1. Precipitation Reactions and Acid-Base Reactions Chemistry 142 B James B. Callis, Instructor Autumn Quarter, 2004 Lecture #11

2. The Solubility of Ionic Compounds in Water The solubility of Ionic Compounds in water depends upon the relative strengths of the electrostatic forces between ions in the ionic compound and the attractive forces between the ions and water molecules in the solvent. There is a tremendous range in the solubility of ionic compounds in water! The solubility of so called “insoluble” compounds may be several orders of magnitude less than ones that are called “soluble” in water, for example: Solubility of NaCl in water at 20 o C = 365 g/L Solubility of MgCl 2 in water at 20 o C = 542.5 g/L Solubility of AlCl 3 in water at 20 o C = 699 g/L Solubility of PbCl 2 in water at 20 o C = 9.9 g/L Solubility of AgCl in water at 20 o C = 0.009 g/L Solubility of CuCl in water at 20 o C = 0.0062 g/L

3. Compare with Table 4.1 of Zumdahl

5. Precipitation Reactions: A solid product is formed When ever two aqueous solutions are mixed, there is the possibility of forming an insoluble compound. Let us look at some examples to see how we can predict the result of adding two different solutions together. Pb(NO 3 ) (aq) + 2 NaI(aq) When we add these two solutions together, the ions can combine in the way they came into the solution, or they can exchange partners. In this case we could have lead nitrate and sodium iodide, or lead iodide and sodium nitrate formed, to determine which will happen we must look at the solubility rules to determine what could form. Rule 3 in Table 4.1 (lecture) indicates that lead iodide will be insoluble, so a precipitate will form! The remainder (sodium nitrate) is soluble. Pb(NO 3 ) 2 (aq) + 2 NaI(aq)

6. Precipitation Reactions: Will a precipitate form? If we add a solution containing potassium chloride to a solution containing ammonium nitrate, will we get a precipitate? KCl(aq) + NH 4 NO 3 (aq) = K + (aq) + Cl - (aq) + NH 4 + (aq) + NO 3 - (aq) By exchanging cations and anions we see that we could have potassium chloride and ammonium nitrate, or potassium nitrate and ammonium chloride. In looking at the solubility rules, Table 4.1: If we mix a solution of sodium sulfate with a solution of barium nitrate, will we get a precipitate? Table 4.1 shows that barium sulfate is insoluble - therefore we expect: Na 2 SO 4 (aq) + Ba(NO 3 ) 2 (aq)

8. Problem 11-1: Predicting Whether a Precipitation Reaction Occurs (see Table 4.1); Writing Equations: a) calcium nitrate and sodium sulfate solutions are added together. Ca(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) Ca 2+ (aq) + 2 NO 3 - (aq) +2 Na + (aq) + SO 4 2- (aq) Molecular Equation Total Ionic Equation Net Ionic Equation Spectator Ions are b) ammonium sulfate and magnesium chloride are added together.

9. Testing the Reactivity of the Cations Ag +, Ba 2+, and Fe 3+ with the Anions Cl -, SO 4 2-, and OH - CationNaCl(aq) (Cl - ) Na 2 SO 4 (aq) (SO 4 2- ) NaOH(aq) (OH - ) Ag + White ppt (AgCl) No ReactionWhite ppt that turns brown (AgOH - > Ag 2 O) Ba 2+ No ReactionWhite ppt (BaSO 4 ) No Reaction Fe 3+ Yellow solution, but no solid No ReactionReddish Brn ppt Fe(OH) 3

10. A Scheme for Selective Precipitation

11. Acid - Base Reactions : Neutralization Rxns. An Acid is a substance that produces H + (H 3 O + ) ions when dissolved in water. A Base is a substance that produces OH - ions when dissolved in water. Acids and Bases are electrolytes, and their strength is categorized in terms of their degree of dissociation in water to make hydronium or hydroxide ions, resp. Strong acids and bases dissociate completely, and are strong electrolytes. Weak acids and bases dissociate only to a small extent (<<100%) and are weak electrolytes. The generalized reaction between an Acid and a Base is: HX(aq) + MOH(aq) MX(aq) + H 2 O(l) Acid + Base = Salt(aq) + Water

