Solutions When table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous the salt is still there, as you can.

Aqueous Reactions: Concentration Double-displacement Reaction Ionic Equation Redox Reaction

Solutions When table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous the salt is still there, as you can tell from the taste, or simply boiling away the water Homogeneous mixtures are called solutions The component of the solution that changes state is called the solute The component that keeps its state is called the solvent if both components start in the same state, the major component is the solvent Tro: Chemistry: A Molecular Approach, 2/e

Describing Solutions Because solutions are mixtures, the composition can vary from one sample to another pure substances have constant composition saltwater samples from different seas or lakes have different amounts of salt So to describe solutions accurately, we must describe how much of each component is present we saw that with pure substances, we can describe them with a single name because all samples are identical Tro: Chemistry: A Molecular Approach, 2/e

Solution Concentration
Qualitatively, solutions are often described as dilute or concentrated Dilute solutions have a small amount of solute compared to solvent Concentrated solutions have a large amount of solute compared to solvent Tro: Chemistry: A Molecular Approach, 2/e

Concentrations—Quantitative Descriptions of Solutions
A more precise method for describing a solution is to quantify the amount of solute in a given amount of solution Concentration = amount of solute in a given amount of solution occasionally amount of solvent Tro: Chemistry: A Molecular Approach, 2/e

Solution Concentration Molarity
Moles of solute per 1 liter of solution Used because it describes how many molecules of solute in each liter of solution Tro: Chemistry: A Molecular Approach, 2/e

Preparing 1 L of a 1.00 M NaCl Solution
Tro: Chemistry: A Molecular Approach, 2/e

because most solutions are between 0 and 18 M, the answer makes sense
Example 4.5: Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution Given: Find: 25.5 g KBr, 1.75 L solution molarity, M Conceptual Plan: Relationships: 1 mol KBr = g, M = moles/L g KBr mol KBr L sol’n M Solution: Check: because most solutions are between 0 and 18 M, the answer makes sense Tro: Chemistry: A Molecular Approach, 2/e

Practice — What Is the molarity of a solution containing 3
Practice — What Is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in mL of solution? Tro: Chemistry: A Molecular Approach, 2/e

Practice — What Is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in mL of solution? Given: Find: 0.20 mol NH3, L solution M 3.4 g NH3, mL solution M Conceptual Plan: Relationships: M = mol/L, 1 mol NH3 = g, 1 mL = L g NH3 mol NH3 M mL sol’n L sol’n Solve: Check: the unit is correct, the number is reasonable because the fraction of moles is less than the fraction of liters Tro: Chemistry: A Molecular Approach, 2/e

Practice — What Is the molarity of a solution containing 3.4 g of NH3 (MM 17.03) in mL of solution? Given: Find: 3.4 g NH3, mL solution M 0.20 mol NH3, L solution M Conceptual Plan: Relationships: M = mol/L, 1 mol NH3 = g, 1 mL = L g NH3 mol NH3 M mL sol’n L sol’n Solve: Check: the unit is correct, the number is reasonable because the fraction of moles is less than the fraction of liters Tro: Chemistry: A Molecular Approach, 2/e 11 11

Using Molarity in Calculations
Molarity shows the relationship between the moles of solute and liters of solution If a sugar solution concentration is 2.0 M, then 1 liter of solution contains 2.0 moles of sugar 2 liters = 4.0 moles sugar 0.5 liters = 1.0 mole sugar 1 L solution : 2 moles sugar Tro: Chemistry: A Molecular Approach, 2/e

Example 4.6: How many liters of 0.125 M NaOH contain 0.255 mol NaOH?
Given: Find: 0.125 M NaOH, mol NaOH liters, L Conceptual Plan: Relationships: 0.125 mol NaOH = 1 L solution mol NaOH L sol’n Solution: Check: because each L has only mol NaOH, it makes sense that mol should require a little more than 2 L Tro: Chemistry: A Molecular Approach, 2/e

Practice — Determine the mass of CaCl2 (MM = 110. 98) in 1. 75 L of 1
Practice — Determine the mass of CaCl2 (MM = ) in 1.75 L of 1.50 M solution Tro: Chemistry: A Molecular Approach, 2/e

Practice — Determine the mass of CaCl2 (MM = 110. 98) in 1. 75 L of 1
Practice — Determine the mass of CaCl2 (MM = ) in 1.75 L of 1.50 M solution Given: Find: 1.50 M CaCl2, 1.75 L mass CaCl2, g mol CaCl2 L sol’n g CaCl2 Conceptual Plan: Relationships: 1.50 mol CaCl2 = 1 L solution; g CaCl2 = 1 mol Solution: Check: because each L has 1.50 mol CaCl2, it makes sense that 1.75 L should have almost 3 moles Tro: Chemistry: A Molecular Approach, 2/e

