1. Chapter 4 Types of chemical reactions and Solution Stoichiometry What are aqueous solutions? Substances dissolved in water  Solvent Why is water considered the most common solvent? It can dissolve many different compounds because it is not a linear molecule  bent at an angle of 105°

2. What type of bond occurs in water? Polar covalent Water is a polar molecule  has an unequal charge distribution What happens to a Salt when dissolved in water? It breaks into cations which are attracted to the negative ends of water and the anions which are attracted to the pos. ends of water This process is called Hydration

3. The nature of aqueous solutions What is a solution? Homogeneous mixture Has 2 parts A. Solute- lesser amount of substance B. Solvent- greater amount of substance How can we characterize a solution? One way is by Electrical Conductivity

4. What is electrical conductivity? Ability to conduct an electric current Electrolyte- substance that when dissolved in water can conduct electricity Can be classified on its ability to conduct electricity

5. ElectrolyteConductivityDegree of Dissociatio n Examples strongHightotalStrong acids and bases weakLow to moderate partialWeak organic acid and bases nonnoneClose to zero Sugar, ethanol

6. The Composition of Solutions The most commonly used expression of concentration is Molarity The moles of solute per volume of solution in liters or M= moles of solute/ liters of solution

7. Examples Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. Calculate the mass of NaCl needed to prepare 175ml of a.500 M NaCl solution. How many ml of solution are necessary if we are to have a 2.48 M NaOH solution that contains 31.52 g of the dissolved solid?

8. Molarity of Ions in Solution Calculate the molarity of all ions in.25 M Ca(ClO) 2. H 2 0(l) Ca(ClO) 2 (s)  Ca 2+ + 2ClO - Molarity of Ca 2+ =.25 M Molarity of ClO - = 2.25 M =.50 M

9. Another Example Determine the molarity of Cl - ions in a solution prepared by dissolving 9.82 g of CuCl 2 in enough water to make 600. ml. 1.Determine the molarity. 2.Write an equation to determine the mole ratio. 3.Set-up a mole problem.

10. Dilutions -adding a solvent ( usually water) to lower the concentration of a solute. M 1 V 1 = M 2 V 2 What volume of 16 M sulfuric acid must be used to prepare 1.5 L of.10 M H 2 SO 4.

11. Types of Chemical Reactions 1.Precipitation reactions( double displacement) 2.Acid- Base reactions 3.Oxidation-reduction reactions

12. Precipitation reactions -2 solutions are mixed, sometimes a precipitate forms What is a precipitate? -an insoluble substance Ba(NO 3 ) 2 (aq) + K 2 CrO 4 (aq)  ?

13. Predict what will happen? 1.Na 2 SO 4 (aq) + Pb(NO 3 ) 2 (aq)  ? 2NaNO 3 (aq) + PbSO 4 (s) 2.3KOH(aq) + Fe(NO 3 ) 3 (aq)  ? 3 KNO 3 (aq) + Fe(OH) 3 (s)

14. Describing Reactions in Solutions 3 Types of Equations 1.Molecular equation K 2 CrO 4 (aq) + Ba(NO 3 ) 2 (aq)  BaCrO 4 (s) + 2KNO 3 (aq) 2.Complete Ionic equation- represents the actual forms of reactants and products 2K + (aq) + CrO 4 - (aq) + Ba +2 (aq) + 2NO 3 - (aq)  BaCrO 4 (s) + 2K + (aq) + 2NO 3 - (aq) Spectator Ions- ions that do not participate directly in the reaction K + and NO 3 - are spectator ions

15. Con. 3.Net Ionic Equation- includes only those components that undergo changes in the reaction Ba 2+ (aq) + CrO 4 2- (aq)  BaCrO 4 (s)

16. Example Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate plus aqueous potassium nitrate. Write the three types of equations. Molecular equation KCl(aq) + AgNO 3 (aq)  AgCl(s) + KNO 3 (aq)

17. Con. Complete ionic K + (aq) + Cl - (aq) + Ag + (aq) + NO 3 - (aq)  AgCl(s) + K + (aq) + NO 3 - (aq) Net ionic Cl - (aq) + Ag + (aq)  AgCl(s)

18. Selective Precipitation -separating cations by precipitating them out one at a time -Use a Solubility chart to make a Flow Chart

