#19 Notes Unit 3: Reactions in Solutions Ch. Reactions in Solutions I. Solvation -the act of dissolving (solute (salt) dissolves in the solvent (water)) Hydration: dissolving in water, the universal solvent. **The partially (-) oxygen pulls apart and surrounds the (+) cation. The partially (+) hydrogen pulls apart and surrounds the (-) anion. II. Electrolytes A. Strong Electrolytes: completely ionize (fall apart) and will conduct electricity in a solution. a) soluble salts**: H 2 O Cr 2 (SO 4 ) 3(s) 2 Cr 3+ (aq) + 3 SO 4 (aq) b) strong acids or bases: strong acids start with H : HCl, HNO 3, H 2 SO 4, HClO 4 {#O - #H 2 are strong acids) strong bases end in OH : NaOH, Ca(OH) 2 (hydroxides) B. Weak Electrolytes: only partially ionize (not all of the molecules fall apart) and conduct electricity poorly in solution. a) slightly/marginally soluble salts, weak acids/bases: HC 2 H 3 O 2(aq) H +1 (aq) + C 2 H 3 O 2 (aq)
+1 NH 3 + H 2 O (l) NH 4 (aq) + OH (aq) Nonelectrolytes: may dissolve in water, but produce no ions H 2 O (sugar: C 6 H 12 O 6(s) C 6 H 12 O 6(aq) ) III. Concentration Molarity = mol solute L solution Ex. 1a) Calculate the molarity of 5.23 g Fe(NO 3 ) 2 in 100.0 cm 3 of solution. 5.23 g Fe(NO 3 ) 2 1 mol = 0.029079 mol 179.855 g 100.0 cm 3 1 ml 1 X10-3 L = 0.100 L 1 cm 3 1 ml M = mol / L = 0.029079 mol = 0.291 M Fe(NO 3 ) 2 0.100 L Ex. 1 b) Calculate the concentration of the ions in part a and write the solvation equation. Fe(NO 3 ) 2(s) Fe 2+ (aq) + 2 NO 3 (aq) 0.291 M Fe(NO 3 ) 2 X1 X2 0.291 M Fe 2+ 0.582 M NO 3 Ex. 2) For 0.250 M Cr 3 (PO 4 ) 2 : Calculate the concentration of the ions and write the solvation equation. Cr 3 (PO 4 ) 2(s) 3 Cr 2+ (aq) + 2 PO 4 3- (aq) 0.250 M Cr 3 (PO 4 ) 2 X3 X2 0.750 M Cr 2+ 3-0.500 M PO 4
#20 Notes III. Concentration Continued: Ex. 3) How do you make 500.0 cm 3 of 4.00 M Fe(NO 3 ) 2 from solid Fe(NO 3 ) 2? M = mol / L 500.0 cm 3 1 ml 1 X10-3 L = 0.500 L 1 cm 3 1 ml 4.00 M = mol 0.500 L 2.00 = mol 2.00 mol Fe(NO 3 ) 2 179.855 g = 360 g Fe(NO 3 ) 2 1mol Add 360. g Fe(NO 3 ) 2 and dilute to 0.500 L with water. Ex. 4) How would you make 1.5 L of 0.50 M NaOH from a 6.0 M stock solution of NaOH? If we dilute orange juice, would we still have the same amount of actual orange stuff (just more water)? Yes, same # of orange particles, same # of mols M = mol / L mol undiluted (strong) = mol diluted (weak) M L = mol M undiluted L undiluted = M diluted L diluted (6.0 M) ( L undiluted ) = (0.50 M) ( 1.5 L) L undiluted = 0.13 1.5 L solution -0.13 L NaOH Add 0.13 L of 6.0 M stock solution to 1.37 L H 2 O. IV. Precipitation Reactions: -are reactions that form solids (precipitate). Solubility Rules ** soluble = aqueous marginally/slightly soluble = solid Solubility Table 1. Nitrate (NO 3 ) salts are soluble. 2. Ammonium (NH 4 +1 ) and Group I Alkali metal (Li +1, Na +1, K +1, Cs +1, Rb +1 ) salts are soluble. 3. Chloride, Bromide, Iodide (Cl, Br, I ) salts are soluble, except for salts with Ag +1, Pb 2+ and Hg 2 2+. 4. Sulfate (SO 4 ) salts are soluble, except Ba 2+, Ca 2+, Pb 2+, and Hg 2 2+. 5. Soluble Hydroxides are LiOH, NaOH and KOH. Marginally soluble: Ba(OH) 2, Sr(OH) 2 and Ca(OH) 2. The rest are slightly soluble. 6. Carbonates (CO 3 ), Sulfides (S ), Chromates (CrO 4 ) and Phosphates (PO 4 3- ) are slightly soluble. (All with ammonium or Group 1 Alkali metals are soluble.)
