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

2. CHAPTER 4 CHEMICAL REACTIONS IN SOLUTION

3. - 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 SOLUTION

4. - 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 (O 2, Ar, etc. in N 2 ), - NO x, SO 2, CO 2 in the atmosphere SOLUTION

5. - A solution in which water (H 2 O) is the solvent NaCl solution: solvent is H 2 O 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) AQUEOUS SOLUTION

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

7. 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 AQUEOUS SOLUTION

8. Solvation Process - Ions in aqueous solution are surrounded by the H 2 O molecules - The O atom in each H 2 O 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) AQUEOUS SOLUTION

9. Molecular Compounds - Most molecular compounds do not form ions in aqueous solution - The molecules disperse throughout the solution - Molecules are surrounded by H 2 O 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) - HNO 3 dissociates into H + (aq) and NO 3 - (aq) AQUEOUS SOLUTION

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

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

12. - 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, NH 4 Cl, KBr, NaNO 3 Strong acids: HCl, HNO 3, H 2 SO 4 Strong Bases: NaOH, KOH, Ca(OH) 2 ELECTROLYTES

13. - 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 (HC 2 H 3 O 2 ), citric acid (C 6 H 8 O 7 ) Weak bases: ammonia (NH 3 ) methylamine, cocaine, morphine ELECTROLYTES

14. - Single arrow is used to represent ionization of strong electrolytes H 2 SO 4 (aq) → H + (aq) + HSO 4 - (aq) - Ions have no tendency of recombining to form H 2 SO 4 - Double arrow is used to represent ionization of weak electrolytes HC 2 H 3 O 2 (aq) ↔ H + (aq) + C 2 H 3 O 2 - (aq) - This implies reaction occurs in both directions - Chemical equilibrium is when there is a balance in both directions ELECTROLYTES

15. NONELECTROLYTES - Substances whose aqueous solutions do not contain ions Examples Many molecular compounds Sucrose (C 12 H 22 O 11 ) ethanol (C 2 H 5 OH)

16. - 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 o C is 204 g/100 mL H 2 O Three factors that affect solubility - Temperature - Pressure - Polarity SOLUBILITY

17. 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 SOLUBILITY

18. The best way to determine the solubility of a substance is by experiment - Most nitrates (NO 3 - ) are soluble - Most salts of alkali metals (Group 1A), ammonium (NH 4 + ), acetates (C 2 H 3 O 2 - ), and perchlorates (ClO 4 - ) are soluble - Most salts containing Cl -, Br -, and I - are soluble Exceptions: salts of Ag +, Hg 2 2+, Pb 2+ SOLUBILITY RULES

19. The best way to determine the solubility of a substance is by experiment - Most sulfates (SO 4 2- ) are soluble Exceptions: BaSO 4, PbSO 4, Hg 2 SO 4, SrSO 4 - Most hydroxides (OH - ) are slightly soluble Hydroxides of Ba 2+, Sr 2+, and Ca 2+ are marginally soluble - Most salts containing S 2-, CO 3 2-, PO 4 3-, CrO 4 2- are insoluble Exceptions: salts of alkali metals and NH 4 + SOLUBILITY RULES

20. - 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 PRECIPITATION REACTIONS

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

22. Example AgNO 3 (aq) + KCl(aq) → white precipitate - Ions present: Ag +, NO 3 -, K +, Cl - - Possible combinations: AgNO 3, AgCl, KCl, KNO 3 - Predict products: AgCl and KNO 3 - KNO 3 is soluble and AgCl is not AgNO 3 (aq) + KCl(aq) → AgCl(s) + KNO 3 (aq)

23. - 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 IONIC EQUATIONS

24. Complete ionic equation - When all ions in both reactants and products are shown AgNO 3 (aq) + KCl(aq) → AgCl(s) + KNO 3 (aq) Ag + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq) → AgCl(s) + K + (aq) + NO 3 - (aq) IONIC EQUATIONS

25. 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) + NO 3 - (aq) + K + (aq) + Cl - (aq) → AgCl(s) + K + (aq) + NO 3 - (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 IONIC EQUATIONS

26. Neutralization Reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H 2 O(l) Complete Ionic Equation H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) → Na + (aq) + Cl - (aq) + H 2 O(l) Net Ionic Equation H + (aq) + OH - (aq) → H 2 O(l) IONIC EQUATIONS

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 ZnCl 2 : 1.00 M Zn 2+ and 2.00 M Cl - 1.00 M Na 2 SO 4 : 2.00 Na + and 1.00 M SO 4 2- 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 SO 4 2- ions in 1.50 L of 0.0150 M Na 2 SO 4 solution 0.0150 M Na 2 SO 4 solution contains: 2 x 0.0150 M Na + ions and 0.0150 M SO 4 2- ions moles Na + = 2 x 0.0150 M x 1.50 L = 0.0450 mol Na + moles SO 4 2- = 0.0150 M x 1.50 L = 0.0225 mol SO 4 2-

32. DILUTION Consider a stock solution of concentration M 1 and volume V 1 If water is added to dilute to a new concentration M 2 and volume V 2 moles before dilution = moles after dilution M 1 V 1 = M 2 V 2 Calculate the volume of 3.50 M HCl needed to prepare 500.0 mL of 0.100 M HCl (3.50 M)(V 1 ) = (0.100 M)(500.0 mL) V 1 = 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 - + AgNO 3 → AgCl (white ppt) + NO 3 - - Determination of sulfates by addition of barium chloride BaCl 2 + SO 4 2- → BaSO 4 (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 + H 2 O 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 H 2 SO 4 solution - Write balanced equation and determine mole ratio - Calculate moles of H 2 SO 4 - Determine moles of KOH - Calculate grams of KOH using molar mass

37. CHEMICAL ANALYSIS (TITRATIONS) 2KOH + H 2 SO 4 → K 2 SO 4 + 2H 2 O 2 mol KOH : 1 mol H 2 SO 4 mol H 2 SO 4 = 0.250 M x 0.02500 L = 0.00625 mol mol KOH = 2 x 0.00625 mol = 0.0125 mol