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UNIT 2: CHEMICAL REACTIONS. 4.1, 4.2 Aqueous Solutions _____________ = _____________ + _______________ Water is polar!!! solutionsolutesolvent O - 3.5.

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Presentation on theme: "UNIT 2: CHEMICAL REACTIONS. 4.1, 4.2 Aqueous Solutions _____________ = _____________ + _______________ Water is polar!!! solutionsolutesolvent O - 3.5."— Presentation transcript:

1 UNIT 2: CHEMICAL REACTIONS

2 4.1, 4.2 Aqueous Solutions _____________ = _____________ + _______________ Water is polar!!! solutionsolutesolvent O H EN Diff so VERY Polar bond   H 2 O strong dipole must be soluble!!

3 Aqueous Reactions © 2012 Pearson Education, Inc. Solutions Solutions are defined as homogeneous mixtures of two or more pure substances. The solvent is present in greatest abundance. All other substances are solutes.

4 Solubility of a substance in water is determined by the AFFINITY of the water molecules for the solute particles For INSOLUBLE cmpds: the attraction of the (+) and (-) ions for each other is stronger than their attraction for POLAR water molecules

5 An ELECTROLYTE is a substance whose aqueous solution conducts electric current. The current is carried by IONS (+) and (-) Key: The more ions in solution, the more current is carried. Thus if NO ions in a solution (just neutral molecules) then NO current can be conducted.

6 STRONG ELECTROLYTES: are excellent “conductors” because ___ solute particle are in the form of ______ allions 1) soluble ionics NaClNH 4 NO 3 KI See Solubility Table! 2) strong acids neutral molecules that “ionize”(react with) H 2 O HI, HBr, HCl,HClO 3, HClO 4, HNO 3, H 2 SO 4 “Is Bright and Clear, No Snow, 3,4” 3) strong bases soluble hydroxides LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 backwards “J”“Can, Sara, Bat” 7 8 all end in “ic”

7 Aqueous Reactions © 2012 Pearson Education, Inc. Acids There are only seven strong acids: Hydrochloric (HCl) Hydrobromic (HBr) Hydroiodic (HI) Nitric (HNO 3 ) Sulfuric (H 2 SO 4 ) Chloric (HClO 3 ) Perchloric (HClO 4 )

8 Aqueous Reactions © 2012 Pearson Education, Inc. Bases The strong bases are the soluble metal salts of hydroxide ion: Alkali metals Calcium Strontium Barium

9 Aqueous Reactions © 2012 Pearson Education, Inc. Dissociation An electrolyte is a substances that dissociates into ions when dissolved in water.

10 Aqueous Reactions © 2012 Pearson Education, Inc. Solutions An electrolyte is a substance that dissociates into ions when dissolved in water. A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

11 Aqueous Reactions © 2012 Pearson Education, Inc. Electrolytes and Nonelectrolytes Soluble ionic compounds tend to be electrolytes.

12 WEAK ELECTROLYTES: are poor “conductors” because ________ solute particles are in the form of _____. very fewions (Almost all solute remains as neutral molecules) 1) weak acidsHFHC 2 H 3 O 2 HNO 2 2) weak (molecular) bases NH 3 CH 3 NH 2

13 NaCl (s) > Na + (aq) + Cl - (aq) “hydrated” or (“solvated”) ions H2OH2O IowaState visual

14 strong electrolyte weak electrolyte non- electrolyte ALL ionsVERY FEW ions NO ions Str Acids Str Bases Soluble Ionics Weak Acids Weak Bases POLAR molecular compounds

15 Aqueous Reactions © 2012 Pearson Education, Inc. Electrolytes and Nonelectrolytes Soluble ionic compounds tend to be electrolytes.

16 Aqueous Reactions © 2012 Pearson Education, Inc. Electrolytes and Nonelectrolytes Molecular compounds tend to be nonelectrolytes, except for acids and bases.

17 Aqueous Reactions © 2012 Pearson Education, Inc. Electrolytes A strong electrolyte dissociates completely when dissolved in water. A weak electrolyte only dissociates partially when dissolved in water.

18 Aqueous Reactions © 2012 Pearson Education, Inc. Strong Electrolytes Are… Strong acids Strong bases Soluble ionic salts

19 BaCl 2(s) dissolves BaCl 2(s) > Ba 2+ (aq) + 2Cl - (aq) H2OH2O “DISSOCIATION” of an ionic compound in water: ionic solid solvated ions SOLUBLE IONIC breaking bonds between ions in the solid then making bonds between each ion and polar H 2 O molecules that surround them

20 Aqueous Reactions © 2012 Pearson Education, Inc. Dissociation When an ionic substance dissolves in water, the solvent pulls the individual ions from the crystal and solvates them. This process is called dissociation.

