Aqueous Reactions © 2015 Pearson Education, Inc. Lecture Presentation Chapter 4 Reactions in Aqueous Solution James F. Kirby Quinnipiac University Hamden,

Chemical Reactions in Aqueous Solution

Three Main Types of Chemical Reactions in Aqueous Solution 1. Precipitation reactions: Result in formation of an insoluble solid (precipitate) that separates from the solution. 2. Acid-base or neutralization reactions: Result in formation of water and an ionic compound (i.e., a salt) due to transfer of hydrogen ions (i.e., protons). 3. Oxidation-reduction (redox) reactions: Result in formation of new compounds due to transfer of electrons among reactants.

Aqueous solutions Solute: substance present in a smaller amount Solvent: substance present in a (much) larger amount Solution: a homogeneous mixture of 2 or more substances

A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount SolutionSolventSolute Soft drink (l) Air (g) Soft Solder (s) H2OH2O N2N2 Pb Sugar, CO 2 O 2, Ar, CH 4 Sn

Aqueous Reactions © 2015 Pearson Education, Inc. Solutions Solutions are defined as homogeneous mixtures of two or more pure substances. The solvent is present in greatest abundance. All other substances are solutes. When water is the solvent, the solution is called an aqueous solution.

Aqueous Reactions © 2015 Pearson Education, Inc. Aqueous Solutions Substances can dissolve in water by different ways:  Ionic Compounds dissolve by dissociation, where water surrounds the separated ions.  Molecular compounds interact with water, but most do NOT dissociate.  Some molecular substances react with water when they dissolve.

NaCl (s)  Na + (aq) + Cl - (aq)

Polar molecule H 2 O molecule

Negative region (the O atom) Positive region (the H atom) Why is water such an effective solvent for ionic compounds? -- ++ ++

Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner.   H2OH2O

Ionic compounds: hydration Negative “pole” of water pointed toward cation; positive “pole” of water pointed toward anion. Clustering of H 2 O molecules around Na + and Cl - ions keeps the ions apart from each other. Polar compounds: hydrogen bonding Negative “pole” of water pointed toward H on -OH groups of polar molecule; positive “pole” of water pointed toward O on -OH group of polar molecule. Clustering of H 2 O molecules around entire molecules keeps the molecules apart from each other.

Aqueous Reactions © 2015 Pearson Education, Inc. Electrolytes and Nonelectrolytes An electrolyte is a substance that dissociates into ions when dissolved in water. A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

Aqueous Reactions © 2015 Pearson Education, Inc. Electrolytes A strong electrolyte dissociates completely when dissolved in water. A weak electrolyte only dissociates partially when dissolved in water. A nonelectrolyte does NOT dissociate in water.

An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity. A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity. nonelectrolyte weak electrolyte strong electrolyte

Substances that, when dissolved in H 2 O, produce solutions that can conduct electricity. A substance that, when dissolved in H 2 O, does not produces a solution that can conduct electricity.

Strong Electrolyte – 100% dissociation NaCl (s) Na + (aq) + Cl - (aq) H2OH2O Weak Electrolyte – not completely dissociated CH 3 COOH CH 3 COO - (aq) + H + (aq) Conduct electricity in solution? Cations (+) and Anions (-)

Strengths of Electrolytes: Complete, Partial, or No Ionization in Water STRONG ELECTROLYTES: Complete Ionization Hydrochloric acid Sodium acetate (salts in general)

WEAK ELECTROLYTES: Partial Ionization Strengths of Electrolytes: Complete, Partial, or No Ionization in Water NON-ELECTROLYTES: No Ionization Acetic acid Glucose Reversible reaction: the reaction can proceed in both directions Forward reaction occurs as fast as back reaction --- equilibrium Refer to Table 4.1, p. 107, Chang 7th ed. For a partial list of electrolytes and nonelectrolytes.

Nonelectrolyte does not conduct electricity? No cations (+) and anions (-) in solution C 6 H 12 O 6 (s) C 6 H 12 O 6 (aq) H2OH2O Strong ElectrolyteWeak ElectrolyteNonelectrolyte HClCH 3 COOH(NH 2 ) 2 CO HNO 3 HFCH 3 OH HClO 4 HNO 2 C 2 H 5 OH NaOHH2OH2OC 12 H 22 O 11 Ionic Compounds

Which solution, NaCl(aq) or CH 3 OH(aq), conducts electricity? a.NaCl(aq) b.CH 3 OH(aq)

Aqueous Reactions © 2015 Pearson Education, Inc. Precipitation Reactions When two solutions containing soluble salts are mixed, sometimes an insoluble salt will be produced. A salt “falls” out of solution, like snow out of the sky. This solid is called a precipitate.

NaCl (s)NaCl (aq) + AgNO 3 (aq) AgCl (s), NaNO 3 (aq) AgCl (s) Precipitation reactions: AgNO 3 (aq) + NaCl (aq)  AgCl (s) + NaNO 3 (aq) Formation of an insoluble product, the precipitate Usually involve ionic compounds Precipitate: insoluble solid that separates from the solution

Precipitation reactions and SOLUBILITY 1. Can we predict if a precipitate will form if we mix 2 solutions or add a compound to a solution? 2. If the answer to No. 1 above is “YES”, how do we do it? The maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature.

