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PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 4

PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 4. DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university. CHAPTER 4 CHEMICAL REACTIONS IN SOLUTION. SOLUTION. - A homogeneous mixture of two or more substances Solvent

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PRINCIPLES OF CHEMISTRY I CHEM 1211 CHAPTER 4

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  1. PRINCIPLES OF CHEMISTRY I CHEM 1211CHAPTER 4 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

  2. CHAPTER 4 CHEMICAL REACTIONS IN SOLUTION

  3. SOLUTION - A homogeneous mixture of two or more substances Solvent - The substance present in the greatest quantity Solute - The other substance(s) dissolved in the solvent

  4. SOLUTION - Solutions can exist in any of the physical states Solid Solution - dental fillings, metal alloys (steel), polymers Liquid Solution - sugar in water, salt in water, wine, vinegar Gas Solution - air (O2, Ar, etc. in N2), - NOx, SO2, CO2 in the atmosphere

  5. AQUEOUS SOLUTION - A solution in which water (H2O) is the solvent NaCl solution: solvent is H2O and solute is NaCl Hydrophilic - Substances that dissolve in water - Water loving (NaCl) Hydrophobic - Substances that do not dissolve well in water - Water fearing (hydrocarbons)

  6. AQUEOUS SOLUTION - Ions make aqueous solutions good conductors of electricity - Solution conductivity indicates the presence of ions

  7. AQUEOUS SOLUTION Ionic Compounds - Form ions in aqueous solution (dissociate into component ions) Example - NaCl solution contains Na+ and Cl- ions NaCl(aq) → Na+(aq) + Cl-(aq) - Each ion is surrounded by water molecules - Good conductor of electricity

  8. AQUEOUS SOLUTION Solvation Process - Ions in aqueous solution are surrounded by the H2O molecules - The O atom in each H2O molecule has partial negative charge (δ-) - Attract positive ions - The H atoms have partial positive charge (δ+) - Attract negative ions - Cations and anions are prevented from recombining - Ions disperse uniformly throughout the solution (homogeneous)

  9. AQUEOUS SOLUTION Molecular Compounds - Most molecular compounds do not form ions in aqueous solution - The molecules disperse throughout the solution - Molecules are surrounded by H2O molecules Example - Sucrose solution contains neutral sucrose molecules - Each molecule is surrounded by water molecules - Poor conductor of electricity - A few molecular compounds form ions in aqueous solution - HCl dissociates into H+(aq) and Cl-(aq) - HNO3 dissociates into H+(aq) and NO3-(aq)

  10. NONAQUEOUS SOLUTION - A solution in which another substance other than water is the solvent Examples Alcohol petroleum ether Pentane Carbon tetrachloride

  11. RATE OF DISSOLUTION The rate at which solutes dissolve can be increased by - Grinding or crushing solute particles (size reduction) - Heating - Stirring or agitation

  12. ELECTROLYTES - Substances whose aqueous solutions contain ions NaCl(aq) → Na+(aq) + Cl-(aq) - Two categories: strong and weak electrolytes Strong Electrolytes - Solutes that completely or nearly completely ionize when dissolved in water Salts: NaCl, NH4Cl, KBr, NaNO3 Strong acids: HCl, HNO3, H2SO4 Strong Bases: NaOH, KOH, Ca(OH)2

  13. ELECTROLYTES - Substances whose aqueous solutions contain ions NaCl(aq) → Na+(aq) + Cl-(aq) - Two categories: strong and weak electrolytes Weak Electrolytes - Only a small fraction of solutes ionize when dissolved in water (exhibit a small degree of ionization) Weak acids: acetic acid (HC2H3O2), citric acid (C6H8O7) Weak bases: ammonia (NH3) methylamine, cocaine, morphine

  14. ELECTROLYTES - Single arrow is used to represent ionization of strong electrolytes H2SO4(aq) → H+(aq) + HSO4-(aq) - Ions have no tendency of recombining to form H2SO4 - Double arrow is used to represent ionization of weak electrolytes HC2H3O2(aq) ↔ H+(aq) + C2H3O2-(aq) - This implies reaction occurs in both directions - Chemical equilibrium is when there is a balance in both directions

  15. NONELECTROLYTES - Substances whose aqueous solutions do not contain ions Examples Many molecular compounds Sucrose (C12H22O11) ethanol (C2H5OH)

  16. SOLUBILITY - A measure of how much of a solute can be dissolved in a solvent at a given temperature - Units: grams/100 mL Example Solubility of sugar in water at 20 oC is 204 g/100 mL H2O Three factors that affect solubility - Temperature - Pressure - Polarity

  17. SOLUBILITY Unsaturated Solution - More solute can still be dissolved at a given temperature Saturated Solution - No more solute can be dissolved at a given temperature Supersaturated Solution - Too much solute has temporarily been dissolved (more than solute solubility) Precipitate - Solute (solid) that falls out of solution

  18. SOLUBILITY RULES The best way to determine the solubility of a substance is by experiment - Most nitrates (NO3-) are soluble - Most salts of alkali metals (Group 1A), ammonium (NH4+), acetates (C2H3O2-), and perchlorates (ClO4-) are soluble - Most salts containing Cl-, Br-, and I- are soluble Exceptions: salts of Ag+, Hg22+, Pb2+

  19. SOLUBILITY RULES The best way to determine the solubility of a substance is by experiment - Most sulfates (SO42-) are soluble Exceptions: BaSO4, PbSO4, Hg2SO4, SrSO4 - Most hydroxides (OH-) are slightly soluble Hydroxides of Ba2+, Sr2+, and Ca2+ are marginally soluble - Most salts containing S2-, CO32-, PO43-, CrO42- are insoluble Exceptions: salts of alkali metals and NH4+