21 HCl solution Str ACID ionizes 100% in H 2 O HCl (g) + H 2 O (l) ---> H 3 O + (aq) + Cl - (aq) “IONIZATION” of an ACID in H 2 O. ALL solute is ions. STRONG ACID

22 NaOH solution Str BASE dissociates 100% in H 2 O NaOH (s) > Na + (aq) + OH - (aq) H2OH2O (ionic) ALL solute is ions STRONG BASE

23 acetic acid solution mostly molecules & very few ions HC 2 H 3 O 2(l) + H 2 O (l) H 3 O + (aq) + C 2 H 3 O 2 - (aq) Weak ACID ionizes only slightly in H 2 O WEAK ACID

24 Methods of Expressing Solution Concentration 1) Mass % = mass of solute x 100 (mass of solute + mass of solvent) Mass %A = mass of component A ratio x 100 total mass of solution “10% MgSO 4(aq) ” means 10.g MgSO g H 2 O = 100.g solution ratio: 10.g solute 100.g solution

25 100% % = 93.20% Problem: How many grams H 2 O are needed to prepare 525g of 6.80%NaCl (aq) ? 525g x 6.80% = 35.7g NaCl solute so 525g solution g NaCl = 489.3g H 2 O solute solvent OR you could find %H 2 O first, then do a proportion: 93.20g = x g 100g 525g x = 489.3g H 2 O

26 2) Volume% = volume of solute x 100 (volume of solute + volume of solvent) Volume%A = volume of component A x 100 total volume of solution again it is a ratio mL of A 100mL solution

27 Problem: How many liters of water are needed to prepare 1.00 liter of 23% by volume CH 3 OH (aq) ? 1.00L x 23% = 0.23L CH 3 OH so 1.00L solution L CH 3 OH = 0.77L H 2 O OR 100% - 23% = 77% 77L = x L 100L 1.00L “Proof” is a volume %. Proof = 2 x volume % alcohol C 2 H 5 OH) also called “EtOH” so “180 proof” whiskey means 90% ethanol by volume!! solute solvent

28 3) Parts per million or parts per billion (ppm) (ppb) used for very small solute concentrations (for mass or volume) ppm(mass) = mass of solute x 10 6 ppb….x 10 9 mass of solution OR ppm = mg solute ppb = µg solute kg solution kg solution “0.1ppm by mass of Pb 2+ ions” in drinking water means 0.1mg Pb 2+ in 1kg water

29 Problem: Find ppm of 0.5g of Ca 2+ ions in 2500g tapwater ppm = 0.5g x g = 200ppm Ca 2+

30 4) MOLARITY (M) = moles of solute Liter of solution mol M L

31 Aqueous Reactions © 2012 Pearson Education, Inc. Molarity Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different. Molarity is one way to measure the concentration of a solution: moles of solute volume of solution in liters Molarity (M) =

32 Aqueous Reactions © 2012 Pearson Education, Inc. Mixing a Solution To create a solution of a known molarity, one weighs out a known mass (and, therefore, number of moles) of the solute. The solute is added to a volumetric flask, and solvent is added to the line on the neck of the flask.

33 Problem: Find Molarity when 53.3grams FeCl 3 are dissolved in 755mL of solution. Now find Molarity of chloride ions in this solution: FeCl 3 Fe Cl - = 53.3g x 1 mol g 0.755L Fe – Cl – 3(35.45) g/mol to convert to moles ……. = 0.435M (means mol/L FeCl 3 ) _ _ _ 3 M 1 to 3 ratio 3(0.435M) = 1.305M Cl - 1 to 1 ratio

34 Problem: How many moles of K + ions are present in 25.0mL of 2.50M K 2 O? moles = Molarity x Volume Liters = (2.50M) (0.0250L) = 0.625mol K 2 O 2 mol K + ions 1 mol K 2 O 0.625mol K 2 O = 1.250mol K + ions x

35 mg Calculate Molarity of 2.0ppm Cr 6+ ions in well water. (find mol/L) (in this very dilute solution, 1mL has a mass of ~1g) 2.0ppm means _______________ ______ = ______ ______ = ______ Convert: numerator ______----> ________ denominator _____----> ________ 2.0mg C r 6+ ions 1kg solution 1kg H 2 O 1L H 2 O 1g H 2 O1mL H 2 O mgmol kgL 2.0mg x gx 1 mol g = mol x x mol 1 L = 3.8 x M 1kg = 1L

36 2. How much LiF is needed to prepare 450.mL of a 1.20M sol’n. (M is ratio of mol solute to L solution so find mol…then mol-->g) _____________ = _______ or mol = M x L 1.20mol LiF 1 L x mol 0.450L 0.54 X = 0.540mol 0.540mol Li – 6.94 F – g/mol x g 1 mol = g LiF

