CHAPTER 4 Reactions in Aqueous Solutions 1 1
Aqueous Solutions aqueous solutions -solute dissolved in water nonelectrolytes - aqueous solutions do not conduct electricity C2H5OH - ethanol 10 10
Aqueous Solutions C6H12O6 - glucose (blood sugar) 11 11
Aqueous Solutions C12H22O11 - sucrose (table sugar) 12 12
Properties of Aqueous Solution Molecular compounds in water (e.g., CH3OH): no ions are formed. If there are no ions in solution, there is nothing to transport electric charge.
Properties of Aqueous Solution Ionic Compounds in Water Ions dissociate in water. In solution, each ion is surrounded by water molecules. Transport of ions through solution causes flow of current.
Properties of Aqueous Solution
Properties of Aqueous Solution Strong electrolytes: completely dissociate in solution. For example: Weak electrolytes: produce a small concentration of ions when they dissolve. These ions exist in equilibrium with the unionized substance.
Properties of Aqueous Solution strong electrolytes - extremely good conductors of electricity HCl, HNO3, etc. strong soluble acids NaOH, KOH, etc. strong soluble bases NaCl, KBr, etc. soluble ionic salts ionize in water essentially 100% 13 13
Properties of Aqueous Solution weak electrolytes CH3COOH, (COOH)2 weak acids NH3, Fe(OH)3 weak bases some soluble covalent salts ionize in water much less than 100% 14 14
Strong and Weak Acids acids generate H+ in aqueous solutions strong acids ionize 100% in water 15 15
Strong and Weak Acids acids generate H+ in aqueous solutions strong acids ionize 100% in water 15 15
entire list of strong water soluble acids Strong and Weak Acids entire list of strong water soluble acids HCl - hydrochloric acid HBr - hydrobromic acid HI - hydroiodic acid HNO3 - nitric acid H2SO4 - sulfuric acid HClO3 - chloric acid HClO4 - perchloric acid 16 16
Strong and Weak Acids weak acids ionize less than 100% in water 10% or less! some common weak acids HF - hydrofluoric acid CH3COOH - acetic acid (vinegar) H2CO3 - carbonic acid (soda water) H2SO3 - sulfurous acid HNO2 - nitrous acid H3PO4 - phosphoric acid 17 17
Strong and Weak Acids weak acids ionize as reversible or equilibrium reactions CH3COOH acetic acid why they ionize less than 100% 18 18
Strong Soluble Bases bases produce OH- ions in solution strong soluble bases ionize 100% in water 19 19
Strong Soluble Bases entire list of strong soluble bases LiOH - lithium hydroxide NaOH - sodium hydroxide KOH - potassium hydroxide RbOH - rubidium hydroxide CsOH - cesium hydroxide Ca(OH)2 - calcium hydroxide Sr (OH)2 - strontium hydroxide Ba (OH)2 - barium hydroxide 20 20
Insoluble Bases ionic but insoluble in water not very basic Cu(OH)2 - copper (II) hydroxide Fe(OH)2 - iron (II) hydroxide Fe(OH)3 - iron (III) hydroxide Zn(OH)2 - zinc (II) hydroxide Mg(OH)2 - magnesium hydroxide 21 21
Weak Bases covalent compounds that ionize slightly in water NH3 - ammonia is most common 22 22
Solubility Rules strong acids strong bases soluble ionic salts completely water soluble strong bases soluble ionic salts Figure 4.3, page 80 of textbook 23 23
Acids, Bases, and Salts Dissociation = pre-formed ions in solid move apart in solution. Ionization = neutral substance forms ions in solution. Acid = substances that ionizes to form H+ in solution (e.g. HCl, HNO3, CH3CO2H, lemon, lime, vitamin C). Bases = substances that react with the H+ ions formed by acids (e.g. NH3, Drano™, Milk of Magnesia™).
Acids, Bases, and Salts
Reactions in Aqueous Solutions molecular equations all reactants & products in molecular or ionic form total ionic equation ions as they exist in solution 24 24
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 25 25
Displacement Reactions Metathesis reactions involve swapping ions in solution: AX + BY AY + BX. Metathesis reactions will lead to a change in solution if one of three things occurs: an insoluble solid is formed (precipitate), weak or nonelectrolytes are formed, or an insoluble gas is formed.
Displacement Reactions
Displacement Reactions
Precipitation Reactions form an insoluble compound crystals molecular equation 26 26
Precipitation Reactions form an insoluble compound crystals molecular equation total ionic reaction 26 26
Precipitation Reactions total ionic reaction 26 26
Precipitation Reactions total ionic reaction net ionic reaction 26 26
Precipitation Reactions net ionic reaction 28 28
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Acids, Bases, and Salts Neutralization occurs when a solution of an acid and a base are mixed: HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq) Notice we form a salt (NaCl) and water. Salt = ionic compound whose cation comes from a base and anion from an acid. Neutralization between acid and metal hydroxide produces water and a salt.
