CHAPTER 5 Chemical Reactions in Aqueous Solution 1 Aqueous Solutions: An Introduction 1. Electrolytes and Extent of Ionization Aqueous solutions consist of a solute dissolved in water. Classification of solutes: Nonelectrolytes solutes that do not conduct electricity in water do not ionize Examples: C H 5 OH ethanol Sugars glucose, etc. Strong and Weak Electrolytes Electrolytes produce ions in solution and conduct electricity Some are Strong electrolytes, Strong electrolytes ionize or dissociate completely. Strong electrolytes approach 100% dissociation in aqueous solutions. NaCl(s) Na Cl - Weak electrolytes ionize or dissociate partially, much less than 100%. HF(l) H F -
Aqueous Solutions: An Introduction strong electrolytes - conduct electricity extremely well in dilute aqueous solutions Examples of strong electrolytes 1. HCl, HNO, etc. strong soluble acids. NaOH, KOH, etc. strong soluble bases. NaCl, KBr, etc. soluble ionic salts ionize in water essentially 100% 4 Aqueous Solutions: An Introduction Classification of solutes weak electrolytes - conduct electricity poorly in dilute aqueous solutions 1. CH COOH, (COOH) weak acids. NH, Fe(OH) weak bases 5 Strong and Weak Electrolytes Nonelectrolytes are covalent compounds that dissolve in water, but do not conduct electricity 6
Strong and Weak Electrolytes 7 Strong Electrolytes Strong Water Soluble Acids HNO ( l) H O HNO 100% ( l) or 100% ( l) H H O NO NO 8 Strong Electrolytes Strong Water Soluble Bases KOH Sr(OH) HO 100% (s) K HO 100% (s) Sr OH - OH - 9
Strong Electrolytes Most Water Soluble Salts NaCl Ca(NO H O 100% (s) Na H O 100% ) ( s) Cl Ca - NO 10 Solubility Solubility is the maximum amount of solute that can dissolve in a given amount of solvent Usually expressed as the amount that dissolves in 100 g solvent Unsaturated Solution contains less than the maximum amount Saturated contains the maximum amount Increasing the temperature makes more dissolve Supersaturated solution 11 Solubility There are rules to determine if something is soluble (dissolves) or insoluble (does not dissolve) 1
Solubility 1 Aqueous Solutions: An Introduction 5. Solubility Guidelines for Compounds in Aqueous Solutions 1) All common compounds of the Group 1 metal ions and the ammonium ion are soluble. Alkalis are soluble Li, Na, K, Rb, Cs, and NH 4 14 Aqueous Solutions: An Introduction ) Common nitrates, acetates, chlorates, and perchlorates are soluble. NO -, CH COO -, ClO -, and ClO - 4 4) Common chlorides. bromides and iodides are soluble. Exceptions AgCl, Hg Cl, & PbCl Common fluorides are water soluble. Exceptions MgF, CaF, SrF, BaF, and PbF 15
Aqueous Solutions: An Introduction 5) Sulfates are soluble. Exceptions PbSO 4, BaSO 4, & HgSO 4 6) Common metal hydroxides are insoluble. Exceptions LiOH, NaOH,, KOH, RbOH & CsOH 16 Aqueous Solutions: An Introduction 7) Carbonates, phosphates, and arsenates are insoluble. CO -, PO - 4, & AsO - 4 Exceptions- IA metals and NH 4 plus Ca to Ba Moderately soluble MgCO 8) Sulfides are insoluble. Exceptions IA metals and NH 4 plus IIA metals 17 Reactions in Aqueous Solutions There are three ways to write reactions in aqueous solutions. 1. Molecular equation Show all reactants & products in molecular or ionic form AgNO NaCl AgCl NaNO ( aq) ( aq) ( s) ( aq). Total ionic equation Show the ions and molecules as they exist in solution Ag NO - Na Cl - AgCl(s) Na NO - 18