Ex. 1a) Write the molecular equation: BaCl 2(aq) + Na 2 SO 4(aq) Steps: 1) Break compounds apart to their Single ions: Ba 2+ Cl & Na +1 SO 4 2) Mix ions to make new neutral compounds: Ba 2+ & SO 4 Na +1 & Cl = BaSO 4 + NaCl BaCl 2(aq) + Na 2 SO 4(aq) BaSO 4(s) + NaCl (aq) 3) Solubility Rules from Table: Rule #4 Rule #2,3 **If no rule, then soluble (aq). 4) Balance equation: BaCl 2(aq) + Na 2 SO 4(aq) BaSO 4(s) + 2 NaCl (aq) Ex. 1b) Write the complete ionic equation: Step 5) Solids and liquids stay together and don t break apart to their ions. Ba 2+ (aq) + 2 Cl (aq) + 2 Na +1 (aq) + SO 4 (aq) BaSO 4(s) + 2 Na +1 (aq) + 2 Cl (aq) Ex. 1c) Write the net ionic equation: Step 6) Cancel ions. Ba 2+ (aq) + SO 4 (aq) BaSO 4(s) Ex. 2) Write the molecular, complete ionic and net ionic equations. NaBr (aq) + Pb(NO 3 ) 2(aq) Na +1 Br & Pb 2+ NO 3 Na +1 & NO 3 Pb 2+ & Br = NaNO 3 + PbBr 2 X2 NaBr (aq) + Pb(NO 3 ) 2(aq) NaNO 3(aq) + PbBr 2(s) Rule #1,2 Rule #3 2 NaBr (aq) + Pb(NO 3 ) 2(aq) 2 NaNO 3(aq) + PbBr 2(s) 2 Na +1 (aq) + 2 Br (aq) + Pb 2+ (aq) + 2 NO 3 (aq) 2 Na +1 (aq) + 2 NO 3 (aq) + PbBr 2(s) 2 Br (aq) + Pb 2+ (aq)) PbBr 2(s)
Ex. 3) Write the molecular, complete ionic and net ionic equations. Na 2 CO 3(aq) + AlI 3(aq) Na +1 CO 3 & Al 3+ I Na +1 & I Al 3+ & CO 3 X2 X3 = NaI + Al 2 (CO 3 ) 3 Na 2 CO 3(aq) + AlI 3(aq) NaI (aq) + Al 2 (CO 3 ) 3(s) Rule #2,3 Rule #6 3 Na 2 CO 3(aq) + 2 AlI 3(aq) 6 NaI (aq) + Al 2 (CO 3 ) 3(s) 6 Na +1 (aq) + 3 CO 3 (aq) + 2 Al 3+ (aq) + 6 I (aq) 6 Na +1 (aq) + 6 I (aq) + Al 2 (CO 3 ) 3(s) 3 CO 3 (aq) + 2 Al 3+ (aq)) Al 2 (CO 3 ) 3(s)
#21 Notes V. Acid/Base Reactions Ex. 1) Write the molecular, complete ionic and net ionic equations. HClO 4(aq) + Ca(OH) 2(s) H +1 ClO 4 & Ca 2+ OH H +1 & OH Ca 2+ and ClO 4 = H 2 O + Ca(ClO 4 ) 2 X2 Water is a pure liquid. HClO 4(aq) + Ca(OH) 2(s) H 2 O (l) + Ca(ClO 4 ) 2(aq) No rule so aqueous. 2 HClO 4(aq) + Ca(OH) 2(s) 2 H 2 O (l) + Ca(ClO 4 ) 2(aq) Solids and liquids stay together. 2 H +1 (aq) + 2 ClO 4 (aq) + Ca(OH) 2(s) 2 H 2 O (l) + Ca 2+ (aq) + 2 ClO 4 (aq) 2 H +1 (aq) + Ca(OH) 2(s) 2 H 2 O (l) + Ca 2+ (aq) Ex. 2) Write the molecular, complete ionic and net ionic equations. HCN (aq) + LiOH (aq) H +1 CN & Li +1 OH **weak acids stay together. HCN is a weak acid. H +1 & OH Li +1 & CN = H 2 O + LiCN HCN (aq) + LiOH (aq) H 2 O (l) + LiCN (aq) HCN (aq) + Li +1 (aq) + OH (aq) H 2 O (l) + Li +1 (aq) + CN (aq) HCN (aq) + OH (aq) H 2 O (l) + CN (aq)