Example: How would you prepare 250. 0 mL of a 1
Example: How would you prepare mL of a 1.00 M solution CuSO45 H2O(MM )? Given: Find: 250.0 mL solution mass CuSO4 5 H2O, g Conceptual Plan: Relationships: 1.00 L sol’n = 1.00 mol; 1 mL = L; 1 mol = g mL sol’n L sol’n g CuSO4 mol CuSO4 Solution: Dissolve 62.4 g of CuSO4∙5H2O in enough water to total mL Check: the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter Tro: Chemistry: A Molecular Approach, 2/e

Practice – How would you prepare 250. 0 mL of 0. 150 M CaCl2 (MM = 110

Practice – How would you prepare 250.0 mL of 0.150 M CaCl2?
Given: Find: 250.0 mL solution mass CaCl2, g Conceptual Plan: Relationships: 1.00 L sol’n = mol; 1 mL = 0.001L; 1 mol = g mL sol’n L sol’n g CaCl2 mol CaCl2 Solution: Dissolve 4.16 g of CaCl2 in enough water to total mL Check: the unit is correct, the magnitude seems reasonable as the volume is ¼ of a liter Tro: Chemistry: A Molecular Approach, 2/e

moles solute in solution 1 = moles solute in solution 2
Dilution Often, solutions are stored as concentrated stock solutions To make solutions of lower concentrations from these stock solutions, more solvent is added the amount of solute doesn’t change, just the volume of solution moles solute in solution 1 = moles solute in solution 2 The concentrations and volumes of the stock and new solutions are inversely proportional M1∙V1 = M2∙V2 Tro: Chemistry: A Molecular Approach, 2/e

Example 4. 7: To what volume should you dilute 0. 200 L of 15
Example 4.7: To what volume should you dilute L of 15.0 M NaOH to make 3.00 M NaOH? Given: Find: V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M V2, L Conceptual Plan: Relationships: M1V1 = M2V2 V1, M1, M2 V2 Solution: Check: because the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO3 to mL? Tro: Chemistry: A Molecular Approach, 2/e

Practice – What is the concentration of a solution prepared by diluting 45.0 mL of 8.25 M HNO3 to mL? Given: Find: V1 = 45.0 mL, M1 = 8.25 M, V2 = mL M2, L Conceptual Plan: Relationships: M1V1 = M2V2 V1, M1, V2 M2 Solution: Check: because the solution is diluted by a factor of 3, the molarity should decrease by a factor of 3, and it does Tro: Chemistry: A Molecular Approach, 2/e

Practice – How would you prepare 200. 0 mL of 0
Practice – How would you prepare mL of 0.25 M NaCl solution from a 2.0 M solution? Tro: Chemistry: A Molecular Approach, 2/e

Practice – How would you prepare 200. 0 mL of 0
Practice – How would you prepare mL of 0.25 M NaCl solution from a 2.0 M solution? Given: Find: M1 = 2.0 M, M2 = 0.25 M, V2 = mL V1, L Conceptual Plan: Relationships: M1V1 = M2V2 M1, M2, V2 V1 Solution: Dilute 25 mL of 2.0 M solution up to mL Check: because the solution is diluted by a factor of 8, the volume should increase by a factor of 8, and it does Tro: Chemistry: A Molecular Approach, 2/e

Solution Stoichiometry
Because molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction Tro: Chemistry: A Molecular Approach, 2/e

Example 4. 8: What volume of 0
Example 4.8: What volume of M KCl is required to completely react with L of M Pb(NO3)2 in the reaction 2 KCl(aq) + Pb(NO3)2(aq)  PbCl2(s) + 2 KNO3(aq)? Given: Find: 0.150 M KCl, L of M Pb(NO3)2 L KCl Conceptual Plan: Relationships: 1 L Pb(NO3)2 = mol, 1 L KCl = mol, 1 mol Pb(NO3)2 : 2 mol KCl L Pb(NO3)2 mol KCl L KCl mol Pb(NO3)2 Solution: Check: because you need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be more than 2x the volume of Pb(NO3)2 Tro: Chemistry: A Molecular Approach, 2/e

Practice — Solution stoichiometry
43.8 mL of M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. What is the molarity of the base? 2 HCl + Ba(OH)2 ® BaCl2 + 2 H2O Tro: Chemistry: A Molecular Approach, 2/e

Practice – 43. 8 mL of 0. 107 M HCl is needed to neutralize 37
Practice – 43.8 mL of M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. What is the molarity of the base? 2 HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2 H2O(aq) Given: Find: 43.8 mL of M HCl, 37.6 mL Ba(OH)2 M Ba(OH)2 Conceptual Plan: Relationships: 1 mL= L, 1 L HCl = mol, 1 mol Ba(OH)2 : 2 mol HCl L HCl mol Ba(OH)2 mol HCl M Ba(OH)2 L Ba(OH)2 Solution: Check: the units are correct, the number makes sense because there are fewer moles than liters Tro: Chemistry: A Molecular Approach, 2/e

