Tuesday September 22 Unit 2: Solutions and Reactions Learning Objectives LO 2.9: The student is able to create or interpret representations that link the concept of molarity with particle views of solutions. LO 2.14: The student is able to apply Coulomb’s Law qualitatively (including using representations) to describe the interactions of ions, and the attractions between ions and solvents to explain the factors that contribute to the solubility of ionic compounds. Today’s Topic: Electrolytes & Properties of Water Format: Notes & Practice Problems

Unit 2: Solutions & Reactions 9/22 Tuesday: Electrolytes & Properties of Water 9/23 Wednesday: Molarity & Preparing Solutions 9/24 Thursday: Acid Base Reactions & Forming Salts 9/25 Friday: Precipitation Reaction & Solubility Rules Quiz 1 9/28 Monday: Redox Equations 9/29 Tuesday: Balancing Redox Equations 9/30 Wednesday: Lab Quiz 2 10/1 Thursday: Lab 10/2 Friday: Unit 2 Exam

Unit 2 Recommended Reading & Problems Chapter & Section Relevant Problems Chapter , 50, 55-63, 67-88, 89, Sections 11.1 / 11.3 / , Section Section , 29-34

Properties of Water  Can dissolve many different substances.  A polar molecule because of its unequal charge distribution.  This molecular shape is called “bent”

Hydration  Positive ends of the water molecules are attracted to the negatively charged anions and the negative ends of the water are attached to the positively charged cations.  This process is called hydration  Hydration of ions causes salts to dissolve by breaking into individual cations and anions.

Solubility  Solubility of ionic substances varies greatly  This is caused by the relative attraction of the ions to each other versus their attraction to the water molecules.  Water can also dissolve non-ionic substances if they are also polar (“like dissolves like”)

What would happen if these items were added to water?  Water dissolves other polar solutes readily because the negative end of the polar solute is attracted to the positive end of the water – hydration occurs  Nonpolar solutes do not have permanent partial negative and partial positive ends so they do not dissolve in water Polar Solute Non-polar Solute

Nature of Aqueous Solutions  Solute – substance being dissolved.  Solvent – liquid water.  Electrolyte – substance that when dissolved in water produces a solution that can conduct electricity.

Electrolytes  Strong Electrolytes – conduct current very efficiently (bulb shines brightly). Completely ionized in water.  Weak Electrolytes – conduct only a small current (bulb glows dimly). A small degree of ionization in water.  Nonelectrolytes – no current flows (bulb remains unlit). Dissolves but does not produce any ions.

Electrolytes Electrolytes are substances that dissolve in water to produce ions in solution. The presence of ions allows a solution to conduct and electrical current. Ions may be produced because the substance that dissolves is ionic (like salts), or because the substance reacts with water to produce ions (as is the case with acids). The more ions that are formed in solution, the stronger the electrolyte. Nonelectrolytes are substances whose aqueous solutions do not contain ions and therefore do not conduct an electrical current

Which would be a better conductor of electricity?

Complete the below table

 Does HA represent a weak acid or a strong acid? Why?  What molecule in the reaction is not used to calculate Ka?  Is the value of a Ka for a weak acid likely to be greater than or less than 1?

Wednesday September 23 Warm-up: Learning Objectives LO 2.8: The student can draw and/or interpret representations of solutions that show the interactions between the solute and solvent. LO 2.9: The student is able to create or interpret representations that link the concept of molarity with particle views of solutions. Today’s Topic: Molarity & Preparing Solutions Format: Quick lecture & Making Solutions Upcoming: Thursday – Acid Base Reactions & Forming Salts Friday – Precipitation Reactions & Solubility Rules & Quiz

Chemical Reactions of Solutions  To perform stoichiometric calculations on solution reactions you must know:  The nature of the reaction (the exact forms the chemicals take when dissolved)  The amounts of chemicals present in the solutions (concentrations)

Molarity Molarity (M) = moles of solute per volume of solution in liters :

Check in exercise A g of sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? ◦ 1.57 M

Concentration of Ions For a 0.25 M CaCl 2 solution: CaCl 2 → Ca Cl – Ca 2+ : 1 × 0.25 M = 0.25 M Ca 2+ Cl – : 2 × 0.25 M = 0.50 M Cl –.

Check in Question Which of the following solutions contains the greatest number of ions? a) mL of 0.10 M NaCl. b) mL of 0.10 M CaCl 2. c) mL of 0.10 M FeCl 3. d) mL of 0.10 M sucrose.