Solubility Rules for Common Ionic Compounds (applies to water at 25 o C) Ag +, Hg 2 2+, Pb 2+ Li +, Na +, K +, Rb +, Cs +, NH 4 +, NO 3 -, C 2 H 3 O 2 -, ClO 4 -, also bicarbonate (HCO 3 - ) and chlorate (ClO 3 - ) Ca 2+, Sr 2+, Ba 2+, Hg 2 2+, Pb 2+, also silver (Ag + ) CO 3 2-, PO 4 3-, C 2 O 4 2-, also chromates (CrO 4 2- ) and sulfites (SO 3 2- ) S 2-, OH -

Aqueous Reactions © 2015 Pearson Education, Inc. Solubility of Ionic Compounds Not all ionic compounds dissolve in water. A list of solubility rules is used to decide what combination of ions will dissolve.

Solubility Rules for Common Ionic Compounds in Water (at 25 o C) Soluble Compounds Exceptions Insoluble CompoundsExceptions All compounds containing alkali metal ions (Li +, Na +, K +, Rb +, Cs + ) and ammonium ion (NH 4 + ) Nitrates (NO 3 - ), acetates (C 2 H 3 O 2 - ), bicarbonates (HCO 3 - ), chlorates (ClO 3 - ), perchlorates (ClO 4 - ) Halides (Cl -, Br -, I - )Halides of silver (Ag + ), mercury(I) (Hg 2 2+ ), lead(II) (Pb 2+ ) Sulfates (SO 4 2- )Sulfates of Group 2 metal ions (Be 2+, Mg 2+, Ca 2+, Sr 2+, Ba 2+ ), silver (Ag + ), mercury(I) (Hg 2 2+ ), lead(II) (Pb 2+ ) Sulfides (S 2- )Sulfides of Groups 1 and 2 metal ions (Li +, Na +, K +, Rb +, Cs +, Be 2 +, Mg 2+, Ca 2+, Sr 2+, Ba 2+ ) and NH 4 + Hydroxides (OH - )Hydroxides of Group 1 metal ions (Li +, Na +, K +, Rb +, Cs + ), Ba 2+, and NH 4 + Carbonates (CO 3 2- ), oxalates (C 2 O 4 2- ), phosphates (PO 4 3- ), chromates (CrO 4 2- ), sulfites (SO 3 2- ) Those of Group 1 metal ions (Li +, Na +, K +, Rb +, Cs + ) and NH 4 +

Solubility Rules for Common Ionic Compounds In water at 25 0 C Soluble CompoundsExceptions Compounds containing alkali metal ions and NH 4 + NO 3 -, HCO 3 -, ClO 3 - Cl -, Br -, I - Halides of Ag +, Hg 2 2+, Pb 2+ SO 4 2- Sulfates of Ag +, Ca 2+, Sr 2+, Ba 2+, Hg 2+, Pb 2+ Insoluble CompoundsExceptions CO 3 2-, PO 4 3-, CrO 4 2-, S 2- Compounds containing alkali metal ions and NH 4 + OH - Compounds containing alkali metal ions and Ba 2+

Precipitation Reactions Precipitate – insoluble solid that separates from solution molecular equation ionic equation net ionic equation Pb 2+ + 2NO 3 - + 2Na + + 2I - PbI 2 (s) + 2Na + + 2NO 3 - Na + and NO 3 - are spectator ions PbI 2 Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) precipitate Pb 2+ + 2I - PbI 2 (s)

Aqueous Reactions © 2015 Pearson Education, Inc. Molecular Equation The molecular equation lists the reactants and products without indicating the ionic nature of the compounds. AgNO 3 (aq) + KCl(aq)  AgCl(s) + KNO 3 (aq)

Aqueous Reactions © 2015 Pearson Education, Inc. Complete Ionic Equation In the complete ionic equation all strong electrolytes (strong acids, strong bases, and soluble ionic salts) are dissociated into their ions. This more accurately reflects the species that are found in the reaction mixture. Ag + (aq) + NO 3 − (aq) + K + (aq) + Cl − (aq)  AgCl(s) + K + (aq) + NO 3 − (aq)

Aqueous Reactions © 2015 Pearson Education, Inc. Net Ionic Equation To form the net ionic equation, cross out anything that does not change from the left side of the equation to the right. The ions crossed out are called spectator ions, K+ and NO 3 −, in this example. The remaining ions are the reactants that form the product—an insoluble salt in a precipitation reaction, as in this example. Ag + (aq) + NO 3 − (aq) + K + (aq) + Cl − (aq)  AgCl(s) + K + (aq) + NO 3 − (aq)

Aqueous Reactions © 2015 Pearson Education, Inc. Writing Net Ionic Equations 1.Write a balanced molecular equation. 2.Dissociate all strong electrolytes. 3.Cross out anything that remains unchanged from the left side to the right side of the equation. 4.Write the net ionic equation with the species that remain.

Molecular and Ionic Equations Spectator ion Spectator ions: Ions that are not involved in the overall reaction

Molecular and Ionic Equations Recap:1. Molecular equation Formulas of all compounds involved are written as though all species existed as whole units. Doesn’t describe accurately what’s going on at the microscopic level. 2. (Complete)Ionic equation Shows dissolved species as free ions. Spectator ions--not involved in overall reaction. 3. Net ionic equation Shows only the species that take part in the reaction.