Practice – 43. 8 mL of 0. 107 M HCl is needed to neutralize 37
Practice – 43.8 mL of M HCl is needed to neutralize 37.6 mL of Ba(OH)2 solution. What is the molarity of the base? 2 HCl(aq) + Ba(OH)2(aq)  BaCl2(aq) + 2 H2O(aq) Given: Find: 43.8 mL of M HCl, 37.6 mL Ba(OH)2 M Ba(OH)2 Conceptual Plan: Relationships: 1 mL= L, 1 L HCl = mol, 1 mol Ba(OH)2 : 2 mol HCl L HCl mol Ba(OH)2 mol HCl M Ba(OH)2 L Ba(OH)2 Solution: Check: the units are correct, the number makes sense because there are fewer moles than liters Tro: Chemistry: A Molecular Approach, 2/e 29 29

What Happens When a Solute Dissolves?
There are attractive forces between the solute particles holding them together There are also attractive forces between the solvent molecules When we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules If the attractions between solute and solvent are strong enough, the solute will dissolve Tro: Chemistry: A Molecular Approach, 2/e

Table Salt Dissolving in Water
Each ion is attracted to the surrounding water molecules and pulled off and away from the crystal When it enters the solution, the ion is surrounded by water molecules, insulating it from other ions The result is a solution with free moving charged particles able to conduct electricity Tro: Chemistry: A Molecular Approach, 2/e

Electrolytes and Nonelectrolytes
Materials that dissolve in water to form a solution that will conduct electricity are called electrolytes Materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes Tro: Chemistry: A Molecular Approach, 2/e

Molecular View of Electrolytes and Nonelectrolytes
To conduct electricity, a material must have charged particles that are able to flow Electrolyte solutions all contain ions dissolved in the water ionic compounds are electrolytes because they dissociate into their ions when they dissolve Nonelectrolyte solutions contain whole molecules dissolved in the water generally, molecular compounds do not ionize when they dissolve in water the notable exception being molecular acids Tro: Chemistry: A Molecular Approach, 2/e

Salt vs. Sugar Dissolved in Water
ionic compounds dissociate into ions when they dissolve molecular compounds do not dissociate when they dissolve Tro: Chemistry: A Molecular Approach, 2/e

Acids Acids are molecular compounds that ionize when they dissolve in water the molecules are pulled apart by their attraction for the water when acids ionize, they form H+ cations and also anions The percentage of molecules that ionize varies from one acid to another Acids that ionize virtually 100% are called strong acids HCl(aq)  H+(aq) + Cl−(aq) Acids that only ionize a small percentage are called weak acids HF(aq)  H+(aq) + F−(aq) Tro: Chemistry: A Molecular Approach, 2/e

Strong and Weak Electrolytes
Strong electrolytes are materials that dissolve completely as ions ionic compounds and strong acids their solutions conduct electricity well Weak electrolytes are materials that dissolve mostly as molecules, but partially as ions weak acids their solutions conduct electricity, but not well When compounds containing a polyatomic ion dissolve, the polyatomic ion stays together HC2H3O2(aq)  H+(aq) + C2H3O2−(aq) Tro: Chemistry: A Molecular Approach, 2/e

Classes of Dissolved Materials

Dissociation and Ionization
When ionic compounds dissolve in water, the anions and cations are separated from each other. This is called dissociation. Na2S(aq)  2 Na+(aq) + S2-(aq) When compounds containing polyatomic ions dissociate, the polyatomic group stays together as one ion Na2SO4(aq)  2 Na+(aq) + SO42−(aq) When strong acids dissolve in water, the molecule ionizes into H+ and anions H2SO4(aq)  2 H+(aq) + SO42−(aq) Tro: Chemistry: A Molecular Approach, 2/e

CaCl2(aq)  Ca2+(aq) + 2 Cl−(aq)
Practice – Write the equation for the process that occurs when the following strong electrolytes dissolve in water CaCl2 HNO3 (NH4)2CO3 CaCl2(aq)  Ca2+(aq) + 2 Cl−(aq) HNO3(aq)  H+(aq) + NO3−(aq) (NH4)2CO3(aq)  2 NH4+(aq) + CO32−(aq) Tro: Chemistry: A Molecular Approach, 2/e

Solubility of Ionic Compounds
Some ionic compounds, such as NaCl, dissolve very well in water at room temperature Other ionic compounds, such as AgCl, dissolve hardly at all in water at room temperature Compounds that dissolve in a solvent are said to be soluble, where as those that do not are said to be insoluble NaCl is soluble in water, AgCl is insoluble in water the degree of solubility depends on the temperature even insoluble compounds dissolve, just not enough to be meaningful Tro: Chemistry: A Molecular Approach, 2/e