Acid-Base Reactions acid + base salt + water recall the acids and bases from before molecular equation 30 30
Acid-Base Reactions acid + base salt + water recall the acids and bases from before molecular equation total ionic equation 30 30
Acid-Base Reactions total ionic equation net ionic equation 31 31
Acid-Base Reactions net ionic equation 32 32
Acid-Base Reactions molecular equation 33 33
Acid-Base Reactions molecular equation total ionic equation 33 33
Acid-Base Reactions total ionic equation net ionic equation 34 34
Acid-Base Reactions net ionic equation 35 35
Molarity: Moles of solute per liter of solution. Solution Composition Solution = solute dissolved in solvent. Solute: present in smallest amount. Water as solvent = aqueous solutions. Change concentration by using different amounts of solute and solvent. Molarity: Moles of solute per liter of solution. If we know: molarity and liters of solution, we can calculate moles (and mass) of solute.
Solution Composition Molarity: Moles of solute per liter of solution.
Concentration of Solutions solution - one substance dissolved in another, commonly one is a liquid concentration - amount of solute dissolved in a solvent “sweet tea” mixed drinks units of concentration 20 14 27 17 3 55 55 3
Molarity (Molar Concentration) molarity = mol solute/L of solution = M Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. 22 16 29 19 5 62 62 5
Molarity (Molar Concentration) molarity = mol solute/L of solution = M Calculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution. 22 16 29 19 5 63 63 5
Molarity (Molar Concentration) Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 . 22 16 29 19 5 64 64 5
Molarity (Molar Concentration) Determine the mass of calcium nitrate required to prepare 3.50 L of 0.800 M Ca(NO3)2 . 22 16 29 19 5 65 65 5
Dilution of Solutions take a concentrated solution and add water to it make tea “less sweet” number of moles of solute remains constant M1V1 = M2V2 works because # of moles is constant If 10 mL of 12 M HCl is added to enough water to give 100 mL of solution, what is concentration of solution? 24 18 31 21 7 68 68 7
Dilution of Solutions take a concentrated solution and add water to it make tea “less sweet” number of moles of solute remains constant M1V1 = M2V2 works because # of moles is constant If 10 mL of 12 M HCl is added to enough water to give 100 mL of solution, what is concentration of solution? (10 mL)(12 M) = M2(100 mL) M2 = 1.2 M HCl 24 18 31 21 7 69 69 7
Dilution of Solutions What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? 25 19 32 22 8 70 70 8
Dilution of Solutions What volume of 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution? 25 19 32 22 8 71 71 8
Stoichiometry with Solutions There are two different types of units: laboratory units (macroscopic units: measure in lab); chemical units (microscopic units: relate to moles). Always convert the laboratory units into chemical units first. Grams are converted to moles using molar mass. Volume or molarity are converted into moles using M = mol/L. Use the stoichiometric coefficients to move between reactants and product.
Stoichiometry with Solutions
Stoichiometry with Solutions What volume of 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4? 26 20 33 23 9 76 76 9
Stoichiometry with Solutions What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate? What mass of aluminum hydroxide precipitates? Al(NO3)3 + 3 NaOH ® Al(OH)3 + 3 NaNO3 27 21 34 24 10 77 77 10
Stoichiometry with Solutions What volume of 0.200 M NaOH will react with 50.0 mL 0f 0.200 M aluminum nitrate? What mass of aluminum hydroxide precipitates? Al(NO3)3 + 3 NaOH ® Al(OH)3 + 3 NaNO3 27 21 34 24 10 78 78 10
Stoichiometry in Solution What mass of Al(OH)3 precipitates in (a)? 79 79
Stoichiometry in Solution What mass of Al(OH)3 precipitates in (a)? 80 80
Titrations
Titrations Suppose we know the molarity of a NaOH solution and we want to find the molarity of an HCl solution. We know: molarity of NaOH, volume of HCl. What do we want? Molarity of HCl. What do we do? Take a known volume of the HCl solution, measure the mL of NaOH required to react completely with the HCl.
Titrations What do we get? Next step? Can we finish? Volume of NaOH. We know molarity of the NaOH, we can calculate moles of NaOH. Next step? We also know HCl + NaOH NaCl + H2O. Therefore, we know moles of HCl. Can we finish? Knowing mol(HCl) and volume of HCl (20.0 mL above), we can calculate the molarity.