Molecular and Ionic Equations How to write ionic and net ionic equations: 1. Write a balanced molecular equation for the reaction. 2. Rewrite the equation to show the dissociated ions that form in solution. Strong electrolytes ionize into cations and anions when dissolved in solution. 3. Identify and cancel spectator ions on both sides of the equation. Result: the net ionic equation. AgNO 3 (aq) + K 2 CrO 4 (aq)  Ag 2 CrO 4 (s) + KNO 3 (aq) 22 2 Ag + (aq) + 2 NO 3 - (aq) + 2 K + (aq) + CrO 4 2- (aq)  Ag 2 CrO 4 (s) + 2 K + (aq) + 2 NO 3 - (aq) 2 Ag + (aq) + CrO 4 2- (aq)  Ag 2 CrO 4 (s) (net ionic equation)

Molecular and Ionic Equations How to write ionic and net ionic equations: 1. Write a balanced molecular equation for the reaction. 2. Rewrite the equation to show the dissociated ions that form in solution. Strong electrolytes ionize into cations and anions when dissolved in solution. 3. Identify and cancel spectator ions on both sides of the equation. Result: the net ionic equation. Pb(NO 3 ) 2 (aq) + NaI (aq)  PbI 2 (s) + NaNO 3 (aq) 22 Pb 2+ (aq) + 2 NO 3 - (aq) + 2 Na + (aq) + 2 I - (aq)  PbI 2 (s) + 2 Na + (aq) + 2 NO 3 - (aq) Pb 2+ (aq) + 2 I - (aq)  PbI 2 (s) (net ionic equation)

Molecular and Ionic Equations How to write ionic and net ionic equations: 1. Write a balanced molecular equation for the reaction. 2. Rewrite the equation to show the dissociated ions that form in solution. Strong electrolytes ionize into cations and anions when dissolved in solution. 3. Identify and cancel spectator ions on both sides of the equation. Result: the net ionic equation. AgNO 3 (aq) + KCl (aq)  AgCl (s) + KNO 3 (aq) Ag + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq)  AgCl (s) + K + (aq) + NO 3 - (aq) Ag + (aq) + Cl - (aq)  AgCl (s) (net ionic equation)

Writing Net Ionic Equations 1.Write the balanced molecular equation. 2.Write the ionic equation showing the strong electrolytes 3.Determine precipitate from solubility rules 4.Cancel the spectator ions on both sides of the ionic equation

Which ions remain in solution after PbI 2 precipitation is complete? a.K + and I – b.Pb 2+ and I – c.K + and NO 3– d.Pb 2+ and NO 3–

Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases 4.3

General Properties of Acids and Bases Acids Bases Sour taste Cause color changes in plant dyes (change litmus from blue to red) React with certain metals (e.g., Zn, Mg, Fe) to produce hydrogen gas: Mg (s) + 2HCl (aq)  MgCl 2 (aq) + H 2 (g) React with carbonates (CO 3 2- ) & bicarbonates (HCO 3 - ) to produce carbon dioxide (CO 2 ) gas: 2HCl(aq) + CaCO 3 (s)  CaCl 2 (aq) + H 2 O(l) + CO 2 (g) HCl(aq) + NaHCO 3 (s)  NaCl(aq) + H 2 O(l) + CO 2 (g) Aqueous acid solutions conduct electricity Bitter taste Cause color changes in plant dyes (change litmus from red to blue) Feel slippery Aqueous base solutions conduct electricity

Acids and Bases Around the House… Acids Bases Aspirin (acetylsalicylic acid) Orange juice (citric acid) Vinegar (dilute solution of acetic acid) Muriatic acid (concentrated hydrochloric acid) Ammonia (concentrated ammonium hydroxide) Washing soda (sodium carbonate) Baking soda (sodium bicarbonate) Drain cleaner (sodium & potassium hydroxides)

Basic Compounds in Antacids 54

Acids and Metals Acids react with metals such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn to produce hydrogen gas and the salt of the metal Molecular equations: 2K(s) + 2HCl(aq)  2KCl(aq) + H 2 (g) metal acid salt hydrogen gas Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) metal acid salt hydrogen gas 55

Learning Check Write a balanced equation for the reaction of magnesium metal with HCl(aq). Label the metal, acid, and salt. 56

Learning Check Write a balanced equation for the reaction of magnesium metal with HCl(aq). Mg(s) + 2HCl(aq)  MgCl 2 (aq) + H 2 (g) metal acid salt 57

Aqueous Reactions © 2015 Pearson Education, Inc. Gas-Forming Reactions  Some metathesis reactions do not give the product expected.  When a carbonate or bicarbonate reacts with an acid, the products are a salt, carbon dioxide, and water. CaCO 3 (s) + 2 HCl(aq)  CaCl 2 (aq) + CO 2 (g) + H 2 O(l) NaHCO 3 (aq) + HBr(aq)  NaBr(aq) + CO 2 (g) + H 2 O(l)

Acids and Carbonates 59 Acids react with carbonates and hydrogen carbonates to produce carbon dioxide gas, a salt, and water 2HCl(aq) + CaCO 3 (s)  CO 2 (g) + CaCl 2 (aq) + H 2 O(l) acid carbonate carbon salt water dioxide HCl(aq) + NaHCO 3 (s)  CO 2 (g) + NaCl (aq) + H 2 O(l) acid bicarbonate carbon salt water dioxide

Learning Check Write a balanced equation for the following reactions: A. MgCO 3 (s) + HBr(aq)  B. HCl(aq) + NaHCO 3 (aq)  60

Solution Write a balanced equation for the following reactions: A. MgCO 3 (s) + 2HBr(aq)  MgBr 2 (aq) + CO 2 (g) + H 2 O(l) carbonate acid salt carbon water dioxide B. HCl(aq) + NaHCO 3 (aq)  NaCl(aq) + CO 2 (g) + H 2 O(l) acid bicarbonate salt carbon water dioxide 61

Neutralization Reactions In a neutralization reaction, an acid reacts with a base to produce salt and water the acid HCl reacts with NaOH to produce salt and water the salt formed is the anion from the acid and cation of the base HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) acid base salt water 62