When Will a Salt Dissolve?
Predicting whether a compound will dissolve in water is not easy The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental results we call this method the empirical method Tro: Chemistry: A Molecular Approach, 2/e

Solubility Rules Compounds that Are Generally Soluble in Water

Solubility Rules Compounds that Are Generally Insoluble in Water

Practice – Determine if each of the following is soluble in water
KOH AgBr CaCl2 Pb(NO3)2 PbSO4 KOH is soluble because it contains K+ AgBr is insoluble; most bromides are soluble, but AgBr is an exception CaCl2 is soluble; most chlorides are soluble, and CaCl2 is not an exception Pb(NO3)2 is soluble because it contains NO3− PbSO4 is insoluble; most sulfates are soluble, but PbSO4 is an exception Tro: Chemistry: A Molecular Approach, 2/e

Precipitation Reactions
Precipitation reactions are reactions in which a solid forms when we mix two solutions reactions between aqueous solutions of ionic compounds produce an ionic compound that is insoluble in water the insoluble product is called a precipitate picture from Tro Intro Chem, 2nd Ed. Tro: Chemistry: A Molecular Approach, 2/e

2 KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2 KNO3(aq)

No Precipitate Formation = No Reaction
KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq) all ions still present,  no reaction Tro: Chemistry: A Molecular Approach, 2/e

Process for Predicting the Products of a Precipitation Reaction
1. Determine what ions each aqueous reactant has 2. Determine formulas of possible products exchange ions (+) ion from one reactant with (-) ion from other balance charges of combined ions to get formula of each product 3. Determine solubility of each product in water use the solubility rules if product is insoluble or slightly soluble, it will precipitate 4. If neither product will precipitate, write no reaction after the arrow Tro: Chemistry: A Molecular Approach, 2/e

Process for Predicting the Products of a Precipitation Reaction
5. If any of the possible products are insoluble, write their formulas as the products of the reaction using (s) after the formula to indicate solid. Write any soluble products with (aq) after the formula to indicate aqueous. 6. Balance the equation remember to only change coefficients, not subscripts Tro: Chemistry: A Molecular Approach, 2/e

Example 4.10: Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride 1. Write the formulas of the reactants K2CO3(aq) + NiCl2(aq)  2. Determine the possible products a) determine the ions present (K+ + CO32−) + (Ni2+ + Cl−)  b) exchange the Ions (K+ + CO32−) + (Ni2+ + Cl−)  (K+ + Cl−) + (Ni2+ + CO32−) c) write the formulas of the products balance charges K2CO3(aq) + NiCl2(aq)  KCl + NiCO3 Tro: Chemistry: A Molecular Approach, 2/e 50

does not apply because NiCO3 is insoluble
Example 4.10: Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride 3. Determine the solubility of each product KCl is soluble NiCO3 is insoluble 4. If both products are soluble, write no reaction does not apply because NiCO3 is insoluble Tro: Chemistry: A Molecular Approach, 2/e

Example 4.10: Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(II) chloride 5. Write (aq) next to soluble products and (s) next to insoluble products K2CO3(aq) + NiCl2(aq)  KCl(aq) + NiCO3(s) 6. Balance the equation K2CO3(aq) + NiCl2(aq)  2 KCl(aq) + NiCO3(s) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Predict the products and balance the equation
KCl(aq) + AgNO3(aq) ® Na2S(aq) + CaCl2(aq) ® (K+ + Cl−) + (Ag+ + NO3−) → (K+ + NO3−) + (Ag+ + Cl−) KCl(aq) + AgNO3(aq) → KNO3 + AgCl KCl(aq) + AgNO3(aq) ® KNO3(aq) + AgCl(s) (Na+ + S2−) + (Ca2+ + Cl−) → (Na+ + Cl−) + (Ca2+ + S2−) Na2S(aq) + CaCl2(aq) → NaCl + CaS Na2S(aq) + CaCl2(aq) ® NaCl(aq) + CaS(aq) No reaction Tro: Chemistry: A Molecular Approach, 2/e

(NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ® NH4C2H3O2(aq) + PbSO4(s)
Practice – Write an equation for the reaction that takes place when an aqueous solution of (NH4)2SO4 is mixed with an aqueous solution of Pb(C2H3O2)2. (NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ® (NH4+ + SO42−) + (Pb2+ + C2H3O2−) → (NH4+ + C2H3O2−) + (Pb2+ + SO42−) (NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ® NH4C2H3O2 + PbSO4 (NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ® NH4C2H3O2(aq) + PbSO4(s) (NH4)2SO4(aq) + Pb(C2H3O2)2(aq) ® 2 NH4C2H3O2(aq) + PbSO4(s) Tro: Chemistry: A Molecular Approach, 2/e

