Chapter 9 Chemical Change

Electrolytes & Conductivity

Electrolytes & Conductivity

Electrolytes & Conductivity

Electrolytes & Conductivity Conducting solutions contain ions. Strong electrolyte: A substance whose solution is a good conductor. Weak electrolyte: A substance whose solution conducts electricity poorly. Nonelectrolyte: A substance whose solution is a nonconductor.

Solutions of Ionic Compounds Sodium Chloride Dissolves in Water The solution that results consists of water molecules, sodium ions, and chloride ions:

Solutions of Ionic Compounds Copper(II) Chloride Dissolves in Water The solution that results consists of water molecules, copper(II) ions, and chloride ions:

Solutions of Ionic Compounds When an ionic compound dissolves in water, its solution consists of water molecules and ions. KOH(s) K+(aq) + OH–(aq) (NH4)2SO4(s) 2 NH4+(aq) + SO42–(aq)

Strong and Weak Acids Acid A molecule or ion that releases a hydrogen ion in water solution. Solution of HCl in water is called hydrochloric acid. All molecules of HCl are converted to ions   HCl + H2O (l)  H3O+(aq.) + Cl-(aq.) To simplify sometime we write the dissociation of HCl as HCl(aq)  H+(aq) + Cl–(aq) For nitric acid: HNO3(aq)  H+(aq) + NO3–(aq)

Strong and Weak Acids Acids are classified as strong or weak. Strong acid All molecules of a strong acid are converted to ions: Only seven common acids are strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4, HClO3.

HAc(aq) H+(aq) + Ac–(aq) Strong and Weak Acids Weak acid Most acids like acetic acid, benzoic acid are weak electrolytes and are considered weak acids. A weak acid is only slightly ionized in solution. HAc(aq) H+(aq) + Ac–(aq) Because of the low concentration of ions, weak acids are poor conductors of electricity.

Strong and Weak Acids

Identifying the Major Species in a Solution Strong and Weak Acids Identifying the Major Species in a Solution Ions are the major species in the solutions of : All soluble ionic compounds Strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4, HClO3. Neutral molecules are the major species in solutions of everything else, primarily Water, Weak acids, Weak bases

Net Ionic Equations Consider the reaction of solutions of silver nitrate and sodium chloride. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) Silver nitrate solution, AgNO3(aq): Silver ions, Ag+(aq) Nitrate ions, NO3–(aq) Water molecules Sodium chloride solution: Sodium ions, Na+(aq) Chloride ions, Cl–(aq)

How are these represented in a conventional equation? Net Ionic Equations How are these represented in a conventional equation? AgNO3(aq) + NaCl(aq) The problem with this conventional representation is that it disguises the true composition of the solutions.

Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) Net Ionic Equations To avoid this misrepresentation, we can write the formulas of the reactants as they actually occur in solution: Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq)

Net Ionic Equations We can perform a similar analysis with the products of the reaction. Silver chloride precipitate: Solid silver chloride, AgCl(s) Sodium nitrate solution: Sodium ions, Na+(aq) Nitrate ions, NO3–(aq) Water molecules

Net Ionic Equations Conventional representation of products: AgCl(s) + NaNO3(aq) Products as they actually occur: AgCl(s) + Na+(aq) + NO3–(aq)

Net Ionic Equations Combining reactants and products: AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq) This is the conventional equation. It is good for stoichiometry problems (Ch. 10), but it lacks in revealing what really happens in solution.

Net Ionic Equations Writing actual reactants and products: Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) AgCl(s) + Na+(aq) + NO3–(aq) This is the total ionic equation. It represents each species as it actually occurs in solution.

Net Ionic Equations Note that some species appear as both reactant and product in the total ionic equation: Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) AgCl(s) + Na+(aq) + NO3–(aq) NO3–(aq) is both reactant and product; Na+(aq) is both reactant and product. Neither ion undergoes a chemical change.

Net Ionic Equations If we eliminate these non-reacting spectator ions, Ag+(aq) + NO3–(aq) + Na+(aq) + Cl–(aq) AgCl(s) + Na+(aq) + NO3–(aq) What remains is the net ionic equation: Ag+(aq) + Cl–(aq) AgCl(s)

Net Ionic Equations A net ionic equation describes what happens in a chemical change: Ag+(aq) + Cl–(aq) AgCl(s) Aqueous silver ion reacts with aqueous chloride ion to yield solid silver chloride. A net ionic equation ignores non-reacting species: Sodium ion Nitrate ion Water molecules

Writing a Net Ionic Equation Net Ionic Equations Writing a Net Ionic Equation Write the conventional equation, including state symbols—(g), (l), (s), and (aq). Balance the equation. Write the total ionic equation by replacing each aqueous (aq) substance that is a strong acid or a soluble ionic compound with its major species. Do not separate a weak acid into ions, even though its state is aqueous (aq). Also, never change solids (s), liquids (l), or gases (g) into ions. Be sure the equation is balanced in both atoms and charge. Write the total ionic equation by removing the spectators from the total ionic equation. Reduce coefficients to lowest terms, if necessary. Be sure the equation is balanced in both atoms and charge.