Neutralization Reactions In neutralization reactions, if we write the strong acid and strong base as ions, we see that H + reacts with OH − to form water, leaving the ions Na + and Cl  in solution: H + (aq) + Cl  (aq) + Na + (aq) + OH  (aq)  Na + (aq) + Cl  (aq) + H 2 O(l) the overall reaction is H 3 O + from the acid and OH  from the base form water: H + (aq) + OH  (aq)  H 2 O(l) 63

Acid-base reactions: HCl (aq) + NaOH (aq)  H 2 O (aq) + NaCl (aq) 0.2 M HCl (aq) 0.2 M NaOH (aq) H 2 O (l), NaCl (aq)

Acid-base reactions: Neutralization reactions Acid and base essentially cancel each other out (neutralize each other) to produce water and a salt: HCl (aq) + NaOH (aq)  H 2 O (l) + NaCl (aq) (acid) (base) (water) (salt) NH 4 OH (aq) + HCl (aq)  H 2 O (l) + NH 4 Cl (aq) (base) (acid) (water) (salt)

Acid-Base Neutralization Reactions acid + base  salt + water H + donor H + acceptor An ionic compound made up of a cation other than H + and an anion other than OH - or O 2- H2OH2O

Aqueous Reactions © 2015 Pearson Education, Inc. Acid-Base Reactions  In an acid–base reaction, the acid (H 2 O above) donates a proton (H + ) to the base (NH 3 above).  Reactions between an acid and a base are called neutralization reactions.  When the base is a metal hydroxide, water and a salt (an ionic compound) are produced.

Aqueous Reactions © 2015 Pearson Education, Inc. Neutralization Reactions When a strong acid (like HCl) reacts with a strong base (like NaOH), the net ionic equation is circled below: HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l) H + (aq) + Cl − (aq) + Na + (aq) + OH − (aq)  Na + (aq) + Cl − (aq) + H 2 O(l) H + (aq) + OH − (aq)  H 2 O(l)

Guide for Balancing Neutralization Reactions 69

Balancing Neutralization Reactions Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid. Step 1Write the reactants and products. Mg(OH) 2 + HNO 3 Step 2Balance the H + in the acid with the OH  in the base. Mg(OH) 2 + 2HNO 3 Step 3Balance the H 2 O with H + and the OH . Mg(OH) 2 + 2HNO 3  salt + 2H 2 O Step 4Write the salt from the remaining ions. Mg(OH) 2 + 2HNO 3  Mg(NO 3 ) 2 + 2H 2 O 70

Learning Check Select the correct group of coefficients for each of the following neutralization equations. 1. HCl(aq) + Al(OH) 3 (aq)  AlCl 3 (aq) + H 2 O(l) A. 1, 3, 3, 1 B. 3, 1, 1, 1 C. 3, 1, 1, 3 2. Ba(OH) 2 (aq) + H 3 PO 4 (aq)  Ba 3 (PO 4 ) 2 (s) + H 2 O(l) A. 3, 2, 2, 2 B. 3, 2, 1, 6 C. 2, 3, 1, 6 71

Solution 1. HCl(aq) + Al(OH) 3 (aq)  AlCl 3 (aq) + H 2 O(l) Step 1Write the reactants and products. HCl + Al(OH) 3 Step 2 Balance the H + in the acid with the OH  in the base. 3HCl + Al(OH) 3 Step 3Balance the H 2 O with H + and the OH . 3HCl + Al(OH) 3  salt + 3H 2 O Step 4Write the salt from the remaining ions. 3HCl(aq) + Al(OH) 3 (aq)  AlCl 3 (aq) + 3H 2 O(l) The answer is C. 3, 1, 1, 3. 72

Solution 2. Ba(OH) 2 (aq) + H 3 PO 4 (aq)  Ba 3 (PO 4 ) 2 (s) + H 2 O(l) Step 1Write the reactants and products. Ba(OH) 2 + H 3 PO 4 Step 2Balance the H + in the acid with the OH  in the base. 3Ba(OH) 2 + 2H 3 PO 4 Step 3Balance the H 2 O with H + and the OH . 3Ba(OH) 2 + 2H 3 PO 4  salt + 6H 2 O Step 4Write the salt from the remaining ions. 3Ba(OH) 2 (aq) + 2H 3 PO 4 (aq)  Ba 3 (PO 4 ) 2 (s) + 6H 2 O(l) The answer is B. 3, 2, 1, 6. 73

Learning Check 74 Write the neutralization reactions for stomach acid, HCl, and the ingredients in Mylanta. Mylanta: Al(OH) 3 and Mg(OH) 2

Solution 75 Write the neutralization reactions for stomach acid, HCl, and the ingredients in Mylanta. Mylanta: For Al(OH) 3 : Step 1Write the reactants and products. Al(OH) 3 + HCl Step 2Balance the H + in the acid with the OH  in the base. Al(OH) 3 + 3HCl Step 3Balance the H 2 O with H + and the OH . Al(OH) 3 + 3HCl  salt + 3H 2 O Step 4Write the salt from the remaining ions. Al(OH) 3 (aq) + 3HCl(aq)  AlCl 3 (aq) + 3H 2 O(l)

Solution 76 Write the neutralization reactions for stomach acid, HCl, and the ingredients in Mylanta. Mylanta: For Mg(OH) 2 : Step 1Write the reactants and products. Mg(OH) 2 + HCl Step 2Balance the H + in the acid with the OH  in the base. Mg(OH) 2 + 2HCl Step 3Balance the H 2 O with H + and the OH . Mg(OH) 2 + 2HCl  salt + 2H 2 O Step 4Write the salt from the remaining ions. 2HCl(aq) + Mg(OH) 2 (aq)  MgCl 2 (aq) + 2H 2 O(l)