Ex. 3) Write the molecular, complete ionic and net ionic equations. HC 2 H 3 O 2(aq) + NaOH (aq) **HC 2 H 3 O 2 is a weak acid. H +1 C 2 H 3 O 2 & Na +1 OH H +1 & OH Na +1 C 2 H 3 O 2 = H 2 O + NaC 2 H 3 O 2 HC 2 H 3 O 2(aq) + NaOH (aq) H 2 O (l) + NaC 2 H 3 O 2(aq) HC 2 H 3 O 2(aq) + Na +1 (aq) + OH (aq) H 2 O (l) + Na +1 (aq) + C 2 H 3 O 2 (aq) HC 2 H 3 O 2(aq) + OH (aq) H 2 O (l) + C 2 H 3 O 2 (aq)
#22 Notes VI. Titration Problems Titration is a lab procedure in which acids and bases are mixed. If the concentration of one (acid or base) is known, the other (base or acid) can be calculated. Remember: HCl (aq) + NaOH (aq) H 2 O (l) + NaCl (aq) Acid Base Water Salt NEUTRAL If there are equal amounts of acid and base, an equivalence point (neutral point) is reached. This can be determined by addition of an indicator, like the cabbage juice. At the equivalence point: mols H +1 = mols OH M = mol/ L M H +1 L H +1 = M OH L OH Ex. 1) What volume of 0.250 M HNO 3 is needed to react completely with 29.4 ml of 0.311M Ca(OH) 2? M H +1 L H +1 = M OH L OH (0.250 M HNO 3 ) (L H +1) = (2 OH ) (0.311 M Ca(OH) 2 ) (29.4 ml) (L H +1) = 73.1 ml HNO 3 Ex. 2) 65 ml of 0.325 M Ca(OH) 2 is added to 80 ml of 0.450 M HNO 3. What is the concentration of excess H +1 or OH ions? M = mol/l OH H +1 (2 OH ) (0.325 M) = mol (0.450 M) = mol (0.065 L) (0.080 L) 0.04225 = mol OH 0.0360 = mol H +1 H +1 + OH H 2 O **H +1 is smaller, so limiting reagent 0.0360 mol 0.04225 mol (Initial) -0.0360 mol -0.0360 mol +0.0360 mol (Change) 0 mol H +1 0.00625 mol OH 0.0360 mol (Final) **0.00625 mol excess (left over) OH Find concentration of excess OH : M = mol /L M = 0.00625 mol OH excess 0.065 L + 0.080 L solution M = 0.043 M OH, basic solution
#23 Notes VII. Oxidation Numbers (for individual atoms): Oxidation Rules: 1) Any free element (without a charge) is zero. Ex.) Zn (s), N 2(g) 2) Monoatomic ions have their charge. Ex.) Zn 2+ (aq), F (aq) 3) Hydrogen is always +1, unless it is alone with a positive metal, then the H is. Ex.) NaH = Na +1 H 4) Oxygen is always -2, unless it is in a peroxide, then O is. Ex.) H 2 O 2 = H +1 2O 2 5) Fluorine is always. 6) The sum of oxidation numbers in a group of elements equals the charge of the group. Ex.) NaCl so NaCl =0, SO 4 so SO 4 =-2 ** Remember charges on Periodic Table Columns, for single element ions. (1 st column = +1, 2 nd column = +2 etc.) Ex.1) Find the charge on each element: a) Na 2 SO 4 Na +1 2 S? O -2 4 1 st column rule #4 **Next, add up charges and solve for S. +2?S -8 = 0 (since a neutral compound) S must equal +6 Na +1 2S +6 O -2 4 b) MgCO 3 Mg +2 C? O -2 3 2 nd column rule #4 +2?C -6 = 0 C must equal 4 Mg +2 C +4 O -2 3
