Chapter 4.1-4.5 Part 2 Types of Chemical Reactions and Solution Stoichiometry PHall AP info. NOT integrated into this presentation - REVISED SEPT 2013.

Chapter 4.1-4.5 Part 2 Types of Chemical Reactions and Solution Stoichiometry PHall AP info. NOT integrated into this presentation - REVISED SEPT 2013

Chapter 4 Table of Contents Copyright © Cengage Learning. All rights reserved 2 4.1 Water, the Common Solven4.1 Water, the Common Solvent 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolyte4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes 4.3 The Composition of Solution4.3 The Composition of Solutions 4.4 Types of Chemical Reaction4.4 Types of Chemical Reactions 4.5 Precipitation Reaction4.5 Precipitation Reactions

Section 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes Return to TOC Copyright © Cengage Learning. All rights reserved 6 Strong Electrolytes – conduct current very efficiently (bulb shines brightly). ex. ionic compounds (NaCl), strong acids (HCl, sulfuric, nitric, and strong bases (KOH, NaOH) Pull up solid sodium chloride video from podcast page to watch. Electrolytes

Section 4.2 The Nature of Aqueous Solutions: Strong and Weak Electrolytes Return to TOC Copyright © Cengage Learning. All rights reserved 6 Weak Electrolytes – conduct only a small current (bulb glows dimly). weak acids (organic acids -acetic, citric, butyric, malic, and weak basis (ammonia) Nonelectrolytes – no current flows (bulb remains unlit). ex. no ions present - sugars, alcohols & ex. molecular compounds that are not acids Electrolytes

Copyright © Houghton Mifflin Company. All rights reserved.4– Electrolyte Behavior Click below to watch visualization. http://college.cengage.com/chemistry/zumdahl/chemistry/7e /assets/students/protected/fae/index.html?layer=act&src=qti workflowflash_4_20.xml&w=750&h=434

An electrolyte is: A substance whose aqueous solution conducts an electric current. ex.ionic compounds & acids & bases A nonelectrolyte is: A substance whose aqueous solution does not conduct an electric current. ex. molecular compounds other than acids and bases Definition of Electrolytes and Nonelectrolytes

The ammeter measures the flow of electrons (current) through the circuit. If the ammeter measures a current, and the bulb glows, then the solution conducts. If the ammeter fails to measure a current, and the bulb does not glow, the solution is non-conducting. Electrolytes vs. Nonelectrolytes

1.Pure water 2.Tap water Sugar solution Sodium chloride solution Hydrochloric acid solution Lactic acid solution Ethyl alcohol solution Pure sodium chloride 1.Pure water 2.Tap water Sugar solution Sodium chloride solution Hydrochloric acid solution Lactic acid solution Ethyl alcohol solution Pure sodium chloride Try to classify the following substances as electrolytes or nonelectrolytes…

ELECTROLYTES: NONELECTROLYTES: Tap water (weak) NaCl solution HCl solution Lactate solution (weak) Pure water Sugar solution Ethanol solution Pure NaCl Answers to Electrolytes Pull up sugar videoclip from podcast page to watch.

Ionic Compounds “Dissociate” NaCl(s)  AgNO 3 (s)  MgCl 2 (s)  Na 2 SO 4 (s)  AlCl 3 (s)  Na + (aq) + Cl - (aq) Ag + (aq) + NO 3 - (aq) Mg 2+ (aq) + 2 Cl - (aq) 2 Na + (aq) + SO 4 2- (aq) Al 3+ (aq) + 3 Cl - (aq) When ionic compounds dissociate completing, they are good electrolytes for conducting electricity.

The reason for this is the polar nature of the water molecule… Positive ions associate with the negative end of the water dipole (oxygen). Negative ions associate with the positive end of the water dipole (hydrogen). Ions tend to stay in solution where they can conduct a current rather than re-forming a solid.

Covalent acids form ions in solution, with the help of the water molecules. For instance, hydrogen chloride molecules, which are polar, give up their hydrogens to water, forming chloride ions (Cl - ) and hydronium ions (H 3 O + ). Some covalent compounds IONIZE in solution

Other examples of strong acids include:  Sulfuric acid, H 2 SO 4  Nitric acid, HNO 3  Hydriodic acid, HI  Perchloric acid, HClO 4 Strong acids such as HCl are completely ionized (or dissociated) in solution. These would be examples of strong electrolytes.

Many of these weaker acids are “organic” acids that contain a “carboxyl” group. The carboxyl group does not easily give up its hydrogen. Weak acids such as lactic acid usually ionize less than 5% of the time.

Other organic acids and their sources include: o Citric acid – citrus fruit o Malic acid – apples o Butyric acid – rancid butter o Amino acids – protein o Nucleic acids – DNA and RNA o Ascorbic acid – Vitamin C This is an enormous group of compounds; these are only a few examples. Because of the carboxyl group, organic acids are sometimes called “carboxylic acids”.