Preparing Solutions

Making Solutions

Dilution The process of adding water to a concentrated or stock solution to achieve the molarity desired for a particular solution. Dilution with water does not alter the numbers of moles of solute present. Moles of solute before dilution = moles of solute after dilution M 1 V 1 = M 2 V 2

Check in Question A 0.50 M solution of sodium chloride in an open beaker sits on a lab bench. Which of the following would decrease the concentration of the salt solution? a)Add water to the solution. b)Pour some of the solution down the sink drain. c)Add more sodium chloride to the solution. d)Let the solution sit out in the open air for a couple of days. e)At least two of the above would decrease the concentration of the salt solution.

Thursday September 24 Warm-up: What is the minimum volume of a 2.00 M NaOH solution needed to make mL of a M NaOH solution?  60.0 mL Learning Objectives LO 3.3: The student is able to use stoichiometric calculations to predict the results of performing a reaction in the laboratory and/or to analyze deviations from the expected results. LO 3.4: The student is able to relate quantities (measured mass of substances, volumes of solutions, or volumes and pressures of gases) to identify stoichiometric relationships for a reaction, including situations involving limiting reactants and situations in which the reaction has not gone to completion. Today’s Topic: Acid Base Reactions & Forming Salts Format: Upcoming: Friday – Precipitation Reactions & Solubility Rules Friday - Quiz Monday – Oxidation & Reduction Reactions

Acid–Base Reactions (Brønsted– Lowry) Acid—proton donor Base—proton acceptor For a strong acid and base reaction: H + (aq) + OH – (aq) H 2 O(l)

Neutralization of a Strong Acid by a Strong Base

Acid-Base Titrations Titration – delivery of a measured volume of a solution of known concentration (the titrant) into a solution containing the substance being analyzed (the analyte). Equivalence point – enough titrant added to react exactly with the analyte. Endpoint – the indicator changes color so you can tell the equivalence point has been reached.

Acid – Base Titrations Go to my website and use the acid-base titration simulation sheet to complete the acid base titration online

Performing Calculations for Acid – Base Reactions 1.List the species present in the combined solution before any reaction occurs, and decide what reaction will occur. 2.Write the balanced net ionic equation for this reaction. 3.Calculate moles of reactants. 4.Determine the limiting reactant, where appropriate. 5.Calculate the moles of the required reactant or product. 6.Convert to grams or volume (of solution), as required.

Check in For the titration of sulfuric acid (H 2 SO 4 ) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of M sulfuric acid to reach the endpoint?

Friday September 24 Warm-Up: For the titration of sulfuric acid (H 2 SO 4 ) with sodium hydroxide (NaOH), how many moles of sodium hydroxide would be required to react with 1.00 L of M sulfuric acid to reach the endpoint? 1.00 mol NaOH Learning Objectives LO 3.1: Students can translate among macroscopic observations of change, chemical equations, and particle views. LO 3.2: The student can translate an observed chemical change into a balanced chemical equation and justify the choice of equation type (molecular, ionic, or net ionic) in terms of utility for the given circumstances. LO 3.10: The student is able to evaluate the classification of a process as a physical change, chemical change, or ambiguous change based on both macroscopic observations and the distinction between rearrangement of covalent interactions and noncovalent interactions. Today’s Topic: Precipitation Reactions & Solubility Rules Upcoming Monday: Oxidation & Reduction Reactions Tuesday: Balancing Oxidation & Reduction Reactions Wednesday-Thursday: Lab Friday: Exam

Quiz

Reactions Observe the changes that take place in the following demonstrations

Precipitation Reaction A double displacement reaction in which a solid forms and separates from the solution. ◦ When ionic compounds dissolve in water, the resulting solution contains the separated ions. ◦ Precipitate – the solid that forms.

Ba 2+ (aq) + CrO 4 2– (aq) → BaCrO 4 (s) The Reaction of K 2 CrO 4 (aq) and Ba(NO 3 ) 2 (aq)

Precipitation of Silver Chloride

Precipitates Soluble – solid dissolves in solution; (aq) is used in reaction equation. Insoluble – solid does not dissolve in solution; (s) is used in reaction equation.

Simple Rules for Solubility 1.Most nitrate (NO 3 - ) salts are soluble. 2.Most alkali metal (group 1A) salts and NH 4 + are soluble. 3.Most Cl -, Br -, and I - salts are soluble (except Ag +, Pb 2+, Hg 2 2+ ). 4.Most sulfate salts are soluble (except BaSO 4, PbSO 4, Hg 2 SO 4, CaSO 4 ). 5.Most OH - are only slightly soluble (NaOH, KOH are soluble, Ba(OH) 2, Ca(OH) 2 are marginally soluble). 6.Most S 2-, CO 3 2-, CrO 4 2-, PO 4 3- salts are only slightly soluble, except for those containing the cations in Rule 2.

