMfVf MiMi = 0.200 x 0.06 4.00 = 0.003 L = 3 mL 3 mL of acid + 57 mL of water= 60 mL of solution

32. Dissolving silver sulfate, Ag 2 SO 4, in water n When silver sulfate dissolves it dissociates into ions. When the solution is saturated, the following equilibrium exists: Ag 2 SO 4 (s)  2 Ag + (aq) + SO 4 2- (aq) n Since this is an equilibrium, we can write an equilibrium expression for the reaction: Ksp = [Ag + ] 2 [SO 4 2- ] Notice that the Ag 2 SO 4 is left out of the expression! Why? Since K is always calculated by just multiplying concentrations, it is called a “solubility product” constant - Ksp.

33. Writing solubility product expressions... n For each salt below, write a balanced equation showing its dissociation in water. n Then write the Ksp expression for the salt. Iron (III) hydroxide, Fe(OH) 3 Nickel sulfide, NiS Silver chromate, Ag 2 CrO 4 Zinc carbonate, ZnCO 3 Calcium fluoride, CaF 2 Try Problems 1 - 8

34. Some K sp Values Note: These are experimentally determined, and may be slightly different on a different Ksp table.

35. Calculating K sp of Silver Chromate n A saturated solution of silver chromate, Ag 2 CrO 4, has [Ag + ] = 1.3 x 10 -4 M. What is the K sp for Ag 2 CrO 4 ? Ag 2 CrO 4 (s)  2 Ag + (aq) + CrO 4 2- (aq) ---- ---- 1.3 x 10 -4 M K sp = [Ag + ] 2 [CrO 4 2- ] K sp = (1.3 x 10 -4 ) 2 (6.5 x 10 -5 ) = 1.1 x 10 -12

36. Calculating the K sp of silver sulfate n The solubility of silver sulfate is 0.014 mol/L. This means that 0.0144 mol of Ag 2 SO 4 will dissolve to make 1.0 L of saturated solution. Calculate the value of the equilibrium constant, K sp for this salt. Ag 2 SO 4 (s)  2 Ag + (aq) + SO 4 2- (aq) --- + 2s + s 2s s K sp = [Ag + ] 2 [SO 4 2- ] = (2s) 2 (s) = (4s 2 )(s) = 4s 3 We know: s = 0.0144 mol/L K sp = 4(0.0144) 3 = 1.2 x 10 -5

37. Calculating solubility, given Ksp n The K sp of NiCO 3 is 1.4 x 10 -7 at 25°C. Calculate its molar solubility. NiCO 3 (s)  Ni 2+ (aq) + CO 3 2- (aq) --- --- + s + s s s K sp = [Ni 2+ ][CO 3 2- ] 1.4 x 10 -7 = s 2 s = = 3.7 x 10 - 4 M

38. Other ways to express solubility... n We just saw that the solubility of nickel (II) carbonate is 3.7 x 10 -4 mol/L. What mass of NiCO 3 is needed to prepare 500 mL of saturated solution? 0.022 g of NiCO 3 will dissolve to make 500 mL solution. Try Problems 9 - 26

39. Calculate the solubility of MgF 2 in water. What mass will dissolve in 2.0 L of water? MgF 2 (s)  Mg 2+ (aq) + 2 F - (aq) ---- ---- + s + 2s s 2s K sp = [Mg 2+ ][F - ] 2 = (s)(2s) 2 = 4s 3 K sp = 7.4 x 10 -11 = 4s 3 s = 2.6 x 10 -4 mol/L

40. Solubility and pH n Calculate the pH of a saturated solution of silver hydroxide, AgOH. Refer to the table in your booklet for the Ksp of AgOH. AgOH (s)  Ag + (aq) + OH - (aq) ---- ---- + s + s s s Ksp = 2.0 x 10 -8 = [Ag + ][OH - ] = s 2 s = 1.4 x 10 -4 M = [OH - ] pOH = - log (1.4 x 10 -4 ) = 3.85 pH = 14.00 - pOH = 10.15

41. The solubility of MgF 2 in pure water is 2.6 x 10 -4 mol/L. What happens to the solubility if we dissolve the MgF 2 in a solution of NaF, instead of pure water? The Common Ion Effect on Solubility

42. Calculate the solubility of MgF 2 in a solution of 0.080 M NaF. MgF 2 (s)  Mg 2+ (aq) + 2 F - (aq) ---- 0.080 M + s + 2s s 0.080 + 2s K sp = 7.4 x 10 -11 = [Mg 2+ ][F - ] 2 = (s)(0.080 + 2s) 2 Since Ksp is so small…assume that 2s << 0.080 7.4 x 10 -11 = (s)(0.080) 2 s = 1.2 x 10 -8 mol/L

43. Explaining the Common Ion Effect The presence of a common ion in a solution will lower the solubility of a salt. n LeChatelier’s Principle: The addition of the common ion will shift the solubility equilibrium backwards. This means that there is more solid salt in the solution and therefore the solubility is lower!

44. Ksp and Solubility n Generally, it is fair to say that salts with very small solubility product constants (Ksp) are only sparingly soluble in water. n When comparing the solubilities of two salts, however, you can sometimes simply compare the relative sizes of their Ksp values. n This works if the salts have the same number of ions! n For example… CuI has Ksp = 5.0 x 10 -12 and CaSO 4 has Ksp = 6.1 x 10 -5. Since the Ksp for calcium sulfate is larger than that for the copper (I) iodide, we can say that calcium sulfate is more soluble.

45. But be careful... Do you see the “problem” here??

46. Mixing Solutions - Will a Precipitate Form? If 15 mL of 0.024-M lead nitrate is mixed with 30 mL of 0.030-M potassium chromate - will a precipitate form? Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq)  PbCrO 4 (s) + 2 KNO 3 (aq)

47. Step 1: Is a sparingly soluble salt formed? We can see that a double replacement reaction can occur and produce PbCrO 4. Since this salt has a very small Ksp, it may precipitate from the mixture. The solubility equilibrium is: PbCrO 4 (s)  Pb 2+ (aq) + CrO 4 2- (aq) Ksp = 2 x 10 -16 = [Pb 2+ ][CrO 4 2- ] If a precipitate forms, it means the solubility equilibrium has shifted BACKWARDS. This will happen only if Qsp > Ksp in our mixture.

48. Step 2: Find the concentrations of the ions that form the sparingly soluble salt. Since we are mixing two solutions in this example, the concentrations of the Pb 2+ and CrO 4 2- will be diluted. We have to do a dilution calculation! Dilution: C 1 V 1 = C 2 V 2 [Pb 2+ ] = [CrO 4 2- ] =

49. Step 3: Calculate Qsp for the mixture. Qsp = [Pb 2+ ][CrO 4 2- ] = (0.0080 M)(0.020 M) Qsp = 1.6 x 10 -4 Step 4: Compare Qsp to Ksp. Since Qsp >> Ksp, a precipitate will form when the two solutions are mixed! Note: If Qsp = Ksp, the mixture is saturated If Qsp < Ksp, the solution is unsaturated Either way, no ppte will form!