4.1 Aqueous Solutions. Chapter 4. Reactions in Aqueous Solution. Electrolytes. Strong Electrolytes. Weak Electrolytes

Chapter 4

Reactions in Aqueous Solution

4.1 Aqueous Solutions

• Solution – homogeneous mixture of 2 or more substances

• Solute – the substance present in a smaller amount (usually solid in Chap. 4)

• Solvent – the substance present in the larger amount (usually liquid in Chap. 4)

• Aqueous solution – solvent is water

Electrolytes

• Electrolyte – substance that gives an aqueous solution that conducts electricity

• Nonelectrolyte – substance that does not produce conducting solution when dissolved in water

• Mobile ions conduct electricity

– electrolytes break apart (dissociate or ionize) into ions when dissolved in water

Electrical Conduction by Electrolyte Solution

M⁺ X^-

Cations (positive ions) migrate to cathode (negative electrode)

Anions (n egative ions) migrate to anode (positive electrode)

e^- e^-

Electrochemistry at electrodes (Chap. 19)

Strong Electrolytes

• Dissociate completely into ions in H₂O

• Salts like NaCl, KNO₃, MgSO₄

KNO3(s) K⁺(aq) + NO3- H2O (aq)

HCl(g) HH₂O ⁺(aq) + Cl^-(aq )

•Strong acids like HCl, HNO₃

Weak Electrolytes

• Ionize partially into ions in H₂O

• Weak acids (HF)

HF(aq) H⁺(aq) + F^-(aq) H2O

NH3(aq) NH4+

(aq) + OH^-(aq) H2O

•Weak bases (NH3)

•Double arrow indicates that the reaction is reversible, i.e., runs both ways

• Ionic compounds are strong electrolytes

• When KCl dissolves:

KCl(s) d K⁺(aq) + Cl^-(aq)

• Ions are surrounded by polar molecules – in general, solvation

– in water, hydration

+ end of dipole points to anion - end of dipole

points to cation

NH₃(aq) NH₄⁺(aq) + OH^-(aq) H₂O

• Reversible reaction - can occur in both directions

• Reactants form products as soon as reaction begins

• Once products are formed, they in turn react to re-form reactants

• Chemical Equilibrium

- when reactants form products as fast as products form reactants, no further net change in concentrations

Reversible Ionization of a Weak Base

•Both NH₃and NH₄⁺exist in rapid equilibrium

= 1% is NH₄⁺(Chap. 15)

Principal Types of Reactions

• Precipitation Reactions (4.2)

• Acid - Base Reactions (4.3)

• Oxidation-Reduction Reactions (4.4)

4.2 Precipitation Reactions

• Reactions that result in the formation of precipitate, insoluble solid that separates from the solution

• Often involve ionic compounds

AgNO₃(aq) + 2NaBr(aq) d AgBr(s) + NaNO₃(aq)

Precipitate

Solubility

• Maximum amount of solute that will dissolve in a given quantity of solvent at a specific temperature

• Table 4.2 classifies substances in one of 3 categories

–insoluble –slightly soluble –soluble

Simple solubility rules

Sol uble salts Exceptio ns Alkal i meta l (Li⁺ to C s⁺),

ammonium ( NH4+) N itrate ( NO3

-), bicar bon ate ( HCO3-), s ulf ate ( SO42-), etc.

Sulfates of Ag⁺, C a²⁺, Sr²⁺, Ba²⁺, Pb²⁺

H alide (Cl^-, Br^-, I^-) Halides of Ag⁺, Hg2 2+, Pb²⁺

In soluble salts Exceptio ns C arbonat e (CO3

2-), phosphat e (PO43-), sulfide (S^2-), etc.

Alkal i metal (Li⁺ to Cs⁺), ammoni um ( NH4+)

H ydrox ide (OH^-) Alkal i metal (Li⁺ to Cs⁺), Ba²⁺

Using the solubility rules, classify as soluble or insoluble.

• Ag2SO4

insoluble

• Li₂S soluble

• Pb(NO₃)₂ soluble

• AgCl insoluble

Ionic equations

• “Molecular” equation: write full formula of each species

AgNO₃(aq) + NaBr(aq) d AgBr(s) + NaNO₃(aq)

Ag⁺(aq) + NO₃^-(aq) + Na⁺(aq) + Br^-(aq) d AgBr(s) + Na⁺(aq) + NO3-(aq)

•Ionic equation: write dissolved species as free ions

• Spectator ions are not involved in the overall reaction: Na⁺(aq) and NO₃^-(aq)

Net ionic equations

Ag⁺(aq) + NO₃^-(aq) + Na⁺(aq) + Br^-(aq) d AgBr(s) + Na⁺(aq) + NO₃^-(aq)

Ag⁺(aq) + Br^-(aq) d AgBr(s)

