1 Topics ? General Properties of Aqueous Solutions ? Precipitation Reactions Acid Base Reactions? - ? Oxidation-Reduction Reactions ? Concentrations of Solutions ? Solution Stoichiometry and Chemical Analysis Figure 4.2: Motion of Ions in Solution Strong and Weak Electrolytes Strong electrolytes completely dissociate in water Wkl l lildii dfWeak e ectro ytes only partially dissoc ate an orm solutions of low ion concentration Molecular Compounds in Water Most polar molecular compounds dissolve in water but no ions are formed example: Non-polar molecular compounds do not dissolve inpp water. example: There are some molecular compounds which when dissolved in water break into ions example: 2 Demonstration Reaction of KI with HgCl 2 Precipitation Reactions (1) Molecular equation: (2) Complete ionic equation:() pq (3) Net ionic equation: Acids (Arrhenius Theory) z Acids are substances that dissociate in aqueous solutions to form H + . z Monoprotic acids: z Polyprotic acids: Bases (Arrhenius Theory) Bases are substances that produce hydroxide ions, OH - , when dissolved in water Figure 4.7: Household Acids and Bases 3 Bronsted Bronsted ?? Lowry Theory Acid: (molecules or ions) donates a proton (H + ) to another molecule or ion in a proton-transfer reaction. Base: (molecules or ions) accepts a proton (H + ) in a proton-transfer reaction Weak Acids and Weak Bases only partially ionized in solution. Weak acids: Weak bases: Molecular Views Comparing Strong Acid HCl and Weak Acid HF in Water Acid ?? Base ReactionsBase Reactions Neutralization reaction acid + base ? water + salt molecular equation: total ionic equation: ionic net ionic equation: Figure 4.10: Reaction of Carbonate With an Acid 4 AcidAcid--Base Reactions with Gas FormationBase Reactions with Net ionic equation: Oxidation and Reduction Oxidation: loss of electrons by an element or an ion Reduction: the gain of electrons by an element or an ion OxidationOxidation--Reduction ReactionsReduction Reactions Oxidation-reduction (redox) reactions involve the transfer of electrons from one species to another Zn(s) + Cu 2+ (aq) ? Zn 2+ (aq) + Cu(s) Oxidation half-reaction: Reduction half-reaction: Oxidizing agent: Reducing agent: Oxidation Numbers Oxidation number increases when the substance is oxidized. S 2+ ? S 4+ + 2e Oxidation number decreases when the substance is reduced. S 4+ + 4e ? S 5 Balancing Simple OxidationOxidation--Reduction Reduction Reactions In a balanced redox equation the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction. Balancing Simple OxidationOxidation--Reduction Reduction Reactions Fe(s) + Cu 2+ (aq) ? Fe 3+ (aq) + Cu(s) Concentrations of Solutions Concentration of a solution: the amount of solute (dissolved substance) in some given amount of solvent, or given amount of solution Molarity: one of several ways to express concentration. It is defined as the number of moles of solute dissolved in one liter of solution. (L)solution of liters (mol) solute of moles )Molarity(M = Example If we dissolve 200.00g NaCl in water and obtain 5.0L of solution, what is the molarity of this solution? M w (NaCl) = 58.45 g/mol) (1) convert mass of NaCl to moles: conv mass of NaCl to moles: (2) calculate the molarity Laboratory Preparation of Solutions Laboratory Preparation of Molar Solution Suppose you are asked to prepare 250mL of 0.12 M solution of CuSO 4 . You are given a bottle of solid CuSO 4 , some distilled water, and a 250mL volumetric flask. (1) - find number of moles of CuSO 4 :find moles of CuS - find mass of CuSO 4 : 6 Dilution Concentrated solution Diluted solution M i = n i / V i M f = n f / V f n i = M i x V i n f = M f x V f n i = n f M i xV i =M f xV f x V = M x V Example How many mL of 5.0 molar solution of HCl you need to prepare 300 mL of 0.1molar solution? Example H 2 S is a foul-smelling compound that can be a component of polluted air. It reacts with aqueous Pb 2+ ion to give black lead sulfide, PbS. H 2 S(aq) + Pb(NO 3 ) 2 (aq) ? PbS(s) + 2HNO 3 (aq) If you have 50.0mL of 0.15M H 2 S and add 40.0mL of 0.2M Pb(NO 3 ) 2 , how many grams of PbS can be formed? *********************************************** Hint: - start with calculation of number of moles of reactants - find out which one is limiting reactant - follow general procedure discussed previously Solution Quantitative Analysis Determination of the amount of a substance or species present in a material. Gravimetric : The amount of a species in a material is determined by converting the species into a product that can be isolated and weighed. Example z A soluble silver compound was analyzed for the percentage of silver by adding sodium chloride solution to precipitate the silver as il hl id If 1 583 f il ds ver c or e. . g o silver compoun gave 1.788 g of silver chloride, what is the mass percent of silver in the compound? z Mw (AgCl) = 143.32 g/mol z Mw(Ag) = 107.87 g/mol. 7 Solution Quantitative Analysis Volumetric analysis A method based on titration (a procedure for dt ii th t f bt Abdetermining the amoun o su s ance y adding a carefully measured volume of a solution with known concentration of B until the reaction of A and B is just complete). Acid ?? Base titrationBase titration Acid ?? Base TitrationBase Titration The equivalence point in the titration is where all the moles of acid(base) have reacted with stoichiometrically equivalent number of moles of base(acid) added from the buret Acid-base indicator is a substance whose color in the acidic solution is different than in basic solution Example Suppose you have a 20.0 mL sample of H 2 SO 4 of unknown concentration. To find the molarity of this solution we will use 0.1 M NaOH. Laboratory procedure:yp - - place 20 mL sample of H 2 SO 4 in a flask - add the acid-base indicator to the flask - - add from the buret 0.1 M NaOH until the indicator changes color - - read the volume of added NaOH (V = 32.56 mL) Solution Suggested steps: (1) write the molecular equation for the reaction (2) calculate number of moles of NaOH used in the titration (3) find the molar ratio n(H 2 SO 4 )/n(NaOH) (4) calculate the number of moles of H 2 SO 4 (5) calculate the molarity of H 2 SO 4 8 Solution dotie Microsoft PowerPoint - chapter 4s.pptx
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