c) HSO 4 (H +1 S? O -2 4) +1?S -8 = S must equal +6 H +1 S +6 O -2 4 d) Ca(NO 2 ) 2 Ca 2+ (N? O -2 2) 2 +2?N(X2) -2(4) = 0 +2 2N -8 = 0 2N = 6 N must equal +3 Ca 2+ (N +3 O -2 2) 2 VIII. Oxidation-Reduction Reactions (Redox) Oxidation- element gets more (+), it loses electrons Reduction- element gets more (-), it gains electrons Ex. 1) Mg + Cu(NO 3 ) 2 Mg(NO 3 ) 2 + Cu Mg oxidized Mg 0 + Cu 2+ (NO 3 ) 2 Mg 2+ (NO 3 ) 2 + Cu 0 Cu reduced Mg oxidized, reducing agent Cu reduced, oxidizing agent Ex. 2) SnCl 4 + Fe SnCl 2 + FeCl 2 Sn reduced Sn 4+ Cl 4 + Fe 0 Sn 2+ Cl 2 + Fe 2+ Cl 2 Fe oxidized Sn reduced, oxidizing agent Fe oxidized, reducing agent
Ex. 3) 2 Na + S Na 2 S Na oxid 2 Na 0 + S 0 Na +1 2S -2 S red Na oxidized, reducing agent S reduced, oxidizing agent Ex. 4) HCl + NaOH NaCl + H 2 O H +1 Cl + Na +1 O -2 H +1 Na +1 Cl + H +1 2O -2 not redox, nothing oxidizes or reduces
#24 Notes IX. Balancing Redox Reactions: 1) Write the half reactions. (one for oxidation, one for reduction) 2) For each half reaction: a) Balance the electrons, by looking at the oxidation number of the elements that are oxidizing or reducing. (Make sure these elements are first balanced!) b) Balance all other atoms except H and O. ( If necessary, put in other compounds from the original reaction.) c) Balance O by adding H 2 O. d) Balance H by adding H +. e) Make sure all elements and charges balance. 3) Multiply the half reactions by numbers so that each has the same number of electrons. 4) Add half reactions and cancel what you can. 5) If in basic solution: a) Notice how many H + are left. Add the same amount of OH - as H +, to both sides of the reaction. b) The OH - and H + that are on the same side make H 2 O. c) Cancel any H 2 O that will cancel. 6) Make sure all charges and atoms balance. Ex. 1) MnO 4 + H 2 SO 3 Mn 2+ + HSO 4 (acidic) (Mn +7 O -2 4) + H +1 2S +4 O -2 3 Mn 2+ + (H +1 S +6 O -2 4) Step 1: S Oxidizes Mn reduces H 2 S +4 O 3 HS +6 O 4 Mn +7 O 4 Mn 2+ Step 2a): Add negative electrons to balance changing oxidation # s. +4 +6 + 2e - +7 + 5e - 2+ H 2 S +4 O 3 HS +6 O 4 + 2e - Mn +7 O 4 + 5 e - Mn 2+ Step 2b): all balanced except H, O Step 2c): Add H 2 O to balance O. H 2 O + H 2 SO 3 HSO 4 + 2e - MnO 4 + 5 e - Mn 2+ + 4 H 2 O Step 2d): Add H + to balance H. H 2 O + H 2 SO 3 HSO 4 + 2e - + 3 H + 8 H + +MnO 4 + 5 e - Mn 2+ + 4 H 2 O Step 3: Multiply to get common # of electrons. 5 (H 2 O + H 2 SO 3 HSO 4 + 2e - + 3 H + ) 2 ( 8 H + + MnO 4 + 5 e - Mn 2+ + 4 H 2 O)