•Net ionic equation: write only species that take part in reaction

Predicting and writing precipitation reactions

1 Write “molecular” equation

2 Dissociate electrolytes d ionic equation 3 Use solubility rules to predict precipitate 4 Cancel spectator ions d net ionic

equation

PbCl

+ Na

SO

1. Write “molecular” equation, using solubility rules to predict precipitate(s)

PbCl₂(aq) + Na₂SO₄(aq)d PbSO₄(s) + 2NaCl (aq) 2. Dissociate electrolytes d ionic equation

Pb^{2 +}(aq) + 2Cl^-(aq) + 2Na⁺(aq) + SO₄^{2 -}(aq)d PbSO₄(s) + 2Na⁺(aq) + 2Cl^-(aq)

3. Cancel spectator ions d net ionic equation Pb^{2 +}(aq) + SO₄^2-(aq)d PbSO₄(s)

More examples (on your own)

• K2SO4+ AgNO3

• (NH₄)₂S + FeSO₄

4.3 Acid-Base Reactions

• Acid-base reactions are very important in chemistry

• Long history

– many models for acids and bases – details in Chaps. 15 and 16

Empirical criteria

• Acid – sour taste

– corrodes metals, often producing H2(g) H₂SO₄(aq) + Fe(s) d^FeSO4(aq) + H₂(g) – turns litmus (plant dye) red

– gives carbon dioxide (g) w/ carbonates 2HCl(aq) + CaCO₃(s) d^CaCl2(aq) +

H₂O(l) + CO₂(g) – electrolyte

Empirical criteria

• Base (alkali)

– bitter taste (NaHCO3= baking soda) – feels slippery (soap)

– turns litmus (plant dye) blue – electrolyte

Arrhenius concept (1887)

• Acid = Proton (H⁺) donor in H2O HCl(aq) dH⁺(aq) + Cl^-(aq) H₂SO₄(aq) dH+(aq) + HSO₄^-(aq)

• Base = Hydroxide (OH^-) donor in H₂O NaOH(aq) d^{Na+(aq) + OH-(aq)}

MgO(s) + H₂O(l) dMg²⁺(aq) + 2 OH^-(aq)

• Neutralization

H⁺(aq) + OH^-(aq)d^H2O(l) Acid + Based^Water

• Brønsted Acid = Proton donor HCl(g) + HF(l) qeH2F⁺(HF) + Cl^-(HF)

• Brønsted Base = Proton acceptor NH3(aq) + H2O(l) qeNH4+(aq) + OH^-(aq)

• Neutralization

NH₃ + HClqeNH4+ + Cl^- Base1 + Acid2qeAcid1 + Base2

• Not necessarily in water

Brønsted-Lowry concept (1923)

Acid1 Base2 Acid2 Base1

Dissociation of HCl in water

H3O⁺= Hydronium ion

H

O

= Hydronium (oxonium) ion

H⁺(aq) + H₂O(l) H3O⁺(aq)

• H+ associates with several H₂O molecules

• e.g. H₇O₃+ or [H(H₂O)_n]⁺(n = 3)

• Often abbreviated H⁺(aq)

NH3(aq) + H2O(l) qeNH4+(aq) + OH-(aq) Base Acid

Weak base in water

• Very little NH₃(=1%) is ionized – typical weak base in water

• H₂O acts as a Brønsted acid in this reaction

• Some substances can act as either acid or base depending on reaction

•Diprotic – supply 2 H⁺in 2 steps H₂S(aq) qe H+(aq) + HS-(aq) SH^-(aq)qe H⁺(aq) + S^2-(aq)

•Triprotic – supply 3 H⁺in 3 steps H₃PO₄(aq)qe H+(aq) + H₂PO₄^-(aq) H₂PO₄^-(aq)qe H+(aq) + HPO₄^{2 -}(aq) HPO₄^2-(aq)qe H⁺(aq) + PO₄^{3 -}(aq)

•Each step is successively weaker

Polyprotic acids

Acid - Base Neutralization Reaction

• Neutralization reaction - reaction between acid and base to produce a salt and water

• Salt - ionic compound w/ cation besides H⁺

• acid + based salt + water

HBr(aq) + KOH(aq)dKBr(aq) + H₂O(l) net: H⁺(aq) + OH^-(aq) dH₂O(l)

Examples of Neutralization Reactions

• HF(aq) + NaOH(aq) d NaF(aq) + H₂O(l)

• 2HNO₃(aq) + Ba(OH)₂(aq) d Ba(NO₃)₂(aq) + 2H₂O(l)

• H₂SO₄(aq) + 2LiOH(aq) d Li₂SO₄(aq) + 2H₂O(l)

4.4 Oxidation-Reduction Reactions

• “Redox reactions”

– electron transfer reactions

• 2K(s) + Cl2(g) d 2KCl(s)

