Role of Water Role of Water Chemical Reactions in Aqueous Solution Role of H 2 O(l) as solvent The polar nature of water molecule Two key features: 1. The distribution of bonding electrons O H covalent bond: the electrons are shared But, O atoms attract the electrons strongly the shared electrons are closer to O atom A slightly negative pole (δ ) develops on O the atom A slightly positive pole (δ + ) develops on H atoms H H O H 2 H 2 O H H No net charge separation Charge separation 0 0 H H δ + δ H O H O Undeformed electron cloud Deformed electron cloud Polar nature of the (O δ- H δ+ ) bond Polar nature of the (O δ- H δ+ ) bond 1
Role of Water Role of H 2 O(l) as solvent The polar nature of water molecule Two key features: 2. Its V shape Two polar (O δ- H δ+ ) bonds are working at an angle of 105.5 Ionic Compounds in Water Role of H 2 O(l) as a polar solvent Many compounds dissolve =>The attraction between water and ions prevail and provide higher stabilization energy than the solid state Ionic Compounds in Water Role of H 2 O(l) as a polar solvent Many compounds dissolve =>The attraction between water and ions prevail and provide higher stabilization energy than the solid state Results a net dipole Many don t =>The attraction between water and ions is insufficient and the solid state provide higher stabilization energy than the dissolved state 2
Ionic Compounds in Water Role of H 2 O(l) as a polar solvent For example, AlCl 3 (aq) Cl H 2 O Al 3+ OH 2 Ionic Compounds in Water Role of H 2 O(l) as a polar solvent But many don t =>The attraction between water and ions is insufficient and the solid state provide higher stabilization energy than the dissolved state Ionic Compounds in Water They are electrolyte dissociate to form ions The solution shows significant conductivity water NaCl(s) Na + (aq) + Cl (aq) Cl H 2 O Cl H 2 O 3
Covalent Compounds in Water They are non-electrolyte do not dissociate to form ions The solution shows no conductivity e.g. Table sugar, grain alcohol etc. They all dissolve in water because.. They all have polar bond (O δ- H δ+ ) bond Solubility Rules Compounds Containing Solubility Exceptions Group IA (Na +, K + ) and NH4 + Soluble Nitrates Soluble Acetates (CH3COO ) Soluble Perchlorates (ClO4 ) Soluble Chlorides, Bromides, Iodides Soluble Ag +, Pb 2+, Hg + Fluorides Soluble Mg 2+, Ca 2+, Ag +, Pb 2+, Sr 2+ Sulfates Soluble Ag +, Ba 2+, Sr 2+, Pb 2+, Hg + Sulfides Insoluble Group IA and NH4 + Carbonates Insoluble Group IA and NH4 + Phosphates Insoluble Group IA and NH4 + Hydroxides Insoluble Group IA and NH4 + Chemical Reactions Chemical reactions describe processes involving chemical change The chemical change involves rearranging matter Converting one or more pure substances into new pure substances Reactants Substances combined in the reaction Products Substances produced in the reaction 4
Types of Reactions Combination Reactions Combination Reactions Reaction Types Combination Decomposition Double Replacement Precipitation Acid-base Chemically a H + -transfer reaction Single Replacement Oxidation-reduction Chemically an e -transfer reaction Gas Forming Combination reactions have the form A + B C Two or more reactants produce a single product 2 H 2 + O 2 2 H 2 O = H = O Other combination reactions: 2 Na(s) + S(s) Na 2 S(s) SO 3 (g) + H 2 O(l) H 2 SO 4 (aq) 2Na (s) + Cl 2 (g) 2NaCl (s) 5