Monday September 28 Warm-Up: Which of the following ions form compounds with Pb 2+ that are generally soluble in water? a) S 2– b) Cl – c) NO 3 – d) SO 4 2– e)Na + Learning Obje ctives LO 3.8: The student is able to identify redox reactions and justify the identification in terms of electron transfer. LO 3.9: The student is able to design and/or interpret the results of an experiment involving a redox titration. Today’s Topic: Oxidation-Reduction Reactions (Redox) Upcoming Tuesday: Balancing Oxidation & Reduction Reactions Wednesday-Thursday: Lab Friday: Exam

Redox Reactions Reactions in which one or more electrons are transferred.

Rules for Assigning Oxidation States 1.Oxidation state of an atom in an element = 0 2.Oxidation state of monatomic ion = charge of the ion 3.Oxygen = -2 in covalent compounds (except in peroxides where it = -1) 4.Hydrogen = +1 in covalent compounds 5.Fluorine = -1 in compounds 6.Sum of oxidation states = 0 in compounds 7.Sum of oxidation states = charge of the ion in ions

Check in Find the oxidation states for each of the elements in each of the following compounds: ◦ K 2 Cr 2 O 7 ◦ CO 3 2- ◦ MnO 2 ◦ PCl 5 ◦ SF 4 K = +1; Cr = +6; O = –2 C = +4; O = –2 Mn = +4; O = –2 P = +5; Cl = –1 S = +4; F = –1

Redox Reactions = transfer of electrons Oxidation Increase in oxidation state Reducing Agent LOSS of electrons Leo (Loss of Electrons is Oxidation) Reduction Decrease in oxidation state Oxidizing Agent GAIN of electrons GER (Gain of Electrons is Reduction)

Check in Which of the following are oxidation-reduction reactions? a)Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) b)Cr 2 O 7 2- (aq) + 2OH - (aq) 2CrO 4 2- (aq) + H 2 O(l) c)2CuCl(aq) CuCl 2 (aq) + Cu(s)

Tuesday September 29 Warm-Up: Identify the oxidizing agent and the reducing agent for the following reaction: Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) Learning Obje ctives LO 3.8: The student is able to identify redox reactions and justify the identification in terms of electron transfer. LO 3.9: The student is able to design and/or interpret the results of an experiment involving a redox titration. Today’s Topic: Balancing Oxidation-Reduction Reactions (Redox) Upcoming Wednesday-Thursday: Lab Friday: Exam

Balancing Redox Reactions by Oxidation States 1.Write the unbalanced equation. 2.Determine the oxidation states of all atoms in the reactants and products. 3.Show electrons gained and lost using “tie lines.” 4.Use coefficients to equalize the electrons gained and lost. 5.Balance the rest of the equation by inspection. 6.Add appropriate states.

Sample Problem Balance the reaction between solid zinc and aqueous hydrochloric acid to produce aqueous zinc (II) chloride and hydrogen gas 1. Unbalanced Equation Zn(s) + HCl(aq) Zn 2+ (aq) + Cl – (aq) + H 2 (g)

Sample Problem Balance the reaction between solid zinc and aqueous hydrochloric acid to produce aqueous zinc (II) chloride and hydrogen gas 2. Oxidation States for each atom Zn(s) + HCl(aq) Zn 2+ (aq) + Cl – (aq) + H 2 (g) 0 +1 –1 +2 –1 0

Sample Problem Balance the reaction between solid zinc and aqueous hydrochloric acid to produce aqueous zinc (II) chloride and hydrogen gas 3. How are electrons gained and lost? Zn(s) + HCl(aq) Zn 2+ (aq) + Cl – (aq) + H 2 (g) 0 +1 –1 +2 –1 0

Sample Problem Balance the reaction between solid zinc and aqueous hydrochloric acid to produce aqueous zinc (II) chloride and hydrogen gas 4. What coefficients are needed to equalize the electrons gained and lost? Zn(s) + 2 HCl(aq) Zn 2+ (aq) + Cl – (aq) + H 2 (g) 0 +1 –1 +2 –1 0

Sample Problem Balance the reaction between solid zinc and aqueous hydrochloric acid to produce aqueous zinc (II) chloride and hydrogen gas 5. What coefficients are needed to balance the remaining elements? Zn(s) + 2 HCl(aq) Zn 2+ (aq) + 2 Cl – (aq) + H 2 (g)