• K loses an electron to become K⁺ 2K d2K⁺+ 2e^-

• Cl₂gains 2 electrons to become 2Cl^- Cl₂+ 2e^-d^2Cl-

• Each step is a half-reaction – explicitly shows electron transfer

GER

Redox mnemonic

• LEO

Loss of Electrons = Oxidation

• GER

Gain of Electrons = Reduction

LEO

Redox Definitions

• 2K(s) + Cl2(g) d 2KCl(s)

• K is oxidized to K⁺ – oxidation half-reaction 2K d2K⁺+ 2e^-

– K is the reducing agent (reductant)

• Cl₂is reduced to 2Cl^- – reduction half-reaction – Cl₂+ 2e^-d^2Cl-

– Cl₂is the oxidizing agent (oxidant)

M + n H⁺ Mⁿ⁺+ n/2 H₂

The Activity Series for Metals

Displacement Reactions

Ca + 2H₂O Ca(OH)₂+ H₂ Pb + 2H₂O Pb(OH)₂+ H₂ M + n H₂O M(OH)_n+ n/2 H₂

Examples:

Short Activity Series

• Li most reactive

• K

• Ba

• Na

• Zn

• H

• Cu

• Hg

• Au least reactive

Oxidation Number

• Reactions of molecular, not just ionic, compounds are redox reactions

P₄(s) + 5O₂(g) d P₄O₁₀(s)

– There are no ionic charges shown, but it is a redox reaction

• Oxidation number (state) – charge an atom would have if e- were transferred completely

P(+5) and O(-2)

– remember: ΣOx# = charge on ion (molecule)

• Chap. 9 explains direction of transfer

Ox# Rules

1. In free elements each atom has Ox # = 0 2. For monatomic ions, Ox # = charge 3. Ox # of oxygen is usually -2 4. Ox # of hydrogen is usually +1

except metal hydrides, Ox # (H) = -1 5. Ox # of fluorine is always -1 6. ΣOx # = charge on molecule or ion 7. Ox # can be fractional

Calculate Ox#s

H3PO4

SO₂ SF6

CO Re₂Cl₈^2- SH3+

Types of Redox Reactions

• Combination Reaction

– substances combine to form 1 product – Fe(s) + O₂(g) d^Fe2O₃(s)

• Decomposition Reaction

– breakdown of compound into components – NH₄NO₂(s) d^N2(g) + H₂O(l)

Displacement Reactions

• Hydrogen displacement

Mg(s) + H₂O(l) d^Mg(OH)2(aq) + H₂(g)

• Metal displacement

– more active metal displaces less active metal (Chap. 19)

Al(s) + Fe₂O₃(s) d^{Fe(l) + Al}2O₃(s)

• Halogen displacement – reactivity F₂> Cl₂> Br₂> I₂ Cl₂(g) + CaBr₂(s) d^Br2(l) + CaCl₂(s)

Disproportionation Reaction

• Element in one oxidation state is simultaneously oxidized and reduced 2CuCl(s) d Cu(s) + CuCl2(s)

• Cu(+1) goes to Cu(0) and Cu(+2)

4.5 Concentration of Solutions

• Amount of solute present in a given quantity of solvent or solution

• Molarity (M) – the number of moles of solute in 1 liter of solution

M = molarity = mol solute L solution

Calculating molarity

What is the molarity of a solution made by dissolving 5.0 g of NaCl in enough water to make 250.0mL of solution?

Conc. = 0.34 M

Conc. NaCl = 0.34 M

• What is the conc. of Na⁺in solution?

[Na⁺] = 0.34 M

• What is the conc. of Cl^-in solution?

[Cl^-] = 0.34 M

Consider a 0.50 M BaCl₂solution.

• [Ba²⁺] = ? 0.50 M

• [Cl^-] = ? 1.0 M

• Molarity (mol/L) can be used to convert – mol solute to L solution

– L solution to mol solute

How many moles of HCl are in 500 mL of 0.30 M HCl?

0.50 L x (0.30 mol/ L) = 0.15 mol

• Convert molarity to mass of solute needed a Transfer solute to volumetric flask b Add enough solvent to dissolve solute c Dilute to mark

Prepare 250 mL of 2.0 M KOH solution

• Need (2.0 mol KOH/ L)(0.25 L) = 0.50 mol KOH

• (0.50 mol KOH)(56 g /mol) = 28 g KOH

• Dissolve 28 g KOH in enough water to make 250 mL of solution

Dilution

• Start with a concentrated stock solution

• Add more solvent to produced solution of lower concentration

• M = mol/ L

• mol contained = M•V

Dilution Formula

• Same # mol solute in each beaker

• mol = M•V in each beaker

• M_iV_i= M_fV_f

i = initial, f = final

Add solvent

How many mL H₂O are required to dilute 205 mL of 1.15 M HCl solution to 0.81 M?

M_iV_i= M_fV_f V_f= M_iV_i/ M_f

V_f= (1.15 M)(0.205 L)/(0.81 M) = 0.291 L VH2O= Vtotal- Vin

VH