Electrolytes

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Electrolytes and Non-electrolytes

• In solid NaCl the ions in the lattice are held in place by strong ionic bonds.
• This means the ions cannot move about in an electric field.
• Therefore, solid NaCl does not conduct electricity.
• When NaCl is added to water, the salt dissociates into Na⁺(aq) and Cl^-(aq).
• In the presence of an electric field, the solution can conduct electricity.
• Strong electrolytes exist in solution entirely (or almost entirely) as ions.
• Non-electrolytes (e.g. sugar) do not conduct electricity at all. (ie. there are no ions in solution.)

Strong and Weak Electrolytes

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Strong and Weak Electrolytes

• Water has a low conductivity so the bulb does not burn in pure water.
• Hydrogen chloride is soluble in water.
• In water, HCl ionizes into H⁺ and Cl^-.
• Since HCl is a strong electrolyte, in water there are no HCl molecules, only H⁺ and Cl^- ions.
• When the HCl(g) is bubbled through the solution, the bulb glows brightly because of the presence of the ions.
• When acetic acid replaces the HCl solution the bulb glows less brightly.
• Acetic acid exists as a mixture of acetic acid molecules and H⁺ and acetate ions in solution.
• Because acetic acid is a mixture of ions and parent molecules in solution, acetic acid is a weak electrolyte.
• Therefore, the acetic acid solution does not cause the bulb to glow brightly.

Dissolution of an Ionic Compound

Fig. 4.3 pg. 124

Properties of Solutes in Aqueous Solution

Ionic Compounds in Water

• When an ionic compound dissolves in water, the ions dissociate.
• This means that in solution, the solid no longer exists as a well ordered arrangement of ions in contact with each other.
• Instead, each ion is surrounded by water molecules.
• The positive ions have the oxygen atoms of water pointing towards the ion, negative ions have the hydrogen atoms of water pointing towards the ion.
• The transport of ions through the solution causes electric current to flow through the solution.
• When a molecular compound dissolves in water (e.g., CH₃OH), there are no ions formed. Therefore, there is nothing in the solution to transport electric charge and the solution does not conduct electricity.

Dissolution of NaCl

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Dissolution of NaCl

• When an ionic solid is placed in water, the ions dissociate.
• If the attractions between the ions and water molecules overcome the ionic attractions in the lattice, the salt is soluble.
• Cations are attracted to the O atoms in water (through lone pairs).
• Anions are attracted to the H atoms in water.
• In solution the ions are separated.

Potassium Iodide and Lead Nitrate

Fig. 4.4 pg. 126

Metathesis Reactions

• Metathesis reactions involve swapping ions in solution:

AX + BY AY + BX.

• Ion swapping will lead to a chemical reaction in solution if one of three things occurs:
  1. an insoluble solid is formed (precipitate),
  2. weak or nonelectrolytes are formed, or
  3. an insoluble gas is formed.

Precipitation Reactions

• A solute is soluble in water if more than 0.01 mol of the substance will dissolve in enough water to make 1 liter of solution.
• A precipitate is an insoluble solid that forms when two solutions are mixed.
  • Consider 2KI(aq) + Pb(NO₃)₂(aq) PbI₂(s) + 2KNO₃(aq)
  • Both KI(aq) + Pb(NO₃)₂(aq) are colorless solutions. When mixed, they form a bright yellow precipitate of PbI₂ and a solution of KNO₃.
  • The final product of the reaction contains solid PbI₂, aqueous K⁺ and aqueous NO₃^- ions.
• The molecular equation lists all the species as molecules:

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

• However, we know that certain substances exist as ions in solution.
• The full ionic equation lists all ions:

2K⁺(aq) +2I^-(aq) + Pb²⁺(aq) + 2NO₃^-(aq) PbI₂(s) + 2K⁺(aq) + 2NO₃^-(aq)

• The net ionic equation cancels those ions that are unchanged:

2I^-(aq) + Pb²⁺(aq) PbI₂(s)

Solubility of Ionic Compounds

Table 4.1 pg. 127

Metathesis Reactions

• In order to determine whether or not a substance will dissolve or form a precipitate, certain rules apply.
• These rules often have important exceptions.

Net Ionic Equations

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Net Ionic Equations

• Write down balanced chemical reaction described in problem.
• Rewrite this reaction in ionic form (ionic equation).
• Cancel out equal quantities of all ions which are identical on both sides of reaction arrow.