17 Extra Information related to terms Solution - a homogeneous mixture of the solute and the solvent Solution= solvent + solute Aqueous (aq)= water solution Tincture = alcohol solution Amalgam = Mercury solution

SolutionsSolutionsSolutionsSolutions solutionsxt

Solution Concentration

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 8 We must know:  The nature of the reaction.  The amounts of chemicals present in the solutions. Chemical Reactions of Solutions

Calculations of Solution Concentration: Molarity Molarity Molarity is the ratio of moles of solute to liters of solution Notes 4.3 The molarity definition is based on the volume of the solution, NOT the volume of water. Vocab. Lesson Incorrect = The solution is 5.0 Molarity. Correct= The solution is 5.0 Molar.

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 10 Exercise A A 500.0-g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? (calculate #moles per L to get molarity)

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 10 Exercise A A 500.0-g sample of potassium phosphate is dissolved in enough water to make 1.50 L of solution. What is the molarity of the solution? (calculate #moles per L to get molarity) K 3 PO 4 is the formula and molar mass 212.27 g/mol (500 g / 212.27 g/mol) / 1.5 L Did you get the following ANSWER? with 3 sig. digits 1.57 M

Preparation of Molar Solutions Exercise B: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution?  Step #1: Ask “How Much?” (What volume to prepare?)  Step #2: Ask “How Strong?” (What molarity?)  Step #3: Ask “What does it weigh?” (Molar mass is?)

Preparation of Molar Solutions Exercise B: How many grams of sodium chloride are needed to prepare 1.50 liters of 0.500 M NaCl solution?  Step #1: Ask “How Much?” (What volume to prepare?)  Step #2: Ask “How Strong?” (What molarity?)  Step #3: Ask “What does it weigh?” (Molar mass is?) 1.500 L0.500 mol 1 L 58.44 g 1 mol = 43.8 g ANSWER How much? x How strong? x What does it weigh?

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 11 For a 0.25 M CaCl 2 solution: CaCl 2 → Ca 2+ + 2Cl –  Ca 2+ : 1 × 0.25 M = 0.25 M Ca 2+  Cl – : 2 × 0.25 M = 0.50 M Cl –. Concentration of Ions Note that there were 2 times as many ions of Cl- formed, so the molarity was doubled due to having twice as many moles like in the titration lab with citric acid which had 3 moles hydrogen ions produced for each mole of the base. Ionic compounds dissociate in solution, multiplying the molarity by the number of ions present.

28 Section 4.3 The Composition of Solutions Return to TOC Additional Examples if you need them Ion Concentration from Solution Concentration Ionic compounds dissociate in solution, multiplying the molarity by the number of ions present What is the Chloride Concentrations [Cl - ] in the following solutions? 2.0M NaCl since NaCl dissolves according to this reaction NaCl => Na + + Cl - the NaCl to Cl - ratio is 1:1, therefore [Cl - ]= 1 x 2.0M =2.0M 1.5M AlCl 3 since AlCl 3 dissolves according to this reaction AlCl 3 = > Al 3+ + 3 Cl - the AlCl 3 to Cl - ratio is 1:3, therefore [Cl - ] = 3 x 1.5M= 4.5M 2.0M CaCl 2 since CaCl 2 dissolves according to this reaction CaCl 2 = > Ca 2+ + 2 Cl - the CaCl 2 to Cl - ratio is 1:2, therefore [Cl - ]= 2 x 2.0 M= 4.0 M

29 Section 4.3 The Composition of Solutions Return to TOC Rearranging the Molarity formula Liters of solution x molarity = moles of solute

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 12 Concept Check 4.3a Which of the following solutions contains the greatest number of ions? a) 400.0 mL of 0.10 M NaCl. b) 300.0 mL of 0.10 M CaCl 2. c) 200.0 mL of 0.10 M FeCl 3. d) 800.0 mL of 0.10 M sucrose.

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 12 Concept Check 4.3a Which of the following solutions contains the greatest number of ions? a) 400.0 mL of 0.10 M NaCl. b) 300.0 mL of 0.10 M CaCl 2. c) 200.0 mL of 0.10 M FeCl 3. d) 800.0 mL of 0.10 M sucrose. 400 x.10 x 2 = 80 mmol 300 x.1 x 3 = 90 mmol 200 x.1 x 4 = 80 mmol 0 (stays as a molecule) = most moles would correspond to most ions if dissociates in solution.