Sample Problem 8.8 Balancing Equations of Acids Write the balanced equation for the neutralization of HCl(aq) and Ba(OH) 2 (s). Solution Step 1Write the reactants and products. HCl(aq) + Ba(OH) 2 (s) → salt + H 2 O(l) Step 2Balance the H + in the acid with the OH – in the base. Placing a coefficient of 2 in front of HCl provides 2H + for the 2OH – in Ba(OH) 2. 2HCl(aq) + Ba(OH) 2 (s) → salt + H 2 O(l) Step 3Balance the H 2 O with the H + and the OH –. Use a coefficient of 2 in front of H 2 O to balance 2H + and 2OH –. 2HCl(aq) + Ba(OH) 2 (s) → salt + 2H 2 O(l) Step 4Write the salt from the remaining ions. Use the ions Ba 2+ and 2Cl – to write the formula of the salt, BaCl 2. 2HCl(aq) + Ba(OH) 2 (s) → BaCl 2 (aq) + 2H 2 O(l) Study Check 8.8 Write the balanced equation for the reaction between H 2 SO 4 (aq) and NaHCO 3 (aq).

Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water 4.3

Acid-Base Definitions: 1. Arrhenius acids & bases: Formulated by the Swedish chemist Svante Arrhenius (1859-1927) in the late 19th century, to classify substances whose properties in aqueous solution were well known. Acids: Substances that ionize in water to produce H + ions. Bases: Substances that ionize in water to produce OH - ions. HCl (aq)  H + (aq) + Cl - (aq) NaOH (aq)  Na + (aq) + OH - (aq)

A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acidbaseacidbase 4.3 A Brønsted acid must contain at least one ionizable proton!

Acid-Base Definitions: 2. Bronsted-Lowry acids & bases: Formulated simultaneously by the Danish chemist Johannes Bronsted (1879-1947), and the British chemist Thomas Lowry (1874-1936), in 1923 to cover substances dissolved in water and other solvents (e.g., alcohols). Acids: Substances that donate H + ions (a.k.a. protons). Bases: Substances that accept H + ions. HC 2 H 3 O 2 (aq) = H + (aq) + C 2 H 3 O 2 - (aq) (acetic acid) (acetate anion) C 2 H 3 O 2 - (aq) + H 2 O (l) = HC 2 H 3 O 2 (aq) + OH - (aq) (acetate anion) (acetic acid)

Aqueous Reactions © 2015 Pearson Education, Inc. Acids The Swedish physicist and chemist S. A. Arrhenius defined acids as substances that increase the concentration of H + when dissolved in water. Both the Danish chemist J. N. Brønsted and the British chemist T. M. Lowry defined them as proton donors.

Aqueous Reactions © 2015 Pearson Education, Inc. Bases Arrhenius defined bases as substances that increase the concentration of OH − when dissolved in water. Brønsted and Lowry defined them as proton acceptors.

Bronsted-Lowry Acids & Bases HC 2 H 3 O 2 (aq) = H + (aq) + C 2 H 3 O 2 - (aq) (acetic acid) (proton) (acetate anion) should be written as… HC 2 H 3 O 2 (aq) + H 2 O(l) = H 3 O + (aq) + C 2 H 3 O 2 - (aq) (acetic acid) (hydronium ion) (acetate anion) Can’t exist alone in solution due to its strong attraction to the negative pole (the O atom) in H 2 O. Consequently, the proton is coordinated to H 2 O molecules (i.e., hydrated) in aqueous solution. Chemists write the hydrogen ion as “H + ” in acid-base reactions as a shorthand, in which the presence of water as solvent is understood.

Bronsted-Lowry Acids & Bases Monoprotic Acids: Those acids which, upon ionization in H 2 O, donate only one proton. Examples: HCl (aq)  H + (aq) + Cl - (aq) HC 2 H 3 O 2 (aq) = H + (aq) + C 2 H 3 O 2 - (aq) HCl is a strong electrolyte--ionizes completely in H 2 O. HC 2 H 3 O 2 is a weak electrolyte--ionizes partially in H 2 O.

Bronsted-Lowry Acids & Bases Polyprotic Acids: Those acids which, upon ionization in H 2 O, donate more than one proton. Example: Sulfuric acid (H 2 SO 4 ) -- a diprotic acid (donates 2 H + upon ionization in H 2 O) H 2 SO 4 (aq)  H + (aq) + HSO 4 - (aq) HSO 4 - (aq) = H + (aq) + SO 4 2- (aq) H 3 PO 4 (aq) = H + (aq) + H 2 PO 4 - (aq) H 2 PO 4 - (aq) = H + (aq) + HPO 4 2- (aq) HPO 4 2- (aq) = H + (aq) + PO 4 3- (aq) Example: Phosphoric acid (H 3 PO 4 ) -- a triprotic acid (donates 3 H + upon ionization in H 2 O) H 2 SO 4 is a strong electrolyte, ionizes completely into H + and HSO 4 -, as represented by the single arrow. HSO 4 - is a weak electrolyte, ionizes partially into H + and SO 4 2-, as represented by the double arrow.