  20. PRECIPITATION REACTIONS - Reactions that result in the formation of an insoluble product - The insoluble product (solid) is known as the precipitate - These products have very low solubility in water - Attraction between the oppositely charged ions is so strong that water molecules cannot separate them - A solute is insoluble if less than 0.01 mol of the solute dissolves in 1 L of solvent

  21. PRECIPITATION REACTIONS To predict solubility - Examine the reactants - Identify the ions present - Predict the products - Identify which are soluble and which are insoluble

  22. PRECIPITATION REACTIONS Example AgNO3(aq) + KCl(aq) → white precipitate - Ions present: Ag+, NO3-, K+, Cl- - Possible combinations: AgNO3, AgCl, KCl, KNO3 - Predict products: AgCl and KNO3 - KNO3 is soluble and AgCl is not AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq)

  23. IONIC EQUATIONS - When all soluble strong electrolytes are shown as ions - Chemical equation is balanced - Soluble compounds (aq) are separated into ions (only strong electrolytes) - Insoluble compounds (s), liquids (l), and gases (g) are NOT separated into ions

  24. IONIC EQUATIONS Complete ionic equation - When all ions in both reactants and products are shown AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq) Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) → AgCl(s) + K+(aq) + NO3-(aq)

  25. IONIC EQUATIONS Net Ionic Equation - When spectator ions are cancelled from the complete ionic equation - Net charge on reactant side must equal net charge on product side Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) → AgCl(s) + K+(aq) + NO3-(aq) Ag+(aq) + Cl-(aq) → AgCl(s) - Some ions appear on both reactant and product sides - These ions play no direct role in the reaction - These ions are called spectator ions

  26. IONIC EQUATIONS Neutralization Reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Complete Ionic Equation H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) → Na+(aq) + Cl-(aq) + H2O(l) Net Ionic Equation H+(aq) + OH-(aq) → H2O(l)

  27. CONCENTRATION OF SOLUTIONS - The amount of solute dissolved in a given quantity of solution MOLARITY (M) - The number of moles of solute per liter of solution - A solution of 1.00 M (read as 1.00 molar) contains 1.00 mol of solute per liter of solution

  28. CONCENTRATION OF SOLUTIONS • Calculate the molarity of a solution made by dissolving 2.56 g of • NaCl in enough water to make 2.00 L of solution • - Calculate moles of NaCl using grams and molar mass • Convert volume of solution to liters • - Calculate molarity using moles and liters

  29. CONCENTRATION OF SOLUTIONS After dissolving 1.56 g of NaOH in a certain volume of water, the resulting solution had a concentration of 1.60 M. Calculate the volume of the resulting NaOH solution - Convert grams NaOH to moles using molar mass - Calculate volume (L) using moles and molarity

  30. CONCENTRATION OF IONS Consider: 1.00 M NaCl: 1.00 M Na+ and 1.00 M Cl- 1.00 M ZnCl2: 1.00 M Zn2+ and 2.00 M Cl- 1.00 M Na2SO4: 2.00 Na+ and 1.00 M SO42- Square brackets are commonly used to represent concentration The concentrations of Na+ and Cl- above may be represented as [Na+] = 1.00 M and [Cl-] = 1.00 M

  31. CONCENTRATION OF IONS Calculate the number of moles of Na+ and SO42- ions in 1.50 L of 0.0150 M Na2SO4 solution 0.0150 M Na2SO4 solution contains: 2 x 0.0150 M Na+ ions and 0.0150 M SO42- ions moles Na+ = 2 x 0.0150 M x 1.50 L = 0.0450 mol Na+ moles SO42- = 0.0150 M x 1.50 L = 0.0225 mol SO42-

  32. DILUTION Consider a stock solution of concentration M1 and volume V1 If water is added to dilute to a new concentration M2 and volume V2 moles before dilution = moles after dilution M1V1 = M2V2 Calculate the volume of 3.50 M HCl needed to prepare 500.0 mL of 0.100 M HCl (3.50 M)(V1) = (0.100 M)(500.0 mL) V1 = 14.3 mL

  33. CHEMICAL ANALYSIS (TITRATIONS) Volumetric Analysis - Analysis by volume - Acid-base titrations Gravimetric Analysis - Analysis by mass - Determination of halides by addition of silver nitrate Cl- + AgNO3 → AgCl (white ppt) + NO3- - Determination of sulfates by addition of barium chloride BaCl2 + SO42- → BaSO4 (white solid) + 2Cl-

  34. CHEMICAL ANALYSIS (TITRATIONS) • Calculate the concentration of NaOH solution if 24.50 mL of this • base is needed to neutralize 12.00 mL of 0.225 M HCl solution • - Write balanced equation and determine mole ratio • - Calculate moles of HCl (convert mL to L) • - Determine moles of NaOH • Calculate molarity of NaOH

  35. CHEMICAL ANALYSIS (TITRATIONS) NaOH + HCl → NaCl + H2O 1 mol NaOH : 1 mol HCl Volume HCl = 12.00 mL = 0.01200 L mol HCl = 0.225 M x 0.01200 mL = 0.00270 mol = mol NaOH

  36. CHEMICAL ANALYSIS (TITRATIONS) • How many grams of KOH are needed to neutralize 25.00 mL of • 0.250 M H2SO4 solution • - Write balanced equation and determine mole ratio • - Calculate moles of H2SO4 • - Determine moles of KOH • - Calculate grams of KOH using molar mass

  37. CHEMICAL ANALYSIS (TITRATIONS) 2KOH + H2SO4 → K2SO4 + 2H2O 2 mol KOH : 1 mol H2SO4 mol H2SO4 = 0.250 M x 0.02500 L = 0.00625 mol mol KOH = 2 x 0.00625 mol = 0.0125 mol

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