Convert the reactions below into net ionic equations:

Reaction between Na₂CO₃(aq) and MgSO₄(aq)

Reaction between Pb(NO₃)₂ (aq) and Na₂S(aq)

Reaction between (NH₄)₃PO₄(aq) and CaCl₂(aq)

Space Filling Models of Acids

Top of Fig. 4.6 pg. 131

Properties of Solutes in Aqueous Solution

Acids

Definitions:

• Dissociation = pre-formed ions in a solid move apart in solution.
• Ionization = neutral substance forms ions in solution.
• Acid = substances which ionizes to form H⁺ in solution.
• Common acids are HCl, HNO₃, CH₃CO₂H (acetic acid or vinegar), lemon, lime, vitamin C.
• Shown here are HCl (top), nitric acid (middle), and acetic acid (bottom). O atoms are shown in red, H in white, Cl in green, N in blue and C in black.
• Bases = substances which react with the H⁺ ions formed by acids.
• Common bases are NH₃ (ammonia), drano, milk of magnesia.

Aqueous Acids

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Introduction to Aqueous Acids

• Acids increase the concentration of [H⁺] when dissolved in water.
• When HCl is added to water it completely dissociates into H⁺ and Cl^-.
• HCl is a strong electrolyte.
• There is no undissociated HCl at the end of the reaction, so HCl is a strong acid.
• HNO₃ is also a strong acid. Therefore, in water it forms H⁺ and NO₃^- with no undissociated HNO₃ left.
• Acetic acid is a weak acid. Therefore, it exists as a mixture of undissociated acetic acid, H⁺, and acetate ions in solution.

Reaction Between Ammonia and Water

Bottom of Fig. 4.6 pg. 131

Properties of Solutes in Aqueous Solution

Strong and Weak Acids and Bases

• Strong acids and bases are strong electrolytes.
• Therefore, they are completely ionized in solution.
• We write this as HCl H⁺ + Cl^-.
• Weak acids and bases are weak electrolytes.
• Therefore, they are partially ionized in solution.
• Since H⁺ is a naked proton, we refer to acids as proton donors and bases as proton acceptors.
• In this figure, we see the proton transfer between NH₃ (a weak base) and water (a weak acid). Since there is a mixture of NH₃, H₂O, NH₄⁺, and OH^- in solution, we write:

Aqueous Bases

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Introduction to Aqueous Bases

• Bases increase the [OH^-] concentration in aqueous solution.
• The most common bases are NaOH, KOH, and Ca(OH)₂.
• When solid NaOH is added to water it completely dissociates into Na⁺ and OH^- in solution.
• Since there is no undissociated NaOH at the end of the reaction, NaOH is a strong base.
• Strong bases are strong electrolytes.
• NH₃ is an example of a base which does not contain OH^-.
• However, when NH₃(g) dissolves in water it forms NH₄⁺ and OH^-.
• Ammonia is a weak base, so there is an equilibrium mixture of NH₃, NH₄⁺, and OH^- in solution.
• Most of the ammonia exists as NH₃ in solution, it is a weak electrolyte.

Strong Acids and Bases

Table 4.2 pg. 132

Strong Acids and Bases

Table 4.2 Common Strong Acids and Bases

Strong Acids	Strong Bases
Hydrochloric, HCl	Group 1A metal hydroxides (LiOH,
Hydrobromic, HBr	NaOH, KOH, RbOH, CsOH)
Hydroiodic, HI	.
Chloric, HClO3	Heavy group 2A metal hydroxides
Perchloric, HClO4	(Ca(OH)2, Sr(OH)2, Ba(OH)2)
Nitric, HNO3	.
Sulfuric, H2SO4	.

Weak and Strong Electrolytes

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Properties of Solutes in Aqueous Solution

Compounds can be classified as strong electrolytes, weak electrolytes, and nonelectrolytes by looking at their solubilities and classifications.

• Water-soluble and ionic = strong electrolyte (NaCl: ionic salt, NH₄Cl: ionic weak acid, NaC₂H₃O₂: ionic weak base).
• Water-soluble strong acid or strong base = strong electrolyte (HCl: ionic strong acid, NaOH: ionic strong base).
• Water-soluble but not ionic weak acid or base = weak electrolyte (NH₃: molecular weak base, HC₂H₃O₂: molecular weak acid).
• Not water-soluble but ionic = weak electrolyte (CaSO₄: "insoluble" ionic).
• Water-soluble but not ionic = nonelectrolyte (CH₃OH: soluble molecular).
• Not water-soluble and not ionic: = nonelectrolyte (C₆H₆: insoluble molecular).

Natural Indicators

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Natural Indicators

• An indicator is a substance that changes color as a function of pH.
• Most indicators are dyes.
• Some indicators can be extracted from natural products.
• Example: the dye which is characteristic of the red color in red cabbage leaves is an indicator.
  • The colorless dye is extracted with methanol.
  • The dye turns yellow/green in base and red in acid.
• Tea is another natural indicator. It changes from brown to yellow/orange in acid. In this experiment, lemon juice is used as the acid.

Dissolution of Mg(OH)₂ by Acid

Fig. 4.8 pg. 135.