Single-Replacement Redox Single-Replacement Reactions Reactants: Element (A) plus a solution of either an acid or an ionic compound (BX) Reaction type: Single-replacement Equation type: A + BX AX + B Products: An ionic compound (usually in solution) (AX) plus an element (B)

Single-Replacement Redox Example: Let us consider the reaction between iron and hydrochloric acid. The full formula equation is:   Fe(s) + 2 HCl(aq.)  FeCl2(aq.) + H2(g)

Single-Replacement Redox The iron atoms, which lost electrons, are said to have been oxidized   Fe(s) – 2e  Fe+2(aq.) and hydrogen ions, the receivers of electrons , are said to have been reduced. 2 H+(aq.) + 2e  H2(g)

Single-Replacement Redox Let us consider the reaction between zinc and copper sulfate. The conventional equation is: Zn(s) + CuSO4 (aq.)  ZnSO4(aq.) + Cu(s)   The net ionic equation is Zn(s) + Cu+2(aq.)  Zn+2 (aq.) + Cu(s) Oxidation: Zn(s) – 2e  Zn+2 (aq.) Reduction: Cu+2(aq.) + 2 e  Cu(s)

Single-Replacement Redox Let us consider the reaction between copper and silver nitrate. The conventional equation is: Cu(s) + 2 AgNO3(aq.)  Cu(NO3)2(aq.) + 2 Ag(s)   The net ionic equation is  Cu(s) + 2 Ag+(aq.)  Cu+2 (aq.) + 2Ag(s) Oxidation: Cu(s) - 2e  Cu+2 (aq.) Reduction: 2 Ag+(aq.) + 2 e  2Ag(s)

Single-Replacement Redox The activity series lists the relative reactivity of species that form positive ions in a single-replacement redox reaction. Note that Cu is above Ag and under Zn on the activity series. This means that Cu replaces Ag in a single-replacement reaction. If solid copper was placed in a zinc chloride solution, there would be no reaction.

Double Replacement Precipitation Reactions These are double-replacement reactions: AX + BY  AY + BX When one product is an insoluble ionic compound, the reaction is called a precipitation reaction and the solid is called a precipitate.

Double Replacement Precipitation Reactions The reaction occurs when the cation from one reactant combines with the anion from the another to form a precipitate. Example 1: reaction of sodium chloride and silver nitrate NaCl(aq.) + AgNO3(aq.)  NaNO3 (aq.) + AgCl(s)   Ag+ (aq.) + Cl- (aq.)  AgCl(s) .

Double Replacement Precipitation Reactions Example 2: reaction of sodium sulfate and barium chloride Na2SO4 (aq) + BaCl2(aq)  BaSO4 (s) + 2 NaCl(aq) Net ionic reaction   Ba2+ (aq.) + SO4 2-(aq.)  BaSO4 (s) .

Double Replacement Precipitation Reactions Solubility rules: The following compounds are insoluble:   Chlorides, bromides, iodides of Ag+, Hg2+2, Pb+2 Sulfates of Ba+2, Sr+2, Ca+2, Pb+2 , Ag+, Hg2+2, Hg+2 Most carbonates, phosphates, hydroxides, sulfides except salt of alkali metals and NH4+

Double Replacement Precipitation Reactions

Double Replacement Precipitation Reactions

Double Replacement Molecule-Formation Reactions A double-replacement reaction AX + BY  AY + BX can lead to the formation of a molecular compound such as water, a weak acid or a weak base.

Double Replacement Molecule-Formation Reactions: Formation of water Example 1: Reaction of a strong acid and a base to form water HCl(aq.) + NaOH(aq.) H2O(l) + NaCl(aq.)   H+(aq.) + OH-(aq.)  H2O(l) Example 2 : Reaction of a weak acid and a base to form water HCH3 COO(aq.) + NaOH(aq.)  NaCH3COO(aq.) + H2O(l) HCH3COO(aq.) + OH-(aq.)  CH3 COO-(aq.) + H2O(l)

Double Replacement Molecule-Formation Reactions: Formation of a weak acid Example 3 : Reaction of hydrochloric acid with sodium benzoate HCl(aq) + NaC7H5O2(aq) HC7H5O2(aq) + NaCl(aq) Total ionic equation: H+(aq) + Cl–(aq) + Na+(aq) + C7H5O2–(aq) HC7H5O2(aq) + Na+(aq) + Cl–(aq) Net ionic equation: H+(aq) + C7H5O2–(aq) HC7H5O2(aq) The H+(aq) combines with the C7H5O2–(aq) to form the molecular product benzoic acid HC7H5O2(aq).