37 3. Find Molarity of a 5.00% Pb(NO 3 ) 2 solution by mass. (find mol/L) D = 1.05g/mL 5.00% Pb(NO 3 ) 2 means ______________ Convert: numerator _____----> ______ denominator _____----> ______ 5.00g Pb(NO 3 ) 2 100g solution g g mol L Pb – N – 2(14.00) O - 6(16.00) g/mol 5.00g x 1 mol = g mol Pb(NO 3 ) 2 100g x mL 1.05g x 1 L = 1000mL L sol’n

38 Pb(NO 3 ) 2(s) Pb 2+ (aq) + 2NO 3 - (aq) NONE of this ALL of this in solution in solution [Pb 2+ ] = _____________ [NO 3 - ] = _____________ [total ion] = _____________ M = mol L = 0.159M 0.159M M (x 2) 0.477M (x 3) H2OH2O

39 1) Find total moles of iodide ions in 250.mL of a 0.750M BaI 2 sol’n. ____________ = _________ 0.750mol 1 L x mol 0.250L BaI 2(s) > Ba 2+ (aq) + 2I - (aq) H2OH2O x = 0.188mol BaI mol 2(0.188)mol 0.376mol of I - ions

40 2. Find [Cl - ] when 9.82g CuCl 2 are dissolved in 600.mL of sol’n. M = ________ X __________ 9.82g 0.600L Cu – Cl – 2(35.45) g/mol g 1 mol = 0.122M CuCl 2 x __________ 0.122M CuCl 2 2 mol Cl - 1 mol CuCl 2 = 0.244mol Cl -

41 3. Which solution of strong electrolytes contains the largest # of chloride ions? A or B 50.0mL of 0.60M MgCl mL of 0.40M NaCl mol = M x L = (0.60M)(0.0500L) = 0.030mol MgCl 2 MgCl 2(s) ---> Mg 2+ (aq) + 2Cl - (aq) 0.030mol2(0.030)mol 0.060mol Cl - = (0.40M)(0.200L) = 0.080mol NaCl NaCl (s) ---> Na + (aq) + Cl - (aq) 0.080mol 0.080mol Cl mol 1 to 2 1 to 1

42 4. How would you prepare: 1.00L of a 2.5M HNO 3 solution from conc. (16M) nitric acid? dilution formula: M conc x V conc = M dil x V dil (16M) (2.5M) ( mL) (x mL) = x = 160mL) 1000mL 2) add H 2 O to line 3 ) cap & mix 1)add 160mL conc acid

43 Aqueous Reactions © 2012 Pearson Education, Inc. Dilution One can also dilute a more concentrated solution by –Using a pipet to deliver a volume of the solution to a new volumetric flask, and –Adding solvent to the line on the neck of the new flask.

44 Aqueous Reactions © 2012 Pearson Education, Inc. Dilution The molarity of the new solution can be determined from the equation M c  V c = M d  V d, where M c and M d are the molarity of the concentrated and dilute solutions, respectively, and V c and V d are the volumes of the two solutions.

45 How would you prepare: 1.00L of a 0.250M NaNO 3 solution from the pure solid? Na – N – O - 3(16.00) 85.00g/mol 0.250M is 0.250mol NaNO 3 in 1Liter sol’n 0.250mol x 85.00g 1mol = g

46 5. Find [Li + ] when 20.0mL of 1.0M LiCl is added to 80.0mL of 2.0M LiBr 1) find moles of each cmpd (solute) 2) find total moles of solute ions 20.0mL of 1.0M LiCl (0.0200L)(1.0M) = 0.020mol LiCl 80.0mL of 2.0M LiBr (0.0800L)(2.0M) = 0.16mol LiBr 0.020mol Li mol Li mol Li + total 100.0mL total volume [Li + ] = mol L = 0.18mol 0.100L = 1.8M

47 TYPES OF REACTIONS IN AQUEOUS SOLUTION 1)precipitation (formation of insoluble solid product) 2)acid-base (formation of water, a neutralization reaction) 3)oxidation-reduction reaction (changing oxidation #s) METATHESIS reaction = DOUBLE REPLACEMENT AB + CD -----> _____( ) + _____( ) make it go! ADCB (s) ppt (l) H 2 O or other molec cmpd (WA) (g) GAS “spectator ions” are (aq) product

48 Aqueous Reactions © 2012 Pearson Education, Inc. Metathesis (Exchange) Reactions Metathesis comes from a Greek word that means “to transpose.” AgNO 3 (aq) + KCl(aq)  AgCl(s) + KNO 3 (aq)

49 Aqueous Reactions © 2012 Pearson Education, Inc. Metathesis (Exchange) Reactions Metathesis comes from a Greek word that means “to transpose.” It appears as though the ions in the reactant compounds exchange, or transpose, ions: AgNO 3 (aq) + KCl(aq)  AgCl(s) + KNO 3 (aq)

50 Aqueous Reactions © 2012 Pearson Education, Inc. Precipitation Reactions When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

51 1) NaCl (aq) + Pb(NO 3 ) 2(aq) ----> Writing equations: 1)The “MOLECULAR” equation is a balanced equation with all formulas & state of matter symbols Ex: ________ ( ) + _______ ( ) PbCl 2 s aq NaNO Pb 2+ Cl - Na + NO 3 -