Dissolution of Mg(OH)₂ with HCl

• Phillip's Milk of Magnesia bottle in Fig. 4.8 a contains milky mixture of Mg(OH)₂ and water.
• Concentrated HCl solution is added from clear bottle in Fig. 4.8 b.
• All Mg(OH)₂ has dissolved in Fig. 4.8 c.
• Mg(OH)₂(s) + 2 HCl(aq) MgCl₂ (aq) + 2 HOH(l)
• Net ionic equation: Mg(OH)₂(s) + 2 H⁺(aq) Mg²⁺(aq) + 2 HOH(l)

Magnesium Hydroxide

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Dissolution of Mg(OH)₂ in Acid

• Milk of magnesia is a suspension of Mg(OH)₂ in water.
• Mg(OH)₂ is relatively insoluble in neutral water.
• In acidic solutions the Mg(OH)₂ dissolves.
• The solid Mg(OH)₂ consists of layers of Mg²⁺ ions sandwiched between OH^- ions.
• In water there is an equilibrium between the solid Mg(OH)₂ and the ions.
• In acidic solution, the hydronium ions (H₃O⁺) react with the OH^- from the Mg(OH)₂ to form water.
• In the process Mg²⁺ ions are free to move about the solution.
• If HCl is used as the acid, the overall chemical equation is:

Mg(OH)₂(s) + 2HCl(aq) MgCl₂(aq) + 2H₂O(l)

• The net ionic equation is:

Mg(OH)₂(s) + 2H₃O⁺(aq) Mg²⁺(aq) + 4H₂O(l)

Electron Transfer Reactions

Right margin pg. 139

Introduction to Oxidation-Reduction Reactions

Oxidation and Reduction

• In all reduction-oxidation (redox) reactions, one species is reduced at the same time another is oxidized.
• The species that causes oxidation is called the oxidizing agent.
• The species that causes reduction is called the reducing agent.
• The oxidizing agent is always reduced and the reducing agent oxidized.
• The substance that is oxidized loses electrons to the substance that is reduced.

Redox I

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Oxidation-Reduction Reactions I

• Consider zinc metal in the presence of oxygen at high temperature.
• The zinc reacts to form ZnO.
• Electrons are transferred from zinc to oxygen.
• So, the zinc becomes Zn²⁺ while the oxygen becomes O^2-.
• Zinc has been oxidized (i.e. its oxidation state has increased) by oxygen (the species causing the oxidation).
• Oxygen has been reduced (i.e. its oxidation state has decreased) by zinc (the reducing agent).
• In general, the loss of electrons is oxidation: Zn Zn²⁺ + 2e^-.
• The gain of electrons is reduction: O + 2e^- O^2-.

Redox II

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Oxidation-Reduction Reactions II

• When Zn is added to dilute acid, H₂(g) and Zn²⁺ ions are formed:

Zn(s) + 2H₃O⁺(aq) Zn²⁺(aq) + H₂(g) + 2H₂O(l)

• H₃O⁺ attacks the zinc atoms and an electron is transferred from the hydronium ion (H₃O⁺) to the zinc metal.
• After two such transfers, zinc is oxidized to Zn²⁺ while H⁺ is reduced to hydrogen gas.
• When zinc is placed in aqueous copper sulfate, a similar reaction occurs: Zn is oxidized by Cu²⁺ to Zn²⁺ while Cu²⁺ is reduced to Cu⁰ by Zn.
• When zinc is oxidized it causes reduction.
• Therefore, zinc is the reducing agent.
• The substances that are reduced, cause oxidation and are the oxidizing agents.

Oxidation of Mg by Acid

Fig. 4.13 pg. 141

Introduction to Oxidation-Reduction Reactions

Oxidation of Metals be Acids and Salts

• It is common for metal to produce hydrogen gas when they react with acids. Shown here is the reaction between Mg and HCl:

Mg(s) + 2HCl(aq) MgCl₂(aq) + H₂(g).

• In the process the metal is oxidized and the H⁺ is reduced.
• It is possible for metals to be oxidized in the presence of a salt:

Fe(s) + Ni(NO₃)₂(aq) Fe(NO₃)₂(aq) + Ni(s).

• The net ionic equation shows the redox chemistry well:

Fe(s) + Ni²⁺(aq) Fe²⁺(aq) + Ni(s).

• In this reaction iron has been oxidized to Fe²⁺ while the Ni²⁺ has been reduced to Ni.

Formation of Silver Crystals

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Formation of Silver Crystals on Copper Wire

• See Fig. 4.15 pg. 125
• Copper wire is placed in a beaker (or test tube).
• A solution of silver nitrate is added to the beaker (test tube).
• Copper reduces Ag⁺ to Ag.
• The Cu is oxidized to Cu²⁺.
• The full molecular equation:

2AgNO₃(aq) + Cu(s) Cu(NO₃)₂(aq) + 2Ag(s)

• The net ionic equation is:

2Ag⁺(aq) + Cu(s) Cu²⁺(aq) + 2Ag(s)

• Note that the solution changes from colorless to blue indicating the presence of copper(II) nitrate.
• Silver crystals form on the Cu wire.