Monoprotic acids HCl H + + Cl - HNO 3 H + + NO 3 - CH 3 COOH H + + CH 3 COO - Strong electrolyte, strong acid Weak electrolyte, weak acid Diprotic acids H 2 SO 4 H + + HSO 4 - HSO 4 - H + + SO 4 2- Strong electrolyte, strong acid Weak electrolyte, weak acid Triprotic acids H 3 PO 4 H + + H 2 PO 4 - H 2 PO 4 - H + + HPO 4 2- HPO 4 2- H + + PO 4 3- Weak electrolyte, weak acid 4.3

STRONG ACIDS Ionize completely…see list HCl  H + + Cl -

Aqueous Reactions © 2015 Pearson Education, Inc. Strong or Weak? Strong acids completely dissociate in water; weak acids only partially dissociate. Strong bases dissociate to metal cations and hydroxide anions in water; weak bases only partially react to produce hydroxide anions.

Weak Acid => an acid that only partially dissociates to H + ions in water. ex) HF (aq) ↔ H + (aq) + F - (aq) note the double arrow indicates that the reaction does not go to completion

Bronsted-Lowry Acids & Bases Monobases: Those bases which, in H 2 O, can accept only one proton. Example: Hydroxide ion (OH - ) H + (aq) + OH - (aq)  H 2 O (l) PO 4 3- (aq) + H + (aq) = HPO 4 2- (aq) HPO 4 2- (aq) + H + (aq) = H 2 PO 4 - (aq) H 2 PO 4 - (aq) + H + (aq) = H 3 PO 4 (aq) Example: Phosphate anion (PO 4 3- ) -- a tribase (can accept 3 H + ) Polybases: Those bases which, in H 2 O, can accept more than one proton. Example: Ammonia (NH 3 ) NH 3 (aq) + H + (aq) = NH 4 + (aq)

In H 2 O, weak Bronsted bases undergo a hydrolysis reaction. This term means that the weak Bronsted base reacts with H 2 O to form an acid and hydroxide ion (OH - ). NH 3 (aq) + H 2 O (l) = NH 4 + (aq) + OH - (aq) PO 4 3- (aq) + H 2 O (l) = HPO 4 2- (aq) + OH - (aq) HPO 4 2- (aq) + H 2 O (l) = H 2 PO 4 - (aq) + OH - (aq) H 2 PO 4 - (aq) + H 2 O (l) = H 3 PO 4 (aq) + OH - (aq) Strong Bronsted bases ionize completely in H 2 O. NaOH (aq)  Na + (aq) + OH - (aq) Ba(OH) 2 (aq)  Ba 2+ (aq) + 2OH - (aq)

Weak Bases: a base that is only partially dissociated to form OH - ions in water. ex) NH 3(aq) + H 2 O (l) ↔ NH 4 + (aq) + OH -

Neutralization Reaction acid + base salt + water All salts are strong electrolytes. HCl (aq) + NaOH (aq) NaCl (aq) + H 2 O H + + Cl - + Na + + OH - Na + + Cl - + H 2 O H + + OH - H 2 O 1) Strong Acid – Strong Base Reaction

2) Weak Acid – Strong Base Two Step Reaction (1) HB (aq) ↔ H + (aq) + B - (aq) (2) H + (aq) + OH - (aq) → HOH (3) Net Ionic: HB (aq) + OH - (aq) → B - (aq) + HOH ex) HF (aq) + OH - (aq) → HOH + F - (aq)

3) Strong Acid – Weak Base Two Step Reaction (1) B + HOH → BH + (aq) + OH - (aq) (2) H + (aq) + OH - (aq) → HOH (3) Net Ionic: H + (aq) + B (aq) → BH + (aq) ex) H + (aq) + NH 3(aq) → NH 4 + (aq)

Oxidation-reduction (redox) reactions: H 2 O 2 (aq) + 2KI (aq) + 2HCl (aq)  I 2 (aq) + 2KCl (aq) I 2 (aq) + 2Na 2 S 2 O 3 (aq)  2NaI (aq) + Na 2 S 4 O 6 (aq) + + + I 2 (aq), KCl (aq), NaI (aq), Na 2 S 4 O 6 (aq) 1 M HCl (aq) 0.1 M H 2 O 2 (aq) Na 2 S 2 O 3 (aq)KI (aq)

Oxidation-Reduction Reactions Cl 2 (g) + 2 Na (s)  2 NaCl (s) 2 Na 0  2 Na + + 2 e - Loss of electrons from Na metal to form Na + cations OXIDATION Cl 2 0 + 2 e -  2 Cl - Gain of electrons from Cl 2 gas to form Cl - anions REDUCTION - 2 e - + 2 e -

Oxidation-Reduction Reactions Cl 2 (g) + 2 Na (s)  2 NaCl (s) 2 Na 0  2 Na + + 2 e - OXIDATION HALF-REACTION Cl 2 0 + 2 e -  2 Cl - REDUCTION HALF-REACTION Cl 2 0 + 2 Na 0  2 Na + + 2 Cl - Add the half-reactions… Electrons cancel each other Na + and Cl - combine to form NaCl OVERALL EQUATION: SUM OF THE HALF-REACTIONS

OILRIGOILRIG xidation s oss (of electrons) ain (of electrons) s eduction

Oxidation-Reduction Reactions Loss of electrons is oxidation. Gain of electrons is reduction. One cannot occur without the other. The reactions are often called redox reactions.