52 Aqueous Reactions © 2012 Pearson Education, Inc. Molecular Equation The molecular equation lists the reactants and products in their molecular form: AgNO 3 (aq) + KCl(aq)  AgCl(s) + KNO 3 (aq)

53 2) The “COMPLETE IONIC” equation shows all reactants and products separated into individual ions as they exist in solution: strong acids, strong bases and soluble ionics (remember SA,SB,SI) + PbCl 2(s) Note: Never separate the ppt, H 2 O or gas 2Na + (aq) + 2Cl - (aq) + Pb 2+ (aq) + 2NO 3 - (aq) -----> 2Na + (aq) + 2NO 3 - (aq)

54 Aqueous Reactions © 2012 Pearson Education, Inc. Ionic Equation In the ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions. This more accurately reflects the species that are found in the reaction mixture: Ag + (aq) + NO 3 − (aq) + K + (aq) + Cl − (aq)  AgCl(s) + K + (aq) + NO 3 − (aq)

55 3) The “NET IONIC EQUATION” is missing the spectator ions, (cancelled) showing only the particles involved in the reaction that made PRODUCTS. (made it go!!) Net ionic equation: 3) Pb 2+ (aq) + 2Cl - (aq) ----> PbCl 2(s)

56 Aqueous Reactions © 2012 Pearson Education, Inc. Net Ionic Equation To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right: Ag + (aq) + NO 3 − (aq) + K + (aq) + Cl − (aq)  AgCl(s) + K + (aq) + NO 3 − (aq)

57 Aqueous Reactions © 2012 Pearson Education, Inc. Net Ionic Equation To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. The only things left in the equation are those things that change (i.e., react) during the course of the reaction: Ag + (aq) + Cl − (aq)  AgCl(s)

58 Aqueous Reactions © 2012 Pearson Education, Inc. Net Ionic Equation To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. The only things left in the equation are those things that change (i.e., react) during the course of the reaction. Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions: Ag + (aq) + NO 3 − (aq) + K + (aq) + Cl − (aq)  AgCl(s) + K + (aq) + NO 3 − (aq)

59 Aqueous Reactions © 2012 Pearson Education, Inc. Writing Net Ionic Equations 1.Write a balanced molecular equation. 2.Dissociate all strong electrolytes. 3.Cross out anything that remains unchanged from the left side to the right side of the equation. 4.Write the net ionic equation with the species that remain.

60 SHORTCUT TO NET IONIC EQUATIONS: 1)BaCl 2(aq) + K 2 SO 4(aq) Net ionic equation: ______________________________________ insoluble soluble spect ions BaSO 4(s) Ba 2+ (aq) + SO 4 2- (aq) BaSO 4(s)

61 2) FeCl 2 (aq) + Na 3 PO 4(aq) Net ionic equation: ______________________________________ Fe 3 (PO 4 ) 2 insoluble soluble spect ions 3Fe 2+ (aq) + 2PO 4 3- (aq) Fe 3 (PO 4 ) 2

62 1)HCl (aq) + KOH (aq) Net ionic equation: ______________________________________ water soluble H + (aq) + OH - (aq) H 2 O (l) spect ions

63 SOLUBILITY RULES Any COMPOUNDS containing these ions are SOLUBLE: “CLAAAN” ! Cl A AA N Chlorates Acetates Ammonium Alkali metals Nitrates ClO 3 - C 2 H 3 O 2 - NH 4 + Li + NO 3 - or K + ClO 4 - CH 3 COO - Na + Rb + Cs + Also all chlorides, bromides and iodides are SOLUBLE except those of Pb 2+, Ag +, Hg 2 2+ “PAH” ! All 8 strong bases (hydroxides) are SOLUBLE.

64 Precipitation Reactions in Aqueous Solution Steps for Determining the Mass of Product (ppt) Formed: 1) Identify the substances present in the mixture of solutions and determine what reaction occurs. 2)Write the balanced molecular equation with state of matter symbols. 3)Calculate moles of each reactant given.(moles = M x V L ) 4)Solve to find the “Limiting Reactant”. 5) Convert moles of product to grams of product.