Solution Formation From Solid

Fig. 4.16 pg. 146

Solution Composition

Molarity

• A solution is made when one substance (the solute) is dissolved in another (the solvent).
• The solute is the substance that is present in smallest amount.
• Solutions in which water is the solvent are called aqueous solutions.
• Solutions can be prepared with different concentrations by adding different amounts of solute to solvent.
• The amount (moles) of solution per liter of solution is the molarity of the solution.
• By knowing the molarity of a quantity in liters of solution, we can easily calculate the number of moles (and, by using molar mass, the mass) of solute.
• In this figure, blue copper(II) sulfate pentahydrate, CuSO₄·5H₂O, is weighed (62.4 g, 0.250 mol) and placed in a 250 mL volumetric flask. A little water is added and the flask swirled to ensure the copper sulfate dissolves. When all the copper sulfate has dissolved, the flask is filled to the mark with water. The molarity of the solution is 0.250 mol / 0.250 L = 1.00 M.

Solution Formation

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Solution Formation from a Solid

• To form a solution of known concentration (standard solution) a certain amount of solid must first be accurately weighed.
• To prepare 250 mL of 1.00 M CuSO₄ solution, 0.250 moles of copper sulfate are required.
• Copper sulfate occurs as the pentahydrate, CuSO₄·5H₂O (FW = 249.7 g/mol). Therefore, 62.40 g of CuSO₄·5H₂O are required.
• The copper sulfate is added to a 250 mL volumetric flask.
• Some water is added and the flask is swirled to dissolve the salt.
• Once all the salt is dissolved, enough water is added to bring the solution to the mark on the volumetric flask.

Dilution

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Solution Formation By Dilution

• See Fig. 4.19 pg.131
• Suppose we want to prepare 250 mL of a 0.100 M solution of copper sulfate starting with a 1.00 M stock solution.
• We carefully pipet 25.00 mL of the stock solution and add it to a 250 mL volumetric flask.
• Some water is added to the volumetric flask and the flask is swirled to ensure good mixing.
• Then enough water is added to make the total volume of the solution 250 mL.

Volumetric Calculations

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Volumetric Calculations

To work problems involving molarity, volume, and grams first calculate number of moles of solute you are working with.

• (vol of solution) x (concentration) = (moles of solute)
• (grams of solute) _ (MM of solute) = (moles of solute)
• (vol of solute) x (density of solute) _ (MM of solute) = (moles of solute)

Next use dimensional analysis to calculate the final quantity you need.

• How many g of K₂Cr₂O₇ in 50.0 mL of 0.850 M solution?
• What is molarity of 2.50 g of (NH₄)₂SO₄ in a 250 mL solution?
• How many mL of 0.387 M CuSO₄ contains 1.00 g of solute?
• How many g of AgNO₃ needed for 100.0 mL of 0.200 M solution?
• How much 6.0 M HNO₃ needed to create 250 mL of 1.0 M HNO₃ ?
• Calculate molarity of 250 mL glycerol (C₃H₈O₃) solution made using 50.000 mL of pure glycerol. Density = 1.2656 g/mL.

Titration

Fig. 4.21 pg. 153

Solution Stoichiometry and Chemical Analysis

Titrations

• A titration is an experiment in which the molarity of a substance is measured by knowing the molarity of another substance.
• Example: Suppose we know the molarity of an NaOH solution and we want to find the molarity of an HCl solution.
  • First let us examine what 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 (20.0 mL, say) and measure the number of mL of NaOH solution required to react completely with the HCl solution.
  • What do we get? Volume of NaOH. Since we already have the 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.

Titration

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Acid-Base Titration

• A titration is an experiment in which the concentration of an unknown acid or base is deduced from the concentration of a standard solution of base or acid.
• Consider the addition of a standard NaOH solution added to a solution of HCl.
• A pH meter is used to measure the acidity of the solution.
• When 0.100 M NaOH solution is added to a 0.100 M solution of HCl, the pH increases from 1.0.
• As more base is added, the pH gradually increases.
• Near the equivalence point (the point defined by stoichiometry as the point at which enough base has been added to neutralize the acid), the pH begins to change dramatically with small additions of base.
• The last few drops of base change the pH from around 3 to 7.
• At the equivalence point the pH is 7.0 because the solution is neutral.
• Addition of one drop of NaOH after equivalence causes the pH to increase to about 10.0.
• Note that the phenolphthalein indicator changes color from colorless to red after the equivalence point.
• The end point of the titration is the point at which we observe a color change.
• The end point is experimentally determined, the equivalence point is determined by stoichiometry.