Double Replacement Molecule-Formation Reactions Example 4 : Reaction to form a weak acid: acetic acid HCl(aq.) + NaCH3 COO(aq.) HCH3COO(aq.) + NaCl(aq.)   H+(aq.) + CH3 COO-(aq.)  HCH3COO(aq.) Example 5: Reaction to form a weak base: ammonia NH3 NH4 Cl(aq.) + NaOH(aq.) NH3(aq.) + H2O(l) + NaCl(aq.) NH4+(aq.) + OH-(aq.)  NH3(aq.) + H2O(l)

Double Replacement Reactions That Form Unstable Products. A double-replacement reaction AX + BY  AY + BX can lead to the formation of a molecular compound that will further decompose into a gas H2CO3(aq)  H2O(l) + CO2(g) H2SO3(aq)  H2O(l) + SO2(g) 2 HNO2(aq)  H2O(l) + N2O3(g

Double Replacement Reactions That Form Unstable Products. Example: The reaction that occurs when hydrochloric acid and sodium carbonate solutions are combined. Conventional double-replacement equation: 2 HCl(aq) + Na2CO3(aq)  H2CO3(aq) + 2 NaCl(aq) Recognize H2CO3(aq) as an unstable product, replace it with the actual products: 2 HCl(aq) + Na2CO3(aq)  H2O(l) + CO2(g) + 2 NaCl(aq) Net ionic equation 2 H+(aq) + CO32–(aq)  H2O(l) + CO2(g)

Double Replacement Reactions That Form Unstable Products. Write the net ionic equation for the reaction that occurs when hydrochloric acid and sodium sulfite solutions are combined. Conventional double-replacement equation: 2 HCl(aq) + Na2SO3(aq) H2SO3(aq) + 2 NaCl(aq) Recognize H2SO3(aq) as an unstable product, replace it with the actual products: 2 HCl(aq) + Na2SO3(aq) H2O(l) + SO2(aq) + 2 NaCl(aq)

Double Replacement Reactions That Form Unstable Products. 2 HCl(aq) + Na2SO3(aq) H2O(l) + SO2(aq) + 2 NaCl(aq) Total ionic equation: 2 H+(aq) + 2 Cl–(aq) + 2 Na+(aq) + SO32–(aq) H2O(l) + SO2(aq) + 2 Na+(aq) + 2 Cl–(aq) Net ionic equation: Eliminate the spectator ions: 2 H+(aq) + SO32–(aq) H2O(l) + SO2(aq) Hydrogen ion combines with sulfite ion to form liquid water and aqueous sulfur dioxide.

Double Replacement Reactions That Form Unstable Products. Three ion combinations yield molecular products that are not the products you would expect. You must be alert and catch them when they appear: 2 H+(aq) + CO32–(aq) H2CO3(aq) H2O(l) + CO2(g) 2 H+(aq) + SO32–(aq) H2SO3(aq) H2O(l) + SO2(aq) NH4+(aq) + OH–(aq) “NH4OH” H2O(l) + NH3(aq)

Double Replacement Reactions with Undissolved Reactants Write the net ionic equation for the reaction that occurs when solid sodium hydroxide is added to a hydrochloric acid solution. Conventional equation: HCl(aq) + NaOH(s)  H2O(l) + NaCl(aq) Total ionic equation H+(aq) + Cl–(aq) + NaOH(s)  H2O(l) + Na+(aq) + Cl–(aq) Net ionic equation H+(aq) + NaOH(s)  H2O(l) + Na+(aq)

Double Replacement Reactions with Undissolved Reactants Sometimes, in a double-replacement reaction, a reactant is a solid. Hydrochloric acid can dissolve insoluble aluminum hydroxide   3 HCl (aq.) + Al(OH)3 (s)  3 H2O (l) + AlCl3 (aq.) When writing ionic equation do not replace solid by its ions 3 H+(aq.) + Al(OH)3 (s)  3 H2O (l) + Al3+ (aq.)

Double Replacement Reaction that will not happen Write the net ionic equation for the reaction that would occur when sodium chloride and potassium nitrate solutions are mixed. Conventional equation: NaCl(aq) + KNO3(aq)  NaNO3 + KCl Reactants and products are strong electrolytes and soluble: NaCl(aq) + KNO3(aq)  NaNO3(aq) + KCl(aq) Total ionic equation: Na+(aq) + Cl–(aq) + K+(aq) + NO3–(aq)  Na+(aq) + NO3–(aq) + K+(aq) + Cl–(aq) All ions are spectators. There is no reaction

Double Replacement Reaction that will not happen Example: Write the net ionic equation for the reaction that occurs when iron(II) sulfate and magnesium nitrate solutions are mixed. Solution: Conventional equation: FeSO4(aq) + Mg(NO3)2(aq) Fe(NO3)2( ) + MgSO4( ) From the solubility table and guidelines, both products are soluble: FeSO4(aq) + Mg(NO3)2(aq) Fe(NO3)2(aq) + MgSO4(aq)

Double Replacement Reaction that will not happen FeSO4(aq) + Mg(NO3)2(aq) Fe(NO3)2(aq) + MgSO4(aq) Total ionic equation: Fe2+(aq) + SO42–(aq) + Mg2+(aq) + 2 NO3–(aq) Fe2+(aq) + 2 NO3–(aq) + Mg2+(aq) + SO42–(aq) Net ionic equation: Eliminate the spectator ions: All ions are spectators! There is no reaction when these two solutions are combined. The conventional equation disguises this fact, but the net ionic equation helps to make it more apparent.

Summary