How many electrons does each oxygen atom gain during the course of this reaction? a. None b.One c.Two d.Four

Identifying Redox Reactions Reducing Agent Causes oxidation Gains one or more electrons Undergoes reduction Oxidation number of atom decreases Oxidizing Agent Causes reduction Loses one or more electrons Undergoes oxidation Oxidation number of atom increases

Cl - Cl Na Na + e-e- e-e- + e - -e - REDUCING AGENT Donates e - to Cl 2, causes Cl to be reduced to Cl - OXIDIZING AGENT Accepts e - from Na, causes Na to be oxidized to Na +

0 0 +1 -1 Cl 2 (g) + 2 Na (s)  2 NaCl (s) Oxidation Number a.k.a. oxidation state --- the number of charges an atom would have in a molecule or an ionic compound if electrons were transferred completely. Formation of an ionic compound: oxidation numbers Formation of a molecular compound: 0 0 +5 -2 P 4 (s) + 5 O 2 (g)  P 4 O 10 (s) oxidation numbers

Oxidation Number Seven rules for assigning oxidation numbers: 1.In FREE elements (uncombined e.g. H 2, P 4, S 8, Ag, Au, Fe), each atom has an oxidation number of ZERO. 2. For ions containing only one atom (monatomic ions), the oxidation number is equal to the charge on the ion. Examples familiar to you: Alkali metals (Group 1) – ox. # is +1. Alkaline earth metals (Group 2) – ox. # is +2. Aluminum ion – ox. # is +3 (all its compounds) 3.The oxidation number of oxygen in most compounds (e.g. MgO and H 2 O) is –2. However, in H 2 O 2 (hydrogen peroxide) and O 2 2- (peroxide anion), the ox. # is –1. 4.The oxidation number of hydrogen is +1, EXCEPT when it is bonded to metals in binary compounds (e.g. LiH and CaH 2 ).

Oxidation Number Seven rules for assigning oxidation numbers: 5.Fluorine has an oxidation number of –1 in ALL its compounds. Other halogens (Cl, Br, I) have NEGATIVE oxidation numbers when they occur as HALIDE ions in their compounds. When combined with oxygen (e.g. in ClO 4 -, HClO 4 ), the halogens have POSITIVE oxidation numbers. 6.In a NEUTRAL molecule or ionic compound, the SUM of the oxidation numbers of ALL the atoms must equal ZERO. In a POLYATOMIC ION, the sum of oxidation numbers of all the elements in the ion must equal the NET CHARGE ON THE ION. 7.Oxidation numbers do not have to be integers. Example: ox. # of O in the superoxide ion, O 2 -,is –1/2. IT’S A WISE IDEA TO REFER TO THE OXIDATION TABLE I GAVE YOU AND BECOME FAMILIAR WITH THE OXIDATION NUMBERS OF THE VARIOUS ELEMENTS, ESPECIALLY THE COMMON OXIDATION NUMBERS.

Oxidation Number Additional information: 1.Metallic elements have only positive oxidation numbers. Nonmetallic elements, however, may have either positive or negative oxidation numbers. 2.The HIGHEST oxidation number an element in Groups 1A through 7A can have is its group number. The halogens, for example, can possess an oxidation number of +7. 3.The transition metals (Groups 1B through 8B) can possess several oxidation numbers.

Oxidation Numbers To determine if an oxidation–reduction reaction has occurred, we assign an oxidation number to each element in a neutral compound or charged entity.

Rules to Assign Oxidation Numbers Elements in their elemental form have an oxidation number of zero. The oxidation number of a monatomic ion is the same as its charge.

Rules to Assign Oxidation Numbers Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. – Oxygen has an oxidation number of −2, except in the peroxide ion, in which it has an oxidation number of −1. – Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.

Rules to Assign Oxidation Numbers Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. – Fluorine always has an oxidation number of −1. – The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, most notably in oxyanions.

Rules to Assign Oxidation Numbers The sum of the oxidation numbers in a neutral compound is zero. The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

Oxide vs Peroxide

Oxidation Number Assign oxidation numbers to all the elements in the following compounds and ions: S 4 O 6 2- KMnO 4 Fe (IO 4 ) 2 PtCl 6 2- Fe(CN) 6 4- NF 3

Oxidation Number CuSO 4 (aq) + Zn (s)  Cu (s) + ZnSO 4 (aq) Cu 2+ (aq) + Zn 0 (s)  Cu 0 (s) + Zn 2+ (aq) Consider the reaction of aqueous copper (II) sulfate with zinc metal: The products are Cu metal and zinc sulfate. Which reactant is oxidized? Zn 0  Zn 2+ + 2e - Which reactant is reduced? Cu 2+ + 2e -  Cu 0 Overall Reaction: Oxidation number of Cu 2+ ions is +2 Oxidation number of Cu metal is 0

Ex) Zn + 2 HCl  ZnCl 2 + H 2 Identify what species is oxidized and reduced? Identify oxidizing and reducing agents.

Ex) 2Al + 3Cl 2  2AlCl 3 Identify what species is oxidized and reduced? Identify oxidizing and reducing agents.

Ex) CH 4 + 2O 2  CO 2 + 2H 2 O Identify what species is oxidized and reduced? Identify oxidizing and reducing agents.

Ex) PbO + CO  Pb + CO 2 Identify what species is oxidized and reduced? Identify oxidizing and reducing agents.

Displacement Reactions Cu(s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag(s) In displacement reactions, ions oxidize an element. In this reaction, silver ions oxidize copper metal: The reverse reaction does NOT occur. Why not?

Why does the solution in Fig. 4.14 turn blue? a. Cu 2+ (aq) ions are blue. b.NO 3 – (aq) ions are blue. c.Ag(s) is blue. d.H 2 O is blue.

Activity Series Elements higher on the activity series are more reactive. They are more likely to exist as ions.

Metal/Acid Displacement Reactions The elements above hydrogen will react with acids to produce hydrogen gas. The metal is oxidized to a cation.