2 KOH(aq) + Mg(NO3)2(aq) ® 2 KNO3(aq) + Mg(OH)2(s)
Ionic Equations Equations that describe the chemicals put into the water and the product molecules are called molecular equations 2 KOH(aq) + Mg(NO3)2(aq) ® 2 KNO3(aq) + Mg(OH)2(s) Equations that describe the material’s structure when dissolved are called complete ionic equations aqueous strong electrolytes are written as ions soluble salts, strong acids, strong bases insoluble substances, weak electrolytes, and nonelectrolytes are written in molecule form solids, liquids, and gases are not dissolved, therefore molecule form 2K+(aq) + 2OH−(aq) + Mg2+(aq) + 2NO3−(aq) ® 2K+(aq) + 2NO3−(aq) + Mg(OH)2(s) Tro: Chemistry: A Molecular Approach, 2/e

Ionic Equations Ions that are both reactants and products are called spectator ions 2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) ® 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s) An ionic equation in which the spectator ions are removed is called a net ionic equation 2 OH−(aq) + Mg2+(aq) ® Mg(OH)2(s) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Write the ionic and net ionic equation
K2CO3(aq) + 2HC2H3O2(aq) ® H2O(l) + CO2(g) + 2KC2H3O2(aq) 2HI(aq) + BaSO3(s) ® BaI2(aq) + H2O(l) + SO2(g) BaBr2(aq) + 2AgNO3(aq) ® 2AgBr(s) + Ba(NO3)2(aq) 2 NH4OH(aq) + CuSO4(aq) ® Cu(OH)2(s) + (NH4)2SO4(aq) Tro: Chemistry: A Molecular Approach, 2/e 57

K2SO4(aq) + 2 AgNO3(aq) ® 2 KNO3(aq) + Ag2SO4(s) sol. salt sol.salt sol. salt insoluble salt 2K+(aq) + SO42−(aq) + 2Ag+(aq) + 2NO3−(aq) ® 2K+(aq) + 2NO3−(aq) + Ag2SO4(s) NIE: 2 Ag+(aq) + SO42−(aq) ® Ag2SO4(s) Tro: Chemistry: A Molecular Approach, 2/e

Na2CO3(aq) + 2 HCl(aq) ® 2 NaCl(aq) + CO2(g) + H2O(l) sol. salt sol.acid sol. salt nonelectrolyte 2Na+(aq) + CO32−(aq) + 2H+(aq) + 2Cl−(aq) ® 2Na+(aq) + 2Cl−(aq) + CO2(g) + H2O(l) NIE: CO32−(aq) + 2 H+(aq) ® CO2(g) + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 59

Weak Electrolytes and Non-electrolytes Does NOT dissociate
Weak acids (acetic acid, nitrous acid, hydrofluoric acid, etc), Weak base (ammonium hydroxide, etc), Insoluble salts largely remain undissociated in water HC2H3O2(aq)  H+(aq) + C2H3O2-(aq) NH4OH(aq)  NH4+(aq) + OH-(aq) PbCl2(s) CaCO3(s) BaSO4(s) Pure gas, liquid, or solid exists as non-dissolved states CO2(g), H2S(g), H2O(l), Al(s), Hg(s) 60

Write the ionic and net ionic equation involving weak electrolyte(s)
Na2CO3(aq) + 2 HC2H3O2(aq) ® 2 NaC2H3O2(aq) + CO2(g) + H2O(l) sol. salt weak acid sol. salt nonelectrolyte 2Na+(aq) + CO32−(aq) + 2HC2H3O2(aq) ® 2Na+(aq) + 2C2H3O2−(aq) + CO2(g) + H2O(l) NIE: CO32−(aq) + 2HC2H3O2(aq) ® 2C2H3O2−(aq) + CO2(g) + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e 61

Write the ionic and net ionic equation involving weak electrolyte(s)
NH4OH(aq) + HC2H3O2(aq) ® NH4C2H3O2(aq) + H2O(l) weak base weak acid sol. salt nonelectrolyte NH4OH(aq) + HC2H3O2 (aq) ® C2H3O2−(aq) + NH4+(aq) + H2O(l) NIE: no spectator ions, same as Complete Ionic Equations Tro: Chemistry: A Molecular Approach, 2/e 62

Acid-Base Reactions Also called neutralization reactions because the acid and base neutralize each other’s properties 2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l) The net ionic equation for an acid-base reaction between strong acid and strong base is H+(aq) + OH(aq)  H2O(l) as long as the salt that forms is soluble in water Tro: Chemistry: A Molecular Approach, 2/e

Acids and Bases in Solution
Acids ionize in water to form H+ ions more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H3O+ most chemists use H+ and H3O+ interchangeably Bases dissociate in water to form OH ions bases, such as NH3, that do not contain OH ions, produce OH by pulling H off water molecules In the reaction of an acid with a base, the H+ from the acid combines with the OH from the base to make water The cation from the base combines with the anion from the acid to make the salt acid + base salt + water Tro: Chemistry: A Molecular Approach, 2/e