65 Problem: What mass of iron(III) hydroxide is produced when 35.0mL of 0.250M Fe(NO 3 ) 3 is mixed with 55.0mL of 0.180M KOH? BAL. EQ: __________ + ________ ----> _________ +________ Fe(NO 3 ) 3 KOH Fe(OH) 3 KNO (0.250M)(0.0350L) (0.180M)(0.0550L) mol mol (aq) (s) (aq) mol Fe(NO 3 ) 3 1 mol Fe(NO 3 ) 3 3 mol KOH = mol KOH Need this much KOH, but don’t have so Fe(NO 3 ) 3 in excess And KOH is LR

66 KOHFe(OH) 3 ppt = mol Fe(OH) 3 x g = 1 mol Fe – O – 3(16.00) H - 3(1.01) g/mol 0.353g Fe(OH) mol mol KOH 3 mol KOH 1 mol Fe(OH) mol

67 Now, calculate the Molarity of all ions left in solution: Fe(NO 3 ) 3 3KOH Fe(OH) 3 3KNO 3 ppt! mol mol mol subscripts coefficients BEFORE AFTER Fe 3+ NO 3 - K + OH - Fe(OH) 3 K + NO mol mol 1 to mol disregard!! mol spectator ions No CHANGE mol mol mol 1 to 1 Since KOH is LR, then only OH- ions got used up as they formed Fe(OH) 3 ppt. Some Fe 3+ ions will be left over.

68 Now to make solving clearer, write “net ionic equation”: & work backwards!! Fe OH > Fe(OH) mol ppt formed [Note: 1:1 ratio between Fe 3+ mol used & Fe(OH) 3 made] Fe 3+ given: Total Vol. Sol’n: mL Fe 3+ used: mL Fe 3+ left: mL = ______L mol used up to make ppt 3( )mol mol ALL USED UP mol mol mol

69 Ions remaining in Sol’n : K + _________mol [K + ] = _______ NO 3 - ________mol [NO 3 - ] = _______ Fe 3+ _________mol [Fe 3+ ] = ________ L M M M

70 __________ + ________ ----> _________ +________ When 200.0mL of 1.0M NaOH is mixed with 300.0mL of 0.50M AlBr 3, find mass of ppt & concentration of all ions left in solution. NaOH AlBr 3 NaBrAl(OH) (1.0M)(0.2000L) (0.50M)(0.3000L) 0.20mol0.15mol (s) (aq) NaOH ---> ppt AlBr 3 ---> ppt 3 mol = 1 mol 0.20mol x mol 1 mol = 1 mol 0.15mol x mol x = 0.067mol pptx = 0.15mol ppt LR = NaOH

71 Mass of ppt: 0.067mol Al(OH) 3 Al – O – 3(16.00) H - 3(1.01) 78.01g/mol x 78.01g = 1 mol g Al(OH) 3

72 Al 3+ given: Total Vol. Sol’n: mL Al 3+ used: mL Al 3+ left: mL = ______L BEFORE AFTER Na + OH - Al 3+ Br - Al(OH) 3 Na + Br mol 0.20 mol 0.15 mol 0.45 mol spectator ions 0.45 mol 0.20 mol mol Al OH > Al(OH) 3 LR mol mol 3(0.067) 0.20mol all used up 0.15mol mol 0.08mol

73 Ions Conc left: : Na + = _____mol [Na + ] = _______ Br - = _____mol [Br - ] = _______ Al 3+ = ____mol [Al 3+ ] = ________ L M M M

74 means it contains “at least” 15% urea (NH 2 ) 2 CO 30% P 2 O 5 15% K 2 O

75 In this experiment, phosphorus in P 2 O 5 will be determined by precipitation of the sparingly soluble salt magnesium ammonium phosphate hexahydrate according to the reactions: 1) P 2 O 5(s) + 3H 2 O (l) -----> 2 H 3 PO 4 (aq) 2) H 3 PO 4(aq) ----> 2H + (aq) + HPO 4 2- (aq) 3) 5H 2 O (l) + Mg 2+ (aq) + NH 4 + (aq) + OH - (aq) + HPO 4 2- (aq) > MgNH 4 PO 4 6H 2 O (s) phosphoric acid non-metallic oxide ionizes, the 1 st H + then the 2 nd H + PPT has captured all P atoms we add water We add NH 3(aq) and MgSO 4(aq)

76 Sample Problem: If a 10.00g sample of soluble plant food yields 10.22g of ppt. MgNH 4 PO 4 6H 2 O (s), calculate the %P and %P 2 O 5 in the sample? 1)solve for grams of P in the sample: using Law of Definite Proportions g MgNH 4 PO 4 6H 2 O = 30.97g P 10.22g ppt x gP X = 1.290g P molar mass ppt = molar mass P actual mass ppt x grams P

77 Thus %P = grams P x 100 g plant food 1.290g P_____ x 100 = 12.90%P Ans! 10.00g plant food 2) solve for grams of P 2 O 5 : using Law of Definite Proportions 2(30.97)g P = g P 2 O g P X g X = 2.956g P 2 O 5

78 %P 2 O 5 = g P 2 O 5 x 100 g plant food 2.956g P 2 O 5 __ x 100 = 29.56% P 2 O 5 Ans! 10.00g plant food Pre-Lab Questions: (on loose leaf & copy questions) 1. The label on a plant food reads Explain what this means. 2. What is %age of phosphorus element in the plant food in Question 1? 3. What is %age of potassium element in the plant food in Question 1? 4. What is the molar mass of MgSO 4 7H 2 O known as Epsom salts?