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 13 Where are we going?  To find the solution that contains the greatest number of moles of ions. How do we get there?  Draw molecular level pictures showing each solution. Think about relative numbers of ions.  How many moles of each ion are in each solution? Let’s Think About It

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 14 The solution with the greatest number of ions is not necessarily the one in which:  the volume of the solution is the largest.  the formula unit has the greatest number of ions. Notice

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 15 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 Dilution

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 16 Concept Check 4.3b 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.

Dilution Exercise C: What volume of stock (11.6 M) hydrochloric acid is needed to prepare 250. mL of 3.0 M HCl solution? M stock V stock = M dilute V dilute (11.6 M)(V stock L) = (3.0 M)(0.250 L) V stock Liters = (3.0 M)(0.250 Liters) 11.6 M = 0.065 L

Dilution Dilution You try: Commercially available concentrated HCl is usually a 12.0 M solution. How does one prepare 100. mL of 1.0 M HCl solution from 12.0 M HCl solution?

ANSWER - DILUTION PROBLEM Example: Commercially available concentrated HCl is usually a 12.0 M solution. How does one prepare 100. mL of 1.0 M HCl solution from 12.0 M HCl solution?  The obvious answer is to dilute the HCl solution. But, what volume of 12.0 M HCl should be diluted? Recall the definition of molarity.  Moles of solute (n) = Molarity (M) x liters of solution (L)  This equation can be written as: moles of solute (n) = M i x V i. The subscript “i” stands for “initial”. www.t2i2edu.com

ANSWER - DILUTION PROBLEM Now, add water to the stock solution. This changes the concentration from M i to M f and the volume from V i to V f. The subscript “f” stands for “final”. Because the number of moles of solute has not changed during dilution, moles of solute (n) = M f x V f, and, M i x V i = M f x V f www.t2i2edu.com

ANSWER - DILUTION PROBLEM M i = 12.0 M V i = To be calculated M f = 1.0 M V f = 100. mL 12.0 M x V i = 1.0 M x 100. mL 8.3 mL of HCl should be put into a 100mL volumetric flask and filled with water to create a 1.0M solution. www.t2i2edu.com

41 Section 4.3 The Composition of Solutions Return to TOC Qualitatively, a dilute solution has a low concentration of solute. AND *A concentrated solution has a high concentration of solute.

Serial Dilution is a multiple dilution which could follow the steps several times. Other methods exists but will not be required on this unit test but may have to be done in college chemistry.

Copyright © Houghton Mifflin Company. All rights reserved.4– How to Make a Dilution This is an optional video for lab and calculating dilutions - Use if needed. For PC - copy this link: http://www.youtube.com/watch?v=MG86IFZi_XM#t=53 Click or copy the link below to go to video online http://www.youtube.com/watch?v=MG86IFZi_XM#t=53 I had to delete embedded file due to size. This will only work with MAC as it’s an embedded file

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 17 Exercise 4.3 D What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution?

Section 4.3 The Composition of Solutions Return to TOC Copyright © Cengage Learning. All rights reserved 17 Exercise 4.3 D What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution? ANSWER: 60.0 mL

Section 4.4 Types of Chemical Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 18 Classification Used in General Chemistry Last Year Precipitation Reactions Acid–Base Reactions Oxidation–Reduction Reactions These are the 3 major classifications our textbook will use or AP Chemistry.

Precipitation Reactions Graphic: Wikimedia Commons User Tubifex

Section 4.5 Precipitation Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 19 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. Precipitation Reaction

Section 4.5 Precipitation Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 20 net equation Ba 2+ (aq) + CrO 4 2– (aq) → BaCrO 4 (s) The Reaction of K 2 CrO 4 (aq) and Ba(NO 3 ) 2 (aq) Ba(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) → BaCrO 4 (s) + 2 KNO 3

Double Replacement Reactions The ions of two compounds exchange places in an aqueous solution to form two new compounds. AX + BY →  AY + BX One of the compounds formed is usually a precipitate (an insoluble solid), an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

Double replacement forming a precipitate… Pb(NO 3 ) 2 (aq) + 2KI(aq)  PbI 2 (s) + 2KNO 3 (aq) Pb 2+ (aq) + 2 NO 3 - (aq) + 2 K + (aq) +2 I - (aq)  PbI 2 (s) + 2K + (aq) + 2 NO 3 - (aq) Pb 2+ (aq) + 2 I - (aq)  PbI 2 (s) Double replacement (ionic) equation Complete ionic equation shows compounds as aqueous ions Net ionic equation eliminates the spectator ions Lead(II) nitrate + potassium iodide →  lead(II) iodide + potassium nitrate

Solubility Rules – Mostly Soluble IonSolubilityExceptions NO 3 - SolubleNone ClO 4 - SolubleNone Na + SolubleNone K+K+K+K+SolubleNone NH 4 + SolubleNone Cl -, I - Soluble Pb 2+, Ag +, Hg 2 2+ SO 4 2- Soluble Ca 2+, Ba 2+, Sr 2+, Pb 2+, Ag +, Hg 2+