How many moles of hydrogen gas would be produced for every mole of magnesium added into the HCl solution? a. None b.One c.Two d.Four

grams of solute = M. V. (molar mass)

Aqueous Reactions © 2015 Pearson Education, Inc. Molarity The quantity of solute in a solution can matter to a chemist. We call the amount dissolved its concentration. Molarity is one way to measure the concentration of a solution: moles of solute volume of solution in liters Molarity (M) =

Aqueous Reactions © 2015 Pearson Education, Inc. Mixing a Solution To create a solution of a known molarity, weigh out a known mass (and, therefore, number of moles) of the solute. Then add solute to a volumetric flask, and add solvent to the line on the neck of the flask.

Dilution Formula M 1. V 1 = M 2. V 2 moles of solute before dilution after dilution

Aqueous Reactions © 2015 Pearson Education, Inc. Dilution One can also dilute a more concentrated solution by –using a pipet to deliver a volume of the solution to a new volumetric flask, and –adding solvent to the line on the neck of the new flask.

Aqueous Reactions © 2015 Pearson Education, Inc. Dilution The molarity of the new solution can be determined from the equation M c  V c = M d  V d, where M c and M d are the molarity of the concentrated and dilute solutions, respectively, and V c and V d are the volumes of the two solutions.

What is the concentration of a solution made by dissolving 0.2500 g of K 2 CrO 4 (potassium chromate) in 500.0 mL of distilled H 2 O? What is the concentration of a new solution made by dilution of 10.00 mL of the above K 2 CrO 4 solution to 1000.0 mL with distilled H 2 O? How many grams of K 2 CrO 4 are contained in the new solution?

How would the volume of standard solution added change if that solution were Ba(OH) 2 (aq) instead of NaOH(aq)? a.Increase by one-half the volume used for titration with NaOH. b.Increase by two the volume used for titration with NaOH. c.Decrease by two the volume used for titration with NaOH. d.Decrease by one-half the volume used for titration with NaOH.

© 2015 Pearson Education, Inc. How would the volume of standard solution added change if that solution were Ba(OH) 2 (aq) instead of NaOH(aq)? a.Increase by one-half the volume used for titration with NaOH. b.Increase by two the volume used for titration with NaOH. c.Decrease by two the volume used for titration with NaOH. d.Decrease by one-half the volume used for titration with NaOH.

Aqueous Reactions © 2015 Pearson Education, Inc. Titration A solution of known concentration, called a standard solution, is used to determine the unknown concentration of another solution. The reaction is complete at the equivalence point.

pH 12.30 0.0 mL HCl added pH 7.00 5.0 mL HCl added pH 3.14 5.2 mL HCl added Titration of 0.200 M NaOH with 0.200 M HCL Start: 5.00 mL of 0.200 M NaOH diluted to 50.0 mL with distilled H 2 O Indicator: Bromothymol blue Y GB pH 6.0 pH 7.0 pH 7.6

See handout

© 2015 Pearson Education, Inc. Which set includes only substances that produce electrolytes in water? a.NaBr, KCl, MgSO 4 b.C 6 H 12 O 6, CH 3 OH, C 6 H 6 c.HCl, NH 3, Cl 2, N 2 d.SiO 2, CaCO 3, H 2 SO 4

© 2015 Pearson Education, Inc. a.an explosion. b.the formation of a gas. c.that the solution boils. d.the formation of a precipitate. Pb(NO 3 ) 2 + 2 KI  PbI 2 + 2 KNO 3 The physical evidence that the above reaction occurs is

© 2015 Pearson Education, Inc. Pb(NO 3 ) 2 + 2 KI  PbI 2 + 2 KNO 3 The physical evidence that the above reaction occurs is a.an explosion. b.the formation of a gas. c.that the solution boils. d.the formation of a precipitate.

© 2015 Pearson Education, Inc. Which pair of compounds will produce a precipitate if solutions of appropriate concentrations are mixed together? a.H 2 SO 4 and NaOH b.HNO 3 and CaCl 2 c.Ba(NO 3 ) 2 and Na 3 PO 4 d.LiCl and SrI 2

© 2015 Pearson Education, Inc. When Zn(s) reacts with HCl(aq) to produce H 2 (g) and ZnCl 2 (aq), the zinc is _____ because it _____ electrons. a.reduced; gains b.reduced; loses c.oxidized; gains d.oxidized; loses

© 2015 Pearson Education, Inc. Al + H +  Al 3+ + H 2 When the oxidation–reduction reaction above is correctly balanced, the coefficients are a.1, 2  1, 1. b.1, 3  1, 2. c.2, 3  2, 3. d.2, 6  2, 3.

Aqueous Reactions © 2015 Pearson Education, Inc. Lecture Presentation Chapter 4 Reactions in Aqueous Solution James F. Kirby Quinnipiac University Hamden,

Similar presentations

Presentation on theme: "Aqueous Reactions © 2015 Pearson Education, Inc. Lecture Presentation Chapter 4 Reactions in Aqueous Solution James F. Kirby Quinnipiac University Hamden,"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Aqueous Reactions © 2015 Pearson Education, Inc. Lecture Presentation Chapter 4 Reactions in Aqueous Solution James F. Kirby Quinnipiac University Hamden,

Similar presentations

Presentation on theme: "Aqueous Reactions © 2015 Pearson Education, Inc. Lecture Presentation Chapter 4 Reactions in Aqueous Solution James F. Kirby Quinnipiac University Hamden,"— Presentation transcript:

Similar presentations

About project

Feedback