19. Solving Stoichiometry Problems for reactions in solutions 1.Write the balanced equation. 2.Set-up 2 mole problems to determine limiting and excess. 3.Change to grams if needed. Example: When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 precipitates. Calculate the mass of PbSO 4 formed when 1.25 L of 0.0500 M Pb(NO 3 ) 2 and 2.00 L of 0.0250 M Na 2 SO 4 are mixed. Na 2 SO 4 (aq) + Pb(NO 3 ) 2(aq)  PbSO 4 (s) + 2NaNO 3 (aq)

20. Acid-Base Reactions Bronsted-Lowry Concept of Acids and Bases - an acid is a proton donor - a base is a proton acceptor Also called a neutralization reaction Produces a salt and water

21. Example 4.12 p. 159 What volume of a 0.100 M HCl solution is needed to neutralize 25.0 ml of.350M NaOH? HCl(aq) + NaOH (aq)  NaCl(s) + H 2 0(l)

22. Acid-Base Titrations Volumetric analysis- a technique for determining the amount of a certain substance by doing a titration Titration- the use of a buret to deliver the measured volume of a known concentration(titrant) into the substance being analyzed(analyte)

23. Con. Equivalence point or Stoichiometric point- the point where enough titrant has been added to react with a substance Often marked by an indicator Indicator- substance added at the beginning of a titration that changes color at or very near the equivalence point Endpoint- point where the color change occurs Ex: phenophthalein(colorless in acid, pink in base)

24. Example 4.14 p. 162 A student carries out an experiment to standardize(determine the exact concentration of) a sodium hydroxide solution. To do this, the student weighs out a 1.3009 g of KHP(KHC 8 H 4 O 4 ). Its molar mass is 204.22g/mol. The student dissolves the KHP in distilled water, adds phenolphthalein as an indicator, and titrates the resulting solution with the sodium hydroxide solution to the endpoint. The difference between the final and initial buret readings is 41.20 ml. Calculate the concentration of the NaOH solution.

25. Oxidation-Reduction Reactions Oxidation States -imaginary charges the atoms would have if shared electrons were divided equally Oxidation- increase in the oxidation state: a loss of electrons Reduction- decrease in oxidation state: gain of electrons

26. A helpful Mnemonic OIL RIG OxidationReductionInvolves LossGain Example: CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) -4 +1(+4) 0 +4 –2(-4) +1(+4) -2 Carbon -4 to +4 oxidized Oxygen 0 to –2 reduced

27. What else is involved in Redox? 1.Oxidizing Agent- electron acceptor 2.Reducing Agent- electron donor In the previous example what is the oxidizing and reducing agent? O 2 is oxidizing agent CH 4 is reducing agent

28. Examples 2Al(s) + 3I 2 (s)  2AlI 3 (s) 2PbS(s) + 3O 2 (g)  2PbO(s) + 2SO 2 (g) What atoms are oxidized/reduced and which compounds are the oxidizing and reducing agents?

29. Balancing Redox equations in Acid Solutions 1.Assign oxidation states 2.Write ½ cells( half-reactions) 3.Balance all elements except H and O 2 4.Balance O by adding H 2 O 5.Balance H by adding H+ 6.Balance Charges by adding e- 7.If necessary, multiply all coefficients in 1 or both ½ cells by an integer to equalize the # of e- 8.Add ½ cells and cancel identical species, check that charges and atoms are balanced

30. Example p. 173 MnO 4 - (aq) + Fe 2+ (aq)  Fe 3+ (aq) + Mn 2+ (aq) acid Example 4.19 p. 175 H + (aq)+ Cr 2 O 7 2- (aq)+C 2 H 5 OH(l)  Cr 3+ (aq) + CO 2 (g) + H 2 O(l)

31. Balancing Redox equations in Basic Solution 1.Assign oxidation states 2.Write ½ cells(half-reactions) 3.Balance all elements except H and O 2 4.Balance O by adding H 2 O 5.Balance H by adding H + 6.Balance charges by adding e - 7.If necessary multiply all coefficients in 1 or both ½ cells by an integer to equalize # of e - 8.Add ½ cells and cancel identical species. Check that charges and atoms are balanced

32. Con. 9.Count # of H + and add same # of OH - to both sides of the equation 10.If H + and OH - are on the same side, form H 2 0. 11.Cancel out H 2 0 molecules on both sides 12.Check that charges and atoms are balanced

33. Balancing Redox in basic solution example I - (aq) + OCl - (aq)  I 2 (s) + Cl - (aq) +H 2 0(l)