Reactions in Aqueous Solutions. Net ionic equation Shows ions that participate in reaction and removes spectator ions. Spectator ions do not participate in the reaction. Ag Cl - AgCl(s) 19 Metathesis Reactions Metathesis reactions occur when two ionic aqueous solutions are mixed and the ions switch partners. AX BY AY BX Metathesis reactions remove ions from solution in ways: 1. form H O - neutralization. form an insoluble solid. Form a gas Ion removal is the driving force of metathesis reactions. 0 Metathesis Reactions 1. Acid-Base (neutralization) Reactions Formation of the nonelectrolyte H O acid base salt water 1
H Metathesis Reactions Molecular equation HBr KOH KBr Total ionic equation Br K OH K Br H O ( l) H O - - - ( aq) ( aq) ( aq) ( aq) ( aq) ( aq) ( l) Net ionic equation H OH - H O ( aq) ( aq) ( l) Metathesis Reactions Molecular equation Ca(OH) HNO Ca(NO ) H O( l) Total ionic equation - - - Ca( aq) OH( aq) H( aq) NO( aq) Ca( aq) NO( aq) HO( l) Net ionic equation OH OH - ( aq ) ( aq ) H or better H H H O - ( aq ) ( aq ) ( l ) O ( l ) Metathesis Reactions Precipitation reactions Precipitation reactions are metathesis reactions in which an insoluble compound is formed. The solid precipitates out of the solution much like rain or snow precipitates out of the air. 4
Precipitation Reactions Molecular equation Ca(NO ) K CO KNO CaCO (s) Total ionic reaction Ca NO K - - ( aq) ( aq) ( aq) ( aq) K Net ionic reaction CO NO CaCO - ( aq) ( aq) ( s) Ca CO CaCO - ( aq) ( aq) ( s) 5 Metathesis Reactions Gas forming reactions are metathesis in which a gas is formed and removed from solution H CO H O(l) CO (g) HSO H O(l) SO (g) HCN(g) NH 4 OH NH (g) H O(l) 6 Metathesis Reactions Molecular equation CaCl Na PO 4( aq) HCl Na SO 7
Aqueous Solutions: An Introduction. Strong and Weak Acids Acids are substances that generate H in aqueous solutions. Strong acids ionize 100% in water. HCl H Cl 100 % - ( g) ( aq ) ( aq ) HNO 100% - H O( l) H O( aq) NO ( aq) HNO or H NO H O - ( aq) ( aq) 8 Aqueous Solutions: An Introduction Strong Acids Formula Name 1. HCl hydrochloric acid. HBr hydrobromic acid. HI hydroiodic acid 4. HNO nitric acid 5. H SO 4 sulfuric acid 6. HClO chloric acid 7. HClO 4 perchloric acid 9 Weak acids ionize significantly less than 100% in water. Common Weak Acids Formula Name 1.HF hydrofluoric acid.ch COOH acetic acid (vinegar).hcn hydrocyanic acid 4.HNO nitrous acid 5.H CO carbonic acid (soda water) 6.H PO 4 phosphoric acid 0
Aqueous Solutions: An Introduction. Reversible Reactions CH COOH acetic acid 7% - HO CHCOO( aq) HO( aq) CH COOH - H 7% CH COOH CH COO ( aq) ( aq) 1 Aqueous Solutions: An Introduction 4. Strong Bases, Insoluble Bases, and Weak Bases Characteristic of common inorganic bases is that they produce OH- ions in solution. Examples of Common Strong Bases 1. LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH) Sr(OH). Ba(OH). Notice that they are all hydroxides of IA and IIA metals Aqueous Solutions: An Introduction Similarly to strong acids, strong bases ionize 100% in water. KOH Ba(OH) K Ba OH - OH -