Decomposition Reactions Decomposition reactions have the form A B + C Single reactant breaks down into two or more products 2 H 2 O 2 (g) 2 H 2 O(l) + O 2 (g) = H = O Decomposition Reactions Further examples: 2 HgO(s) 2 Hg(l) + O 2 (g) This reaction was used by Joseph Priestley in the discovery of oxygen in 1774 CaCO 3 (s) CaO(s) + CO 2 (g) This reaction is used industrially to produce both lime (CaO) and CO 2 from limestone (CaCO 3 ) Air bags: 2 NaN 3 (s) 2 Na (s) + 3 N 2 (g) Chapter 5 Double Replacement Double replacement reactions are also called metathesis reactions or partner swapping reactions They have the form AX + BY BX + AY + HF + NaOH NaF + H 2 O = H = O = Na = F 6
Double Replacement Double replacement reactions often take place in water and are of two basic types: Acid base neutralization reactions: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) (acid) (base) (salt) (water) H 2 SO 4 (aq) + 2 NaOH(aq) Na 2 SO 4 (aq) + H 2 O(l) Precipitation reactions (solid forms): Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2 NaNO 3 (aq) Here the barium sulfate precipitates out of solution Mixtures and Solutions Solute the component of a solution that is dissolved in another substance Solvent the medium in which a solute dissolved to form a solution Solution a homogenous mixture in which the components are evenly distributed in each other Aqueous Any solution in which water is the solvent Precipitate Insoluble product of a reactant Mixtures and Solutions Electrolyte All ionic compounds that are soluble in water and conduct electricity Strong electrolyte Weak electrolyte Non-electrolyte 7
Mixtures and Solutions Precipitation Reactions Ionic Equations Precipitation reactions K 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq) When an ion appears on both sides of a chemical equation it can be canceled out Spectator Ions Writing Equations Molecular Total ionic Net Ionic Equation CuCl 2 (s) Cu 2+ (aq) + 2Cl - (aq) 100 % dissociation (strong electrolyte) Net ionic CH 3 COOH(aq) CH 3 COO - (aq)+ H + (aq) <5% ionized (weak electrolyte) http://www.dlt.ncssm.edu/tiger/flash/moles/doubledisp_reaction-precipitation.html 8
Ionic Equations Mg(ClO 4 ) 2 (aq) + K 2 CO 3 (aq) MgCO 3 (s) + KClO 4 (aq) Total: Cancel: Net Ionic: Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + NaNO 3 (aq) Total: Cancel: Net ionic: Ionic Equations KNO 3 (aq) + CaCl 2 (aq) Total: Cancel: Net Ionic: Whimbey A tip to good problem solving Break up the problem into parts Identify the information (that can be used to calculate and/or derive other unknown quantities) Recognize the process to do such derivation 9
Molar Mass Molar mass = Mass in grams of one mole of any element, numerically equal to its atomic weight Molar mass of molecules can be determined from the chemical formula and molar masses of elements Each H 2O molecule contains 2 H atoms and 1 O atom Each mole of H 2 O molecules contains 2 moles of H and 1 mole of O One mole of O atoms corresponds to 15.9994 g Two moles of H atoms corresponds to 2 x 1.0079 g Sum = molar mass = 18.0152 g H 2 O per mole Molar Mass Calculate the molar mass of the following Magnesium nitrate, Mg(NO 3 ) 2 1 Mg = 24.3050 2 N = 2x 14.0067 = 28.0134 6 O = 6 x 15.9994 = 95.9964 Molar mass of Mg(NO 3 ) 2 = 148.3148 g Calcium carbonate, CaCO 3 1 Ca = 40.078 1 C = 12.011 3 O = 3 x 15.9994 Molar mass of CaCO 3 = 100.087 g Iron(II) sulfate, FeSO 4 Molar mass of FeSO 4 = 151.909 g Problem 1: Molar Mass The solid magnesium sulfate has seven water of hydration and is written as formula of magnesium sulfate, 7H 2 O. A) Calculate the formula weight of magnesium sulfate and provide its molar mass. B) Calculate the weight percentages of i) sulfur and ii) water in the salt. 10