12. Selected Acids and Bases Acids Bases Strong (100% of H + ) Strong (100% of OH - ) Hydrochloric, HCl Sodium hydroxide, NaOH Hydrobromic, HBr Potassium hydroxide, KOH Hydroiodoic, HI Calcium hydroxide, Ca(OH) 2 Nitric acid, HNO 3 Strontium hydroxide, Sr(OH) 2 Sulfuric acid, H 2 SO 4 Barium hydroxide, Ba(OH) 2 Perchloric acid, HClO 4 Weak (low % of H + ) Weak (low % yield of OH - ) Hydrofluoric, HF Ammonia, NH 3 Phosphoric acid, H 3 PO 4 Acetic acid, CH 3 COOH (or HC 2 H 3 O 2 )

13. Problem 11-2: Writing Balanced Equations for Neutralization Reactions - I Problem: Write balanced chemical reactions (molecular, total ionic, and net ionic) for the following Chemical reactions: a) calcium hydroxide(aq) and hydroiodic acid(aq) b) lithium hydroxide(aq) and nitric acid(aq) c) barium hydroxide(aq) and sulfuric acid(aq) Plan: These are all strong acids and bases, therefore they will make water and the corresponding salts. Solution: a) Ca(OH) 2 (aq) + 2HI(aq) CaI 2 (aq) + 2H 2 O(l) Ca 2+ (aq) + 2 OH - (aq) + 2 H + (aq) + 2 I - (aq) Ca 2+ (aq) + 2 I - (aq) + 2 H 2 O(l) 2 OH - (aq) + 2 H + (aq) 2 H 2 O(l)

14. Problem 11-2: Writing Balanced Equations for Neutralization Reactions - II b) LiOH(aq) + HNO 3 (aq) c) Ba(OH) 2 (aq) + H 2 SO 4 (aq)

16. Finding the Concentration of Acid from an Acid - Base Titration Volume (L) of base (difference in buret readings) needed to titrate Moles of base needed to titrate Moles of acid which were titrated M (mol/L) of original acid soln. M (mol/L) of base molar ratio volume (L) of acid

17. Potassium Hydrogenphthalate KHC 8 H 4 O 4 C O OK+K+ C O H O K+K+ O C O O C O H+H+ = “KHP” = a common acid used to titrate bases (M = 204.2 g/mol)

18. Problem 11-3: Finding the Concentration of Base from an Acid - Base Titration - I Problem: A titration is performed between sodium hydroxide and KHP (204.2 g/mol) to standardize the base solution, by placing 50.00 mg of solid potassium hydrogenphthalate in a flask with a few drops of an indicator. A buret is filled with the base, and the initial buret reading is 0.55 mL; at the end of the titration the buret reading is 33.87 mL. What is the concentration of the base? Plan: Use the molar mass of KHP to calculate the number of moles of the acid, from the balanced chemical equation, the reaction is equal molar, so we know the moles of base, and from the difference in the buret readings, we can calculate the molarity of the base. Solution: The rxn is: KHP(aq) + OH - (aq) --> KP - (aq) + H 2 O(l) or HKC 8 H 4 O 4 (aq) + OH - (aq) KC 8 H 4 O 4 - (aq) + H 2 O(l)

19. Problem 11-3: Finding the Concentration of Base from an Acid - Base Titration - II moles KHP = 50.00 mg KHP Volume of base = Final buret reading - Initial buret reading = one mole of H + = one mole of OH - ; therefore ____________ moles of KHP will titrate _____________ moles of NaOH. molarity of NaOH = = moles L molarity of base =

22. Answers to Problems in Lecture #11 1.(a) Net ionic equation: Ca 2+ (aq) + SO 4 2- (aq) = CaSO 4 (s) (b) No reaction occurs. 2.(b) Total ionic reaction: Li + (aq) + OH - (aq) + H + (aq) + NO 3 - (aq) = Li + (aq) + NO 3 - (aq) + H 2 O(l) (c) Total ionic reaction: Ba 2+ (aq) + 2 OH - (aq) + 2 H + (aq) + SO 4 2- (aq) = BaSO 4 (s) + 2 H 2 O(l) 3. 0.07349 M