Aqueous Solutions: An Introduction Weak bases are covalent compounds that ionize slightly in water. Ammonia is most common weak base NH NH H O NH OH ( g) ( l) 4( aq) - 4 The Arrhenius Theory Neutralization reactions are the combination of H (or H O ) with OH - to form H O. Strong acids are acidic substances that ionize 100% in water. List of aqueous strong acids: HCl, HBr, HI, H SO 4, HNO, HClO 4, HClO Strong bases are basic substances that ionize 100% in water. List of aqueous strong bases: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH), Sr(OH), Ba(OH), Na O, K O, etc. 5 The Arrhenius Theory Strong acid-strong base reaction The formula unit equation is: HCl aq NaOH NaCl H ( ) ( aq) ( aq) ) The total ionic equation is: O (l - - H( aq ) Cl( aq) Na( aq) OH( aq) Na( aq) Cl( aq) H O (l ) The net ionic equation is: - H OH H O (l ( aq) ( aq) ) 6
Naming Some Inorganic Compounds Salts are formed by the reaction of the acid with a strong base. Acid Salt HNO NaNO nitrous acid sodium nitrite HNO NaNO nitric acid sodium nitrate H SO Na SO sulfurous acid sodium sulfite 7 Naming Some Inorganic Compounds There are two other possible acid and salt combinations. Acids that have 1 more O atom than the ic acid are given the prefix per. Acids that have one less O atom than the ous acid are given the prefix hypo. 8 Naming Some Inorganic Compounds Illustrate this series of acids and salts with the Cl ternary acids and salts. Acid Na Salt HClO NaClO hypochlorous acid sodium hypochlorite HClO NaClO chlorous acid sodium chlorite HClO NaClO chloric acid sodium chlorate HClO 4 NaClO 4 perchloric acid sodium perchlorate 9
Acid-Base Reactions in Aqueous Solutions There are four acid-base reaction combinations that are possible: 1. Strong acids strong bases. Weak acids strong bases. Strong acids weak bases 4. Weak acids weak bases Let us look at one example of each acid-base reaction. 40 Acid-Base Reactions in Aqueous Solutions 1. Strong acids - strong bases forming soluble salts This is one example of several possibilities hydrobromic acid calcium hydroxide The molecular equation is: HBr Ca(OH) CaBr H O (l) Net: H OH- ( aq) H O (l) 41 Acid-Base Reactions in Aqueous Solutions 1. Strong acids-strong bases forming insoluble salts There is only one reaction of this type: sulfuric acid barium hydroxide The molecular equation is: H SO 4 Ba(OH) BaSO 4(s) H O (l) 4
Acid-Base Reactions. Weak acids - strong bases forming soluble salts nitrous acid sodium hydroxide The molecular equation is: HNO NaOH NaNO H O (l) The total ionic equation is: HNO Na OH - Na NO - H O (l) The net ionic equation is: HNO OH - NO - H O (l) 4 Acid-Base Reactions. Strong acids - weak bases- forming soluble salts nitric acid ammonia The molecular equation is: HNO NH NH 4 NO The net equation is: H NH NH 4 44 Acid-Base Reactions 4. Weak acids - weak bases forming soluble salts acetic acid ammonia The molecular equation is: CH COOH NH NH 4 CH COO The total ionic equation is: CH COOH NH NH 4 CH COO - The net ionic equation is: CH COOH NH NH 4 CH COO - 45
Acidic Salts and Basic Salts Polyprotic acids with less than the stoichiometric amount of base. 1:1 ratio. H SO 4 NaOH NaHSO 4 H O (l) The acidic salt sodium hydrogen sulfate is formed. 1: ratio. H SO 4 NaOH Na SO 4 H O (l) The normal salt sodium sulfate is formed. 46 Soultions A solution is a mixture of two or more substances dissolved in another. Solute is the substance present in the smaller amount. Solvent is the substance present in the larger amount. In aqueous solutions, the solvent is water. 47 Types of Solutions 48
Concentrations of Solutions Common unit of concentration: molarity = number of moles M = L number of moles of solute liters of solution 49 Concentrations of Solutions 50 Concentrations of Solutions Example 9-5: Calculate the molarity of a solution that contains 1.5 g of sulfuric acid in 1.75 L of solution.? mol HSO4 1.5g HSO4 = L sol'n 1.75 L sol'n 1molHSO4 98.1g H SO 4 51