Calculate Problem 2: Molar Mass A) the molar masses of iron(ii)sulphate and iron(iii)sulphate B) the weight of % iron in both salts Solutions Solution: a homogenous mixture in which the components are evenly distributed in each other Solute: the component of a solution that is dissolved in another substance Solvent: the medium in which a solute dissolved to form a solution Aqueous: any solution in which water is the solvent Solutions The properties and behavior of solutions often depend not only on the type of solute but also on the concentration of the solute. Concentration: ti the amount of solute dissolved d in a given quantity of solvent or solution many different concentration units (%, ppm, g/l, etc) often expressed as Molarity 11
Solution Concentrations Solution Concentrations Solution Concentrations Molarity = moles of solute per liter of solution Designated by a capital M (mol/l) 6.0 M HCl 6.00 moles of HCl per liter of solution. 9.0 M HCl 9.0 moles of HCl per liter of solution. Molarity can be used as a conversion factor. The definition of molarity contains 3 quantities: Molarity = moles of solute volume of solution in liters If you know two of these quantities, you can find the third. Determine the molarity of each solution 2.50 L of solution containing 1.25 mol of solute Molarity = moles of solute volume of solution in liters 225 ml of solution containing 0.486 mole of solute 100. ml of solution containing 2.60 g of NaCl Strategy: g mol molarity 12
Problem 3: Solution Concentrations Calculate the molarity of A)225 ml of solution containing 0.486 mole of aluminum chloride. B)100. ml of solution containing i) 2.6 moles of iron(iii) chloride and ii) 5.83 moles of iron(ii) chloride. Problem 4: Solution Concentrations A)Calculate how many gram of HCl is needed to make a liter of 1.5 molar solution.. B) Calculate the molarity of 117g of sodium chloride dissolved in 5 L solution Example: How many moles of HCl are present in 2.5 L of 0.10 M HCl? Molarity = moles of solute volume of solution in liters Given: Find: 25Lofsoln 2.5 0.10M HCl mol HCl Use molarity as a conversion factor Mol HCl = 2.5 L soln x 0.10 mol HCl 1 L of soln = 0.25 mol HCl 13
Example: What volume of a 0.10 M NaOH solution is needed to provide 0.50 mol of NaOH? Given: 0.50 mol NaOH 010MN 0.10 NaOH Find: vol soln Use M as a conversion factor Vol soln = 0.50 mol NaOH x 1 L soln 0.10 mol NaOH = 5.0 L solution Solutions of exact concentrations are prepared by dissolving the proper amount of solute in the correct amount of solvent to give the desired final volume Determine the proper amount of solute How is the final volume measured accurately? Example: How many grams of CuSO 4 are needed to prepare 250.0 ml of 1.00 M CuSO 4? Given: 250.0 ml solution Find: 1.00 M CuSO 4 g CuSO 4 Conversion factors: Molarity, molar mass Strategy: ml L mol grams 14
Solution Concentrations Given: 250.0 ml solution, 1.00 M CuSO 4 Find: g CuSO 4 Strategy: ml L mol grams gcuso4= 250.00 ml soln x 1L x 100mol 1.00 1000 ml 1 L soln Describe how to prepare the following: a) 500. ml of 1.00 M potassium dichromate (K 2 Cr 2 O 7 ) Steps involved in preparing solutions from pure solids x 159.6 g CuSO4 1 mol b) 100. ml of 3.00 M glucose(c 6 H 12 O 6 ). = 39.9 g CuSO4 15
Steps involved in preparing solutions from pure solids Calculate the amount of solid required Weigh out the solid Place in an appropriate volumetric flask Fill flask about half full with water and mix. Fill to the mark with water and invert to mix. Solutions of exact concentrations can also be prepared by diluting a more concentrated solution of the solute to the desired concentration Many laboratory chemicals such as acids are purchased as concentrated solutions (stock solutions). 12 M HCl 12 M H 2 SO 4 More dilute solutions are prepared by taking a certain quantity of the stock solution and diluting it with water. 16