79 ACID-BASE REACTIONS are ______________________reactions. The “driving force” behind neutralization reactions is the ___________________ or a__________________. !!!! ACID + BASE -----> SALT + H 2 O _____ + _____ -----> _____ + _____ neutralization formation of water weak electrolyte HA BOH BABA H2OH2O

80 Definitions: Arrhenius acid _____________________________ Acid increases _________________________________ Ex: recall: HCl (g) + H 2 O (l) -----> H 3 O + (aq) + Cl - (aq) simplified: Arrhenius base _____________________________ Ex: recall: Ca(OH) 2(s) > Ca 2+ (aq) + 2OH - (aq) produces H 3 O + ions in water [ H + ] (molarity of H + ions) in solution HCl (aq) -----> H + (aq) + Cl - (aq) releases OH - ions in water H2OH2O

81 Bronsted-Lowry acid ______________________________ Bronsted-Lowry base _____________________________ Ex: recall: NH 3(g) + H 2 O (l) -----> NH 4 + (aq) + OH - (aq) _____ ______ ______ _______ is a proton donor (H + ) is a proton acceptor (H + ) WEAK BASE BLB BLAConj Base Conj Acid

82 Aqueous Reactions © 2012 Pearson Education, Inc. Acids The Swedish physicist and chemist S. A. Arrhenius defined acids as substances that increase the concentration of H + when dissolved in water. Both the Danish chemist J. N. Brønsted and the British chemist T. M. Lowry defined them as proton donors.

83 Aqueous Reactions © 2012 Pearson Education, Inc. Bases Arrhenius defined bases as substances that increase the concentration of OH − when dissolved in water. Brønsted and Lowry defined them as proton acceptors.

84 + Br - (aq) STRONG ACID-STRONG BASE Molec. EQ: Complete Ionic EQ: Net Ionic EQ: HBr (aq) + KOH (aq) ----> ________ + _____ (aq) (l) H2OH2O KBr H 2 O (l) H + (aq) + Br - (aq) + K + (aq) + OH - (aq) ---> + K + (aq) (spect will cancel) SA,SB,SI get pulled apart SAME for every SA/SB reaction.

85 Aqueous Reactions © 2012 Pearson Education, Inc. Acid-Base Reactions In an acid–base reaction, the acid donates a proton (H + ) to the base.

86 Aqueous Reactions © 2012 Pearson Education, Inc. Neutralization Reactions Generally, when solutions of an acid and a base are combined, the products are a salt and water: CH 3 COOH(aq) + NaOH(aq)  CH 3 COONa(aq) + H 2 O(l)

87 ________ + _____ WEAK ACID-STRONG BASE Molec. EQ: Complete Ionic EQ: Net Ionic EQ: HF (aq) + NaOH (aq) ----> (aq) (l) H2OH2O NaF + F - (aq) H 2 O (l) HF (aq) + Na + (aq) + OH - (aq) ---> + Na + (aq) WA + F - (aq) H 2 O (l) HF (aq) + OH - (aq) --->

88 STRONG ACID-WEAK BASE Molec. EQ: Complete Ionic EQ: Net Ionic EQ: ________ + _________ HNO 3(aq) + Mg(OH) 2(aq) ----> (aq) (l) H2OH2OMg(NO 3 ) H 2 O (l) 2H + (aq) + 2NO 3 - (aq) + Mg(OH) 2(aq) ---> + Mg 2+ (aq) + 2NO 3 - (aq) 2H 2 O (l) + Mg(OH) 2(aq) ---> + Mg 2+ (aq) 2H + (aq)

89 FINDING MOLARITY OF IONS 5.25g of Ba metal is placed in enough water to make 45.0mL. Find [OH - ] _________ + _______ > ______ + _________ Find moles Ba reacted: Find moles OH - : Find Molarity of OH - : M = mol L Ba (s) H 2 O (l) H 2(g) Ba(OH) 2(aq) 2 1 mol > 1 mol Ba 2+ ions 2 mol OH - ions 5.25g Ba x 1 mol = g mol Ba x 2 mol OH - = 1 mol Ba mol OH [OH - ] = 1.70M

90 TITRATION: is an analytical technique whereby a solution of known concentration and volume (titrant) is reacted with a solution of known volume (analyte) to determine its concentration.

91 Aqueous Reactions © 2012 Pearson Education, Inc. Titration Titration is an analytical technique in which one can calculate the concentration of a solute in a solution.