Concentrations of Solutions Example 9-6: Calculate the molarity of a solution that contains 1.5 g of sulfuric acid in 1.75 L of solution.? molhso L sol'n 4 1.5g HSO4 1molHSO4 = 1.75 L sol'n 98.1g H SO 0.078mol HSO = L = 0.078M H SO 4 4 4 5 Concentrations of Solutions Example 9-7: Determine the mass of calcium nitrate required to prepare.50 L of 0.800 M Ca(NO ). You do it! 5 Concentrations of Solutions Example 9-7: Determine the mass of calcium nitrate required to prepare.50 L of 0.800 M Ca(NO ). 0.800 mol Ca(NO )? g Ca(NO ) =. 50 L L 164 g Ca(NO ) = 459 g Ca(NO ) 1 mol Ca(NO ) 54
Calculations Involving Molarity Example: If 100.0 ml of 1.00 M NaOH and 100.0 ml of 0.500 M H SO 4 solutions are mixed, what will the concentration of the resulting solution be? 55 Calculations Involving Molarity Example: If 10.0 ml of 1.00 M KOH and 100.0 ml of 0.500 M H SO 4 solutions are mixed, what will be the concentration of KOH and K SO 4 in the resulting solution? 56 Calculations Involving Molarity Example: What volume of 0.750 M NaOH solution would be required to completely neutralize 100 ml of 0.50 M H PO 4? 57
Dilution of Solutions To dilute a solution, add solvent to a concentrated solution. The number of moles of solute in the two solutions remains constant. The relationship M 1 V 1 = M V is appropriate for dilutions, but not for chemical reactions. 58 Dilution of Solutions Example: If 10.0 ml of 1.0 M HCl is added to enough water to give 100. ml of solution, what is the concentration of the solution? 1.0 M M 1V1 = M V 10.0 ml = M 100.0 ml M 1.0 M 10.0 ml = 100.0 ml = 1.0 M 59 Dilution of Solutions Example: What volume of 18.0 M sulfuric acid is required to make.50 L of a.40 M sulfuric acid solution? 60
Dilution of Solutions Example: What volume of 18.0 M sulfuric acid is required to make.50 L of a.40 M sulfuric acid solution? M 1 V 1 = M V M V V1 = M 1.50 L.40 M V1 = 18.0 M = 0. L or ml 61 Solutions in Chemical Reactions Combine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution. 6 Using Solutions in Chemical Reactions Example: What volume of 0.500 M BaCl is required to completely react with 4. g of Na SO 4? Na SO4 BaCl BaSO4 NaCl 6
Using Solutions in Chemical Reactions Example: What volume of 0.500 M BaCl is required to completely react with 4. g of Na SO 4? Na SO BaCl BaSO NaCl? L BaCl 1mol BaCl 1mol Na SO 4 1mol Na SO4 = 4. gna SO4 14 g Na SO 4 1L BaCl 0.500 mol BaCl 4 4 = 0.0608 L 64 Using Solutions in Chemical Reactions Example: (a)what volume of 0.00 M NaOH will react with 50.0 ml 0f 0.00 M aluminum nitrate, Al(NO )? ( NO ) NaOH Al( OH) Al NaNO You do it! 65 Using Solutions in Chemical Reactions Example: (a)what volume of 0.00 M NaOH will react with 50.0 ml 0f 0.00 M aluminum nitrate? Al ( NO ) NaOH Al(OH) 1 L? ml NaOH = 50.0 ml Al(NO ) sol' n 1000 ml 0.00 mol Al(NO ) sol' n mol NaOH 1 L Al(NO ) sol' n 1 mol Al(NO ) 1 L NaOH = 0.150 0.00 mol NaOH L or 150 NaNO ml NaOH sol' n 66
Using Solutions in Chemical Reactions (b)what mass of Al(OH) precipitates in (a)? 67 Using Solutions in Chemical Reactions (b) What mass of Al(OH) precipitates in (a)? 1 L? g Al(OH) = 50.0 ml Al(NO) sol' n 1000 ml 0.00 mol Al(NO) 1 mol Al(OH) 78.0 g Al(OH) 1 L Al(NO ) sol' n 1mol Al(NO ) 1 mol Al(OH) = 0.780 g Al(OH) 68 Using Solutions in Chemical Reactions Titrations are a method of determining the concentration of an unknown solutions from the known concentration of a solution and solution reaction stoichiometry. 69