A given volume of a stock solution contains a specific number of moles of solute. 25 ml of 6.0 M HCl contains 0.15 mol HCl (How do you know this???) If 25 ml of 6.0 M HCl is diluted with 25 ml of water, the number of moles of HCl present does not change. Still contains 0.15 mol HCl Moles solute moles solute = before dilution after dilution Although the number of moles of solute does not change, the volume of solution does change. The concentration of the solution will change since Molarity = mol solute volume solution When a solution is diluted, the concentration of the new solution can be found using: M c V c = moles = M d V d Where, M c = concentration of concentrated solution (mol/l) Vc = volume of concentrated t solution M d = concentration of diluted solution (mol/l) V d = volume of diluted solution 17
Example: What is the concentration of a solution prepared by diluting 25.0 ml of 6.00 M HCl to a total volume of 50.0 ml? Given: V c =250mL 25.0 M c = 6.00 M V d = 50.0 ml Find: M d M c x V c = M d x V d M c V c = M d V d 6.00 M x 25.0 ml = M d x 50.0 ml M d = 6.00 M x 25.0 ml = 3.00 M 50.0 ml Note: V c and V d do not have to be in liters, but they must be in the same units. Describe how to prepare 500. ml of 0.250 M NaOH solution using a 6.00 M NaOH solution. Given: M c = 6.00 M M d = 0.250 M V d = 500.0 ml Find: V c M c V c = M d V d What volume of 2.30 M NaCl should be diluted to give 250. ml of a 0.90 M solution? A) 152.3 ml B) 97.8 ml C) 639 ml D) 252.3 ml 18
Review If you dissolve 9.68 g of potassium chloride in 1.50 L, what is the final molar concentration? A) 0.087 B) 0.129 C) 0.194 D) 0.968 How many grams of sodium sulfate are contained in (dissolved in) 45.0 ml of 3.00 M solution? A) 0.124 B) 12.9 C) 42.6 D) 19.2 Review What volume of 2.06 M potassium permanganate contains 322 g of the solute? Conversion Factors Number of particles Moles Mass Avogadro s number Molar 6.022 x 10 23 mass 19
Number of particles Conversion Factors Avogadro s number 6.022 x 10 23 Moles Solution of a certain concentration Mass Molar mass Volume of solution in L Whimbey Problem Solving Problem solver: The person will solve the assigned problems using pencil, paper, calculator, and any other required items. During the entire process of solving the solver will articulate what she/he is doing out loud. Listener: The person listens to the solver and prompts solver to keep talking. Prompts such as What next?, Does that sound right?, and Talk to me. are appropriate. NEVER give the right answer, only prompt the solver. Source: Beyond Problem Solving and Comprehension by Whimbey and Lochhead, 1984 Molar Conversions Determine the following: The moles of FeCl 3 in a 50.0 g sample The mass of MgCl 2 in a 2.75 mole sample 20
Molar Conversions Determine the following: How many FeCl 3 units are there in a 50.0 g sample? Molar Conversions Determine the following: Molarity of 50.0 g FeCl 3 solution in 500 ml? Acids and Bases Acid a substance that, when dissolved in water, increases the concentration of hydrogen ions (H+) in solution How many MgCl 2 units are there in a 2.75 mole sample? How many grams of MgCl 2 are there in a 250 ml of 2.75 M mole sample? Base a substance that, t when dissolved d in water, increases the concentration ti of hydroxide ions (OH ) in solution Acid base neutralization reactions: HCl (aq) + NaOH(aq) NaCl(aq) + H 2 O(l) (acid) (base) (salt) (water) H 2 SO 4 (aq) + 2 NaOH(aq) Na 2 SO 4 (aq) + H 2 O(l) 21