92 M H+ x V acid = M OH- x V base moles of H + = moles of OH - H + (aq) + OH - (aq) > H 2 O (l) Neutralization equation: MOLARITY acid ion x VOLUME acid = MOLARITY base ion x VOLUME base

93 1. Find molarity of HCl if 36.7mL of acid is needed to titrate 43.2mL of 0.236M Ca(OH) 2 Note: HCl ---> H + + Cl - Ca(OH) > Ca OH - M H+ x V acid = M OH- x V base [H + ] = [HCl] [OH - ] = 2[Ca(OH) 2 ] [OH - ] = 2(0.236M) = 0.472M (0.472M)(43.2mL) (36.7mL) = (x) x = (0.472M)(43.2mL) (36.7mL) x = 0.556M = [H + ] = [HCl]

94 2. What volume of 0.80M Ba(OH) 2 is needed to neutralize 50.0mL of a 0.20M H 3 PO 4 solution? Note: H 3 PO > 3H + + PO 4 3- Ba(OH) > Ba OH - M H+ x V acid = M OH- x V base [H + ] = 3[H 3 PO 4 ] [H + ] = 3(0.20M) = 0.60M [OH - ] = 2[Ba(OH) 2 ] [OH - ] = 2(0.80M) = 1.60M (x) (1.60M) (0.60M)(50.0mL) = x = 19mL Ba(OH)

95 FINDING MOLARITY OF ALL IONS LEFT IN SOLUTION When 12.5g KOH is added to 450.0mL of 0.250M H 2 SO 4, find concentration of all ions left in solution following this neutralization reaction. Is solution acidic? basic/ or neutral? 1) Initially, find moles of base added & moles of acid present: H 2 SO 4(aq) + KOH (s) -----> H 2 O (l) + K 2 SO 4(aq) ( )( ) moles moles M0.4500L g KOH K – O – H g/mol x 1 mol = 56.11g 0.223mol 2 spect

96 2) Find LR and moles H 2 O (l) formed: H 2 SO 4 H 2 OKOH H 2 O 1 mol = 2 mol 0.113mol x mol 2 mol = 2 mol 0.223mol x mol x = 0.226mol H 2 O x = 0.223mol H 2 O LR = KOH

97 3) Before & After moles of all ions: H + SO 4 2- K + OH > H 2 O K + SO ) Net Ionic EQ: 2(0.113) mol mol mol mol mol mol mol H + (aq) + OH - (aq) H 2 O (l) mol mol mol ALL used up LR H + given: Total Vol. Sol’n: mL = H + used: H + left: mol mol mol L

98 _1_1 5) now find Molarity of all ions left is solution spectators: K + _____mol = [K + ] = L SO 4 2- _____mol = [SO 4 2- ] = L excess ion when forming water: H + ______mol = [H + ] = L Since acid ion is left over, the solution is therefore ______ Formula:pH = - log [H + ] so pH = ____ Acidic solutions: pH < 7 Basic solutions:pH > 7 Neutral solutions:pH = M M M acidic = - log 0.007M _1_1

99 Aqueous Reactions © 2012 Pearson Education, Inc. Oxidation-Reduction Reactions An oxidation occurs when an atom or ion loses electrons. A reduction occurs when an atom or ion gains electrons. One cannot occur without the other.

100 Aqueous Reactions © 2012 Pearson Education, Inc. Oxidation Numbers To determine if an oxidation–reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.

101 OXIDATION-REDUCTION REACTIONS “REDOX” OIL-RIG OXIDATION is REDUCTION is A “redox” reaction always involves a RULES for ASSIGNING OXIDATION NUMBERS: 1)Any free element is __ 2)Monatomic ions is loss of electrons & an increase in oxidation #. gain of electrons & a decrease in oxidation #. change in oxidation #s. 0 H 2 Fe C P 4 o o o o the charge on the ion Na + Pb 4+ N 3- Br

102 3) H in a molecular cmpd is ___ H in a metal hydride is ___ 4) O in a compound is ___ except peroxides is ___ 5) F in a compound is always ____ In ionic compounds, it is the ion’s charge anyway, but in a molecular compound, F atom gets a -1 because it has the highest EN of all atoms (ie greatest attraction for electrons in a shared pair) NaH FeH Na 2 O 2 H 2 O NF BrCl IBr highest EN halogen gets the -1

103 6) sum of oxidation #s in a compound is ___ 7) sum of oxidation #s in a polyatomic ion is its _________ 0 charge CrO 4 2- = Cr = +6 NO 2 - = N = +3

104 REACTION TYPES. Yes, ONLY IF oxid #s change after the reaction! 1) Synthesis: E + E ---> C E + C ---> C C + C ---> C __________ ___________ ___________ 2) Decomposition: C ---> E + E C ---> E + C C ---> C + C __________ __________ ____________ 3) Single Replacement: E + C ----> E + C _______________ Yes Usually Not Yes Usually Not Yes always

105 4) Double Replacement (metathesis): C + C ----> C + C ____________ 5) Combustion: C x H y + O > CO 2 + H 2 O __________________ (2 ions switching places) NO Yes always -?

106 A REDUCING AGENT causes __________ in another reactant, but itself undergoes __________. Metal elements love to _____ electrons (________) and form (+) ions, so they are good reducing agents. 2Na (s) + O 2(g) -----> 2Na 2 O (s) oxidation reduction lose oxidize oxid red Oxid. Agent: Red. Agent: Na O2O2

107 An OXIDIZING AGENT causes __________ in another reactant, but itself undergoes __________. Non-metal elements love to _____ electrons (_________) and form (-) ions, so they are good oxidizing agents. oxidation reduction reducegain

108 Note: Atoms in POLYATOMIC IONS with HIGH oxidation #s are ______________________________________________ good reducers therefore strong oxidizing agents KMnO 4 = 0= Mn = +7 +1

109 It is possible to get a “non-integer” oxidation #: Means Fe 3 O = 0= /3

110 Broader Definition:OXIDATION is the ______________ REDUCTION is the _______________ Ex: PbO (s) + CO (g) ---> Pb (s) + CO 2(g) PbO _______________ and _______________ CO _______________ and _______________ Common method for reducing metal oxides to get the metal element isolated. gain of oxygen loss of oxygen reduces from +2 to 0 oxidizes from +2 to +4 loses oxygen gains oxygen

111 6FeCl 2(aq) + 6HCl (aq) + NaClO 3(aq) --> 6FeCl 3(aq) + NaCl (aq) +3H 2 O (l) 6Fe 2+ (aq) + 12Cl - (aq) + 6H + (aq) + 6Cl - (aq) + Na + (aq) + ClO 3 - (aq) -----> 6Fe 3+ (aq) + 18Cl - (aq) + Na + (aq) + Cl - (aq) + 3H 2 O (l) Net ionic EQ: 6Fe 2+ (aq) + 6H + (aq) + ClO 3 - (aq) --->6Fe 3+ (aq) + Cl - (aq) + 3H 2 O (l) Oxidation: Reduction: 6Fe > 6Fe e - ClO e - --> Cl - Everything balances: elements, e - s & charges (+12) + (+6) + (-1) = (+18) + (-1) + (0) (+17) = (+17)

112 We can consider a balanced net ionic equation as the sum of an oxidation ½ reaction + a reduction ½ reaction

113 today tomorrow

114 BALANCE by the Half-Reaction Method (see steps!) MnO C 2 O > Mn 2+ + CO 2 oxidation reduction MnO >Mn 2+ C 2 O > CO 2 + 4H 2 O 8H + + (-2) = (0) + 2e - (+8) + (-1) = (+2) + (0) 5e (+7) = (+2) 2 5 (in acidic solution) 5C 2 O > 10CO e - 10e H + + 2MnO > 2Mn H 2 O 5C 2 O H + + 2MnO > 10CO 2 + 2Mn H 2 O

115 Redox Balancing in BASIC solution Al + MnO > MnO 2 + Al(OH) Al-->Al(OH) 4 - MnO >MnO 2 4H 2 O++ 2H 2 O4H H + +4OH - 4OH OH - 4H 2 O (-4) + (0) + (0) = (-1) + (0) + 3e - (0) + (-1) = (0) +(0) + (-4) 4OH - + 4H 2 O+Al + 4H 2 O + MnO > Al(OH) H 2 O+ MnO 2 + 2H 2 O +4OH - + 2H 2 O Al + 2H 2 O + MnO > Al(OH) MnO 2 oxidation reduction 1)

116 Cl > Cl - + OCl oxidation reduction 2) Cl 2 ---> OCl - Cl > Cl H 2 O+ +4H + + 4OH - 4OH - + 4H 2 O (-4) + (0) + (0) = (-2) + (0) (0) = (-2) + 2e - 2e - + 4OH - + 2H 2 O ++ Cl > 2OCl - + 4H 2 O + 2Cl - 2H 2 O 4OH - + 2Cl > 2OCl - + 2H 2 O+ 2Cl - Cl 2 hypochlorite ion

117 SINGLE-REPLACEMENT REACTIONS all are redox!!! E + C ----> E + C When Element is a METAL: spect. ion_____ ____ + _____ ----> ____ + _____ The metal element _________________________________ When Element is a NON-METAL: spect. ion_____ ____ + _____ ----> ____ + _____ The non-metal element ______________________________ _ MABAMBMB B-B- M oxidizes ↑ while metal ion A + reduces↓ A+A+ N AN B N reduces ↓ while nonmetal ion B - ↑ oxidizes

118 Problem: Zn (s) + AgNO 3(aq) ---> _____ + _________ E o oxid = _____ E o red = _____ E o cell = _____ Complete ionic EQ: ____________________________________________________ Net Ionic EQ: Ag (s) Zn(NO 3 ) 2(aq) V 0.80V 1.56V yes Zn (s) + 2Ag + (aq) + 2NO 3 - (aq) -----> 2Ag (s) + Zn 2+ (aq) + 2NO 3 - (aq) Zn (s) + 2Ag + (aq) -----> 2Ag (s) + Zn 2+ (aq) Spect ion: _____NO 3 -

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