Step 4: Add ½ reactions. 5 H 2 O + 5 H 2 SO 3 5 HSO 4 + 10e - + 15 H + ) 16 H + + 2 MnO 4 + 10 e - 2 Mn 2+ + 8 H 2 O cancel e -, reduce H + & H 2 O 5 H 2 SO 3 + H + + 2 MnO 4 5 HSO 4 + 2 Mn 2+ + 3 H 2 O Ex. 2) As 2 O 3 + NO 3 H 3 AsO 4 + NO (acidic) As +3 2O -2 3 + (N +5 O -2 3) H +1 3As +5 O -2 4 + N +2 O -2 As oxidizes As +3 2O 3 H 3 As +5 O 4 N reduces N +5 O 3 N +2 O ** Balance As before doing electrons, so then multiply charges by 2, since there are 2 Arsenics!! +6 +10 +4e- 3e- +5 +2 As +3 2O 3 2 H 3 As +5 O 4 + 4e - 3e - + N +5 O 3 N +2 O 5 H 2 O + As 2 O 3 2 H 3 AsO 4 + 4e - 3e - + NO 3 NO + 2 H 2 O 5 H 2 O + As 2 O 3 2 H 3 AsO 4 + 4e - + 4 H + 4 H + + 3e - + NO 3 NO + 2 H 2 O 3 ( 5 H 2 O + As 2 O 3 2 H 3 AsO 4 + 4e - + 4 H + ) 4 ( 4 H + + 3e - + NO 3 NO + 2 H 2 O) 15 H 2 O + 3 As 2 O 3 6 H 3 AsO 4 + 12e - + 12 H + 16 H + + 12e - + 4 NO 3 4 NO + 8 H 2 O cancel e -, reduce H 2 O & H + 7 H 2 O + 3 As 2 O 3 + 4 H + + 4 NO 3 6 H 3 AsO 4 + 4 NO
Ex.3 ) MnO 4 MnO 4 + MnO 2 (acidic) (Mn +6 O -2 4) (Mn +7 O -2 4) + Mn +4 O -2 2 Mn oxidizes Mn +6 O 4 Mn +7 O 4 Mn reduces Mn +6 O 4 Mn +4 O 2 +6 +7 +1e - 2e - +6 +4 Mn +6 O 4 Mn +7 O 4 + 1e - 2e - + Mn +6 O 4 Mn +4 O 2 MnO 4 MnO 4 + 1e - MnO 4 MnO 4 + 1e - 2e - + MnO 4 MnO 2 + 2 H 2 O 4 H + + 2e - + MnO 4 MnO 2 + 2 H 2 O 2( MnO 4 MnO 4 + 1e - ) 1 (4 H + + 2e - + MnO 4 MnO 2 + 2 H 2 O) 2 MnO 4 2 MnO 4 + 2e - 4 H + + 2e - + MnO 4 MnO 2 + 2 H 2 O 3 MnO 4 + 4 H + 2 MnO 4 + MnO 2 + 2 H 2 O Pull down ions or compounds from the original reaction as needed to balance.
#25 Notes: Balancing continued. Ex. 4) NO 2 + Al NH 3 + AlO 2 (basic) (N +3 O -2 2) + Al 0 N -3 H +1 3 + (Al +3 O -2 2) Al oxidizes Al 0 Al +3 O 2 N reduces N +3 O 2 N -3 H 3 0 +3 +3e - 6e - +3-3 Al 0 Al +3 O 2 + 3e - 6e - + N +3 O 2 N -3 H 3 2 H 2 O + Al AlO 2 + 3e - 6e - + NO 2 NH 3 + 2 H 2 O 2 H 2 O + Al AlO 2 + 3e - + 4 H + 7 H + + 6e - + NO 2 NH 3 + 2 H 2 O 2 ( 2 H 2 O + Al AlO 2 + 3e - + 4 H + ) 1 ( 7 H + + 6e - + NO 2 NH 3 + 2 H 2 O) 4 H 2 O + 2 Al 2 AlO 2 + 6e - + 8 H + 7 H + + 6e - + NO 2 NH 3 + 2 H 2 O 2 H 2 O + 2 Al + NO 2 2 AlO 2 + H + + NH 3 1 H + Step 5: Note how many H +. Add this amount of OH - to both sides, canceling the H +. 2 H 2 O + 2 Al + NO 2 2 AlO 2 + H + + NH 3 + 1 OH - +1 OH - ** H + + OH - H 2 O OH + 2 H 2 O + 2 Al + NO 2 2 AlO 2 + H 2 O + NH 3 Cancel H 2 O, if necessary. OH + 1 H 2 O + 2 Al + NO 2 2 AlO 2 + NH 3
Ex. 5) Do Ex. 3) as if it were BASIC. 3 MnO 4 + 4 H + 2 MnO 4 + MnO 2 + 2 H 2 O Note the number of H +, and add this amount of OH - to both sides. 3 MnO 4 3 MnO 4 + 4 H + 2 MnO 4 + MnO 2 + 2 H 2 O + 4 OH - + 4 OH - ** H + + OH - H 2 O + 4 H 2 O 2 MnO 4 + MnO 2 + 2 H 2 O + 4 OH - (so 4 H + + 4 OH - 4 H 2 O) Cancel/Reduce H 2 O: 3 MnO 4 + 2 H 2 O 2 MnO 4 + MnO 2 + 4 OH - *End of Notes* (Assignments #26-27 are Review Assignments. There are no notes for these assignments.)