Common Acids Tro: Chemistry: A Molecular Approach, 2/e

Common Bases Tro: Chemistry: A Molecular Approach, 2/e

HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)

Example: Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 1. Write the formulas of the reactants HNO3(aq) + Ca(OH)2(aq)  2. Determine the possible products a) determine the ions present when each reactant dissociates or ionizes (H+ + NO3−) + (Ca2+ + 2OH−)  b) exchange the ions, H+ combines with OH− to make H2O(l) (H+ + NO3−) + (Ca2+ + OH−)  (Ca2+ + NO3−) + H2O(l) c write the formula of the salt (H+ + NO3−) + (Ca2+ + OH−)  Ca(NO3)2 + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e

Example: Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 3. Determine the solubility of the salt Ca(NO3)2 is soluble 4. Write an (s) after the insoluble products and an (aq) after the soluble products HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + H2O(l) 5. Balance the equation 2 HNO3(aq) + Ca(OH)2(aq)  Ca(NO3)2(aq) + 2 H2O(l) Tro: Chemistry: A Molecular Approach, 2/e

Example: Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 6. Dissociate all aqueous strong electrolytes to get complete ionic equation not H2O 2 H+(aq) + 2 NO3−(aq) + Ca2+(aq) + 2 OH−(aq)  Ca2+(aq) + 2 NO3−(aq) + H2O(l) 7. Eliminate spectator ions to get net-ionic equation 2 H+(aq) + 2 OH−(aq)  2 H2O(l) H+(aq) + OH−(aq)  H2O(l) Tro: Chemistry: A Molecular Approach, 2/e

HCl(aq) + Ba(OH)2(aq) ® H2SO4(aq) + Sr(OH)2(aq) ® (H+ + Cl−) + (Ba2+ + OH−) → (H+ + OH−) + (Ba2+ + Cl−) HCl(aq) + Ba(OH)2(aq) → H2O(l) + BaCl2 2 HCl(aq) + Ba(OH)2(aq) ® 2 H2O(l) + BaCl2(aq) (H+ + SO42−) + (Sr2+ + OH−) → (H+ + OH−) + (Sr2+ + SO42−) H2SO4(aq) + Sr(OH)2(aq) → H2O(l) + SrSO4 H2SO4(aq) + Sr(OH)2(aq) → 2 H2O(l) + SrSO4 H2SO4(aq) + Sr(OH)2(aq) ® 2 H2O(l) + SrSO4(s) Tro: Chemistry: A Molecular Approach, 2/e

Titration Often in the lab, a solution’s concentration is determined by reacting it with another material and using stoichiometry – this process is called titration In the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed. At this point, called the endpoint, the reactants are in their stoichiometric ratio. the unknown solution is added slowly from an instrument called a burette a long glass tube with precise volume markings that allows small additions of solution Tro: Chemistry: A Molecular Approach, 2/e

Acid-Base Titrations The difficulty is determining when there has been just enough titrant added to complete the reaction the titrant is the solution in the burette In acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity the chemical is called an indicator At the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH also known as the equivalence point Tro: Chemistry: A Molecular Approach, 2/e

Titration Tro: Chemistry: A Molecular Approach, 2/e

Titration The titrant is the base solution in the burette.
As the titrant is added to the flask, the H+ reacts with the OH– to form water. But there is still excess acid present so the color does not change. At the titration’s endpoint, just enough base has been added to neutralize all the acid. At this point the indicator changes color. Tro: Chemistry: A Molecular Approach, 2/e

12.54 mL of 0.100 M NaOH Given: 10.00 mL HCl
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Write down the given quantity and its units Given: mL HCl 12.54 mL of M NaOH Tro: Chemistry: A Molecular Approach, 2/e

Write down the quantity to find, and/or its units
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given: mL HCl 12.54 mL of M NaOH Write down the quantity to find, and/or its units Find: concentration HCl, M Tro: Chemistry: A Molecular Approach, 2/e

Collect needed equations and conversion factors
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given: mL HCl 12.54 mL of M NaOH Find: M HCl Collect needed equations and conversion factors HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)  1 mole HCl = 1 mole NaOH 0.100 M NaOH 0.100 mol NaOH  1 L sol’n Tro: Chemistry: A Molecular Approach, 2/e

Write a conceptual plan
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given: mL HCl 12.54 mL of M NaOH Find: M HCl Rel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L Write a conceptual plan mL NaOH L NaOH mol NaOH mol HCl mL HCl L HCl Tro: Chemistry: A Molecular Approach, 2/e

Apply the conceptual plan
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given: mL HCl 12.54 mL of M NaOH Find: M HCl Rel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Apply the conceptual plan = 1.25 x 103 mol HCl Tro: Chemistry: A Molecular Approach, 2/e