Titrations Acid-base Titration Terminology 1. **Titration A method of determining the concentration of one solution by reacting it with a solution of known concentration.. **Primary standard A chemical compound which can be used to accurately determine the concentration of another solution. Examples include KHP and sodium carbonate. 70 Titrations Acid-base Titration Terminology 1. **Standard solution A solution whose concentration has been determined using a primary standard.. **Standardization The process in which the concentration of a solution is determined by accurately measuring the volume of the solution required to react with a known amount of a primary standard. 71 Titrations **Indicator A substance that exists in different forms with different colors depending on the concentration of the H in solution. Examples are phenolphthalein and bromothymol blue. 5. **Equivalence point The point at which stoichiometrically equivalent amounts of the acid and base have reacted. 6. **End point The point at which the indicator changes color and the titration is stopped. 7
The Mole Method and Molarity HC HC CH C C CH Potassium hydrogen phthalate is a very good primary standard. It is often given the acronym, KHP. KHP has a molar mass of 04. g/mol. OA very common mistake is O for Cstudents to see the acronym CH C KHP OH and think KOH that this compound is O HC C O Cmade of potassium, hydrogen, CH C and OH phosphorous. OH HC KHP C O- K H O acidic H 7 The Mole Method and Molarity Example: Calculate the molarity of a NaOH solution if 7. ml of it reacts with 0.4084 g of KHP. NaOH KHP NaKP H O 1 mol KHP? mol NaOH = 0.4084 g KHP 1 mol KHP? mol NaOH = 0.4084 g KHP 04. g KHP 04. g KHP 1 mol NaOH 1 mol NaOH = 0.0000 mol NaOH 1 mol KHP = 0.0000 mol NaOH 0.00001 mol NaOH KHP? M NaOH = = 0.07 M NaOH 0.07 L NaOH 74 Using Solutions in Chemical Reactions Example: What is the molarity of a KOH solution if 8.7 ml of the KOH solution is required to react with 4. ml of 0. M HCl? KOH HCl KCl HO 75
Using Solutions in Chemical Reactions Example: What is the molarity of a KOH solution if 8.7 ml of the KOH solution is required to react with 4. ml of 0. M HCl? KOH HCl KCl HO 4. ml 0. M HCl = 9.6 mmol HCl 1mmol KOH 9.6 mmol HCl = 9.6 mmol KOH 1mmol HCl 9.6 mmol KOH = 0.49 M KOH 8.7 ml KOH 76 Using Solutions in Chemical Reactions Example: What is the molarity of a barium hydroxide solution if 44.1 ml of 0.10 M HCl is required to react with 8. ml of the Ba(OH) solution? 77 Using Solutions in Chemical Reactions Example: What is the molarity of a barium hydroxide solution if 44.1 ml of 0.10 M HCl is required to react with 8. ml of the Ba(OH) solution? Ba(OH) HCl BaCl H O 0.007 mol Ba(OH) 0.08 L Ba(OH) (0.0441L HCl)(0.10 M HCl) = 0.00454 mol HCl 1mol Ba(OH) 0.00454 mol HCl mol HCl = 0.007 mol Ba(OH) = 0.059M Ba(OH) 78
The Mole Method and Molarity Example: Calculate the molarity of a sulfuric acid solution if. ml of it reacts with 0.1 g of Na CO. NaCO HSO 4 NaSO 4 CO HO 79 The Mole Method and Molarity Example: An impure sample of potassium hydrogen phthalate, KHP, had a mass of 0.884 g. It was dissolved in water and titrated with 1.5 ml of 0.100 M NaOH solution. Calculate the percent purity of the KHP sample. NaOH KHP NaKP H O 80