Solubility Rules – Mostly Insoluble IonSolubilityExceptions CO 3 2- Insoluble Group IA and NH 4 + PO 4 3- Insoluble Group IA and NH 4 + OH - Insoluble Group IA and Ca 2+, Ba 2+, Sr 2+ S 2- Insoluble Groups IA, IIA, and NH 4 +

Solubility Chart: Common salts at 25  C S = Soluble I = Insoluble P = Partially Soluble X = Other

Section 4.5 Precipitation Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 22 Soluble – solid dissolves in solution; (aq) is used in reaction. Insoluble – solid does not dissolve in solution; (s) is used in reaction. Insoluble and slightly soluble are often used interchangeably. Precipitates

Section 4.5 Precipitation Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 23 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. Simple Rules for Solubility

Solubility Trends  The solubility of MOST solids increases with temperature.  The rate at which solids dissolve increases with increasing surface area of the solid.  The solubility of gases decreases with increases in temperature.  The solubility of gases increases with the pressure above the solution.

Therefore… Solids tend to dissolve best when: o Heated o Stirred o Ground into small particles Gases tend to dissolve best when: o The solution is cold o Pressure is high

Solubility Chart

68 Section 4.5 Precipitation Reactions Return to TOC Example for setting up precipitation problem

Saturation of Solutions  A solution that contains the maximum amount of solute that may be dissolved under existing conditions is saturated.  A solution that contains less solute than a saturated solution under existing conditions is unsaturated.  A solution that contains more dissolved solute than a saturated solution under the same conditions is supersaturated.

Section 4.5 Precipitation Reactions Return to TOC Copyright © Cengage Learning. All rights reserved 24 Concept Check 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 +

71 Section 4.5 Precipitation Reactions Return to TOC Predicting Precipitation Products Videoclip 4:12 minutes Pull up the Predicting Precipitation Products Videoclip to watch.

Gravimetric Analysis Gravimetric analysis: a chemical analysis method based on the measurement of masses. Can be used in combination with precipitation reactions to determine the amount of a species present in solution. First isolate the precipitate by filtration and drying and subsequently weigh it. Using this mass and stoichiometry relationships, we can determine the amount (mass or moles) of species present in solution. www.t2i2edu.com

73 Chapter 4 Table of Contents Homework Ch. 4 part 1 Sect.4.1-4.5+ pg. 171-173 #17ab,18a-d+, 21a-c, 23a-d, 26, 27, 30a, 31, 33, 36(a-h), 38(a-d), 40, 41b, 47, 49, & 53 (36 parts) due Monday - October 7th. Keep in mind you will have other assignments on Thursday also, so try to get as much done now as you can. CW: Take your balanced equation w/sheet and enter into sciencegeek.net AP Chem Review: http://www.sciencegeek.net/Chemistry/taters/EquationBalancing.htm Show me score above & complete the following quiz & show me scores. http://www.sciencegeek.net/APchemistry/APtaters/ReactionIdentification.htm Class Assignments - Tues Oct. 1, 2013 and next slide

74 Chapter 4 Table of Contents CW: Go over some previous problems. Notes 4.1-4.5+ including videoclips Check balanced eqn sheet on computer - show score Quiz - Types of chemical rxns (based on last year classifications) Homework Ch. 4 part 1 Sect.4.1-4.5+ pg. 171-173 #17ab,18a-d+, 21a-c, 23a-d, 26, 27, 30a, 31, 33, 36(a-h), 38(a-d), 40, 41b, 47, 49, & 53 (36 parts) due Monday - October 7th. Keep in mind you will have other assignments on Thursday also, so try to get as much done now as you can. HW: Lab report Mass vs. Mole Relationships due Thursday HW: Study for ch. 3 quiz on Thursday HW today due with limiting reactants w/sheet by end of block. TEST ch.3-4 tentatively set for October 11th 74 Current & Upcoming Assignments - TUES - OCT 1, 2013

75 Section 4.5 Precipitation Reactions Return to TOC END OF SLIDES FOR SECTION 4.1-4.5

Chapter 4.1-4.5 Part 2 Types of Chemical Reactions and Solution Stoichiometry PHall AP info. NOT integrated into this presentation - REVISED SEPT 2013.

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Chapter 4.1-4.5 Part 2 Types of Chemical Reactions and Solution Stoichiometry PHall AP info. NOT integrated into this presentation - REVISED SEPT 2013.

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Presentation on theme: "Chapter 4.1-4.5 Part 2 Types of Chemical Reactions and Solution Stoichiometry PHall AP info. NOT integrated into this presentation - REVISED SEPT 2013."— Presentation transcript:

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