Apply the conceptual plan
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given: mL HCl 12.54 mL NaOH Find: M HCl Rel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Apply the conceptual plan Tro: Chemistry: A Molecular Approach, 2/e

Check the solution HCl solution = 0.125 M
Example 4.14: The titration of mL of HCl solution of unknown concentration requires mL of M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given: mL HCl 12.54 mL NaOH Find: M HCl Rel: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L M = mol/L CP: mL NaOH → L NaOH → mol NaOH → mol HCl; mL HCl → L HCl & mol  M Check the solution HCl solution = M The units of the answer, M, are correct. The magnitude of the answer makes sense because the neutralization takes less HCl solution than NaOH solution, so the HCl should be more concentrated. Tro: Chemistry: A Molecular Approach, 2/e

Practice − What is the concentration of NaOH solution that requires 27
Practice − What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL of M H2SO4? H2SO NaOH ® Na2SO4 + 2 H2O 50.0 mL 27.5 mL M ? M Tro: Chemistry: A Molecular Approach, 2/e 83

Practice — What is the concentration of NaOH solution that requires 27
Practice — What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL of M H2SO4? 2 NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2 H2O(aq) Given: Find: 50.0 mL of M H2SO4, 27.5 mL NaOH M NaOH Conceptual Plan: Relationships: 1 mL= 0.001L, 1 LH2SO4 = mol, 2mol NaOH : 1mol H2SO4 L H2SO4 mol H2SO4 mol NaOH M NaOH L NaOH Solution: Check: Tro: Chemistry: A Molecular Approach, 2/e 84

Practice — What is the concentration of NaOH solution that requires 27
Practice — What is the concentration of NaOH solution that requires 27.5 mL to titrate 50.0 mL of M H2SO4? 2 NaOH(aq) + H2SO4(aq)  Na2SO4(aq) + 2 H2O(aq) Given: Find: 50.0 mL of M H2SO4, 27.5 mL NaOH M NaOH Conceptual Plan: Relationships: 1 mL= 0.001L, 1 LH2SO4 = mol, 2mol NaOH : 1mol H2SO4 L H2SO4 mol H2SO4 mol NaOH M NaOH L NaOH Solution: Check: the units are correct, the number makes because the volume of NaOH is about ½ the H2SO4, but the stoichiometry says you need twice the moles of NaOH as H2SO4 Tro: Chemistry: A Molecular Approach, 2/e 85 85

Gas-Evolving Reactions
Some reactions form a gas directly from the ion exchange K2S(aq) + H2SO4(aq)  K2SO4(aq) + H2S(g) Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water K2SO3(aq) + H2SO4(aq)  K2SO4(aq) + H2SO3(aq) H2SO3  H2O(l) + SO2(g) Tro: Chemistry: A Molecular Approach, 2/e

NaHCO3(aq) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)

Compounds that Undergo Gas-Evolving Reactions

Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves 1. Write the formulas of the reactants Na2CO3(aq) + HNO3(aq)  2. Determine the possible products a) determine the ions present when each reactant dissociates or ionizes (Na+ + CO32−) + (H+ + NO3−)  b) exchange the anions (Na+ + CO32−) + (H+ + NO3−)  (Na+ + NO3−) + (H+ + CO32−) c) write the formula of compounds balance the charges Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3 Tro: Chemistry: A Molecular Approach, 2/e

Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l)
Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves 3. Check to see if either product is H2S - No 4. Check to see if either product decomposes – Yes H2CO3 decomposes into CO2(g) + H2O(l) Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l) Tro: Chemistry: A Molecular Approach, 2/e

Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l)
Example 4.14: When an aqueous solution of sodium carbonate is added to an aqueous solution of nitric acid, a gas evolves 5. Determine the solubility of other product NaNO3 is soluble 6. Write an (s) after the insoluble products and an (aq) after the soluble products Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l) 7. Balance the equation Tro: Chemistry: A Molecular Approach, 2/e

HCl(aq) + Na2SO3(aq) ® H2SO4(aq) + CaS(aq) ® (H+ + Cl−) + (Na+ + SO32−) → (H+ + SO32−) + (Na+ + Cl−) HCl(aq) + Na2SO3(aq) → H2SO3 + NaCl HCl(aq) + Na2SO3(aq) → H2O(l) + SO2(g) + NaCl 2 HCl(aq) + Na2SO3(aq) ® H2O(l) + SO2(g) + 2 NaCl(aq) (H+ + SO42−) + (Ca2+ + S2−) → (H+ + S2−) + (Ca2+ + SO42−) H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4 H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4(s) Tro: Chemistry: A Molecular Approach, 2/e

Other Patterns in Reactions
The precipitation, acid-base, and gas-evolving reactions all involve exchanging the ions in the solution Other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactions also known as redox reactions many involve the reaction of a substance with O2(g) 4 Fe(s) + 3 O2(g)  2 Fe2O3(s) Tro: Chemistry: A Molecular Approach, 2/e