Acids and Bases as Electrolytes Acids and Bases Acid-Base Reactions Strong acids and bases are strong electrolytes Weak acids and bases are weak electrolytes HNO 3 (aq) + NaOH(aq) Total: Cancel: Net Ionic: 22
Acid-Base Reactions Acid-Base Titrations Problem on Acid-Base Titration H 2 SO 4 (aq) + KOH(aq) Total: Cancel: Net Ionic: Standardized solution A solution whose concentration is known Acid-base indicator A chemical which changes color when the ph of the solution changes Equivalence point - when an exact equivalent of acid is added to a base or vice-a-versa End point - when the indicator indicates the complete neutralization of an acid by a base or vicea-versa Sample problem 4.6 23
Acid-Base Reactions Are Proton (H + ) Transfer Reaction HCl (aq) H + (aq) + Cl (aq) H + OH 2 Cl H 2O How is the HCl molecule in gas phase, where there is no water? Acid-Base Reactions Are Proton (H + ) Transfer Reaction Proton (H + ) Transfer HCl (g) + H 2 O (l) H 3 O + (aq) + Cl (aq) H 2 O H2 O H Acid-Base Reactions Are Proton (H + ) Transfer Reaction When NaOH(aq) is added Proton (H + ) Transfer H 3 O + (aq) + Cl (aq) + [Na + (aq) + OH (aq)] H 2 O(l) + Cl (aq) + Na + (aq) + HOH (l)] H + OH H + 2 Cl Cl H 2O HCl (g) H 2 O H + OH 2 OH 2 Solvated hydronium cation Getting rid of spectator ions H 3 O + (aq) + OH (aq) H 2 O(l) + HOH (l) or,2h 2 O (l) Or, H + (aq) + OH (aq) H 2 O(l) 24
Molar Conversions Concentrations of Acids Concentrations of Acids How many moles of H + ions are present in 2.5 moles of H 2 SO 4? Given: 2.5 moles H 2 SO 4 Find: moles H + Mol H + = 2.5 mol H 2 SO 4 x 2 mol H + Mol H + = 5.0 mol H + Conversion factor: molar ratio 1 mol H 2 SO 4 The ph scale ph = -log [H + ] ph of vinegar = -log (1.6 x 10-3 M) = - (-2.80) = 2.80 ph of pure water = -log (1.0 x 10-7 M) = - (-7.00) = 7.00 ph of blood = -log (4.0 x 10-8 M) = - (-7.40) = 7.40 ph of ammonia = -log (1.0 x 10-11 M) = - (-11.00) = 11.00 Lower the concentration of H +, higher the ph acidic substance, ph< 7 Basic substance, ph >7 neutral, ph = 7 On serial dilution of acid solution, the ph increases because the concentration of H + ions decreases with dilution Concentration ph 0.1M HCl 1 0.01 M HCl 2 0.001 M HCl 3 The H + concentration of a solution of known ph can be calculated using the following equation: [H + ] = 10 -ph 25
Concentrations of Acids Calculate ph of 0.0065 M HCl solution. Review What volume of 8.00 M sulfuric acid should be diluted to produce 0.500 L of 0.250 M solution? Redox Reactions Redox reactions are also called Single replacement or substitution reactions They have the form A + BX B + AX, where A and B are elements and BX and AX are compounds Calculate the concentration of H + ion in a solution of ph 7.5. + What s the ph of the final solution? H 2 + CuO = H = O H 2 O + Cu = Cu 26
Redox Reactions Other redox reactions 3 C(s) + 2 Fe 2 O 3 (s) 4 Fe(s) + 3 CO 2 (g) Cu(NO 3 ) 2 (aq) + Zn(s) Zn(NO 3 ) 2 (aq) + Cu(s) Single replacement reactions are also oxidationreduction (REDOX) reactions. Oxidation Redox Reactions Are Electron (e ) Transfer Reactions The oxidation number of an atom is the charge that atom would have if the compound was composed of ions. Oxidation of Mg (0) to Mg (II): 2Mg (s) + O 2 (g) 2MgO (s) Reduction of Fe (III) to Fe(0): Fe 2 O 3 (s) + 3CO (g) 2Fe (s) + 3 CO 2 (g) Oxidation Number Reduction OXIDATION NUMBER Reduction of Ag (I) to Ag and oxidation of Cu (0) to Cu (II): 2Ag + (aq) + Cu(s) 2 Ag (s) + Cu 2+ (aq) 27
SF 6 and FeS Oxidation Number Some Reactions Produce Gas Reactions leading to the formation of an insoluble gas Some Reactions Produce Gas Examples Na 2 CO 3 (aq) + 2HCl (aq) 2NaCl (aq) + CO 2 (g) +H 2 O (l) 28