Titrations What is the molarity of a KOH solution if 38.7 mL of KOH solution is required to react with 43.2 mL of 0.223 M HCl? 29 23 36 26 12 84 84 12
Titrations What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution? 29 23 36 26 12 85 85 12
Titrations What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution? 29 23 36 26 12 88 88 12
Titrations What is the molarity of a barium hydroxide solution if 44.1 mL of 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution? 29 23 36 26 12 89 89 12
Oxidation Number Rules The oxidation number of any free, uncombined element is zero. The oxidation number of an element in a simple (monatomic) ion is the charge on the ion. In the formula for any compound, the sum of the oxidation numbers of all elements in the compound is zero. In a polyatomic ion, the sum of the oxidation numbers of the constituent elements is equal to the charge on the ion.
Oxidation Number Rules Hydrogen, H, in combined form, has the oxidation number +1, usually. Oxygen, O, in combined form, has the oxidation number -2, usually.
Oxidation Numbers assign oxidation numbers to each element in these compounds NaNO3 K2Sn(OH)6 H3PO4 37 37
Oxidation Numbers assign oxidation numbers to each element in these compounds NaNO3 K2Sn(OH)6 H3PO4 +1 +5 -2 37 37
Oxidation Numbers assign oxidation numbers to each element in these compounds NaNO3 K2Sn(OH)6 H3PO4 +1 +5 -2 +1 +4 -2 +1 37 37
Oxidation Numbers assign oxidation numbers to each element in these compounds NaNO3 K2Sn(OH)6 H3PO4 +1 +5 -2 +1 +4 -2 +1 +1 +5 -2 37 37
Oxidation Numbers assign oxidation numbers to each element in these polyatomic ions SO32- HCO3- Cr2O72- 39 39
Oxidation Numbers assign oxidation numbers to each element in these polyatomic ions SO32- HCO3- Cr2O72- +4 -2 39 39
Oxidation Numbers assign oxidation numbers to each element in these polyatomic ions SO32- HCO3- Cr2O72- +4 -2 +1+4 -2 39 39
Oxidation Numbers assign oxidation numbers to each element in these polyatomic ions SO32- HCO3- Cr2O72- +4 -2 +1+4 -2 +6 -2 39 39
Redox Reactions oxidation - loss of electrons oxidation number increases reduction - gain of electrons oxidation number decreases or reduces! oxidizing agents - chemical species that gain electrons and oxidize some other species electron thieves they are reduced 40 40
Redox Reactions reducing agents - chemical species that lose electrons and reduce some other species victims of electron thieves they are oxidized How do we balance redox reactions? 41 41
Balancing Redox Reactions Half-Reaction Method 36 37 38 38 38
Half Reaction Method Balancing redox reactions by half reaction method rules: Write the unbalanced reaction Break into 2 half reactions - 1 for oxidation 1 for reduction Mass balance each half reaction by adding appropriate stoichiometric coefficents in acidic solutions we can add H+ or H2O in basic solutions we can add OH- or H2O 37 38 39 39 39
Half Reaction Method Charge balance the half reactions by adding appropriate numbers of electrons Multiply each half reaction by a number to make the number of electrons added equal to the number of electrons consumed Add the two half reactions Eliminate any common terms 38 39 40 40 40
Half Reaction Method Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. 39 40 41 41 41
Half Reaction Method Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. 40 41 42 42 42
Half Reaction Method Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. 41 42 43 43 43
Half Reaction Method Example: Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. 42 43 44 44 44
Half Reaction Method Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 43 44 45 45 45
Half Reaction Method Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 43 44 46 46 45
Half Reaction Method Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 44 45 47 47 46
Half Reaction Method Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 45 46 48 48 47
Half Reaction Method Example: Dichromate ions oxidize iron (II) ions to iron (III) ions and are reduced to chromium (III) ions in acidic solution. Write and balance the net ionic equation for the reaction. 46 47 49 49 48
Half Reaction Method Example: In basic solution hydrogen peroxide oxidizes chromite ions, Cr(OH)4-, to chromate ions, CrO42-. Write and balance the net ionic equation for this reaction. 47 48 50 50 49
Half Reaction Method 48 49 51 51 50
Half Reaction Method Example: When chlorine is bubbled into basic solution, it forms chlorate ions and chloride ions. Write and balance the net ionic equation. 49 50 52 52 51
Half Reaction Method 50 51 53 53 52
Synthesis Question Nylon is made by the reaction of hexamethylene diammine with adipic acid 90
Synthesis Question in a 1 to 1 mole ratio. The structure of nylon is: where the value of n is typically 450,000. On a daily basis, a DuPont factory makes 1.5 million pounds of nylon. How many pounds of hexamethylene diamine and adipic acid must they have available in the plant each day? 91
Group Activity Manganese dioxide, potassium hydroxide and oxygen react in the following fashion: A mixture of 272.9 g of MnO2, 26.6 L of 0.250 M KOH, and 41.92 g of O2 is allowed to react as shown above. After the reaction is finished, 347.6 g of K2MnO4 is separated from the reaction mixture. What is the per cent yield of this reaction? 92