Combustion as Redox 2 H2(g) + O2(g)  2 H2O(g)

Redox without Combustion 2 Na(s) + Cl2(g)  2 NaCl(s)
2 Na  2 Na+ + 2 e Cl2 + 2 e  2 Cl Tro: Chemistry: A Molecular Approach, 2/e

Reactions of Metals with Nonmetals
Consider the following reactions: 4 Na(s) + O2(g) → 2 Na2O(s) 2 Na(s) + Cl2(g) → 2 NaCl(s) The reactions involve a metal reacting with a nonmetal In addition, both reactions involve the conversion of free elements into ions 4 Na(s) + O2(g) → 2 Na+2O2– (s) 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Tro: Chemistry: A Molecular Approach, 2/e

Oxidation and Reduction
To convert a free element into an ion, the atoms must gain or lose electrons of course, if one atom loses electrons, another must accept them Reactions where electrons are transferred from one atom to another are redox reactions Atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced Leo Ger 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na → Na+ + 1 e– oxidation Cl2 + 2 e– → 2 Cl– reduction Tro: Chemistry: A Molecular Approach, 2/e

Electron Bookkeeping For reactions that are not metal + nonmetal, or do not involve O2, we need a method for determining how the electrons are transferred Chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction even though they look like them, oxidation states are not ion charges! oxidation states are imaginary charges assigned based on a set of rules ion charges are real, measurable charges Tro: Chemistry: A Molecular Approach, 2/e

Rules for Assigning Oxidation States
Rules are in order of priority 1. free elements have an oxidation state = 0 Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g) 2. monatomic ions have an oxidation state equal to their charge Na = +1 and Cl = −1 in NaCl 3. (a) the sum of the oxidation states of all the atoms in a compound is 0 Na = +1 and Cl = −1 in NaCl, (+1) + (−1) = 0 Tro: Chemistry: A Molecular Approach, 2/e

3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = −2 in NO3–, (+5) + 3(−2) = −1 4. (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 in NaCl 4. (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl2 Tro: Chemistry: A Molecular Approach, 2/e

5. in their compounds, nonmetals have oxidation states according to the table below nonmetals higher on the table take priority Tro: Chemistry: A Molecular Approach, 2/e

Example: Determine the oxidation states of all the atoms in a propanoate ion, C3H5O2–
There are no free elements or free ions in propanoate, so the first rule that applies is Rule 3b (C3) + (H5) + (O2) = −1 Because all the atoms are nonmetals, the next rule we use is Rule 5, following the elements in order: H = +1 O = −2 (C3) + 5(+1) + 2(−2) = −1 (C3) = −2 C = −⅔ Note: unlike charges, oxidation states can be fractions! Tro: Chemistry: A Molecular Approach, 2/e

Practice – Assign an oxidation state to each element in the following
Br2 K+ LiF CO2 SO42− Na2O2 Br = 0, (Rule 1) K = +1, (Rule 2) Li = +1, (Rule 4a) & F = −1, (Rule 5) O = −2, (Rule 5) & C = +4, (Rule 3a) O = −2, (Rule 5) & S = +6, (Rule 3b) Na = +1, (Rule 4a) & O = −1 , (Rule 3a) Tro: Chemistry: A Molecular Approach, 2/e

Oxidation and Reduction Another Definition
Oxidation occurs when an atom’s oxidation state increases during a reaction Reduction occurs when an atom’s oxidation state decreases during a reaction CH O2 → CO2 + 2 H2O − – −2 oxidation reduction Tro: Chemistry: A Molecular Approach, 2/e

Oxidation–Reduction Oxidation and reduction must occur simultaneously
if an atom loses electrons another atom must take them The reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized The reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl2 is the oxidizing agent Tro: Chemistry: A Molecular Approach, 2/e

Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O
Example: Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions: Reducing agent Oxidizing agent Fe + MnO4− + 4 H+ → Fe3+ + MnO H2O +7 −2 +1 +3 +4 −2 +1 −2 Oxidation Reduction Tro: Chemistry: A Molecular Approach, 2/e

Practice – Assign oxidation states, determine the element oxidized and reduced, and determine the oxidizing agent and reducing agent in the following reactions: Sn4+ + Ca → Sn Ca2+ F2 + S → SF4 Ca is oxidized, Sn4+ is reduced Ca is the reducing agent, Sn4+ is the oxidizing agent −1 S is oxidized, F is reduced S is the reducing agent, F2 is the oxidizing agent Tro: Chemistry: A Molecular Approach, 2/e

Solutions When table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous the salt is still there, as you can.

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Solutions When table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous the salt is still there, as you can.

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Presentation on theme: "Solutions When table salt is mixed with water, it seems to disappear, or become a liquid – the mixture is homogeneous the salt is still there, as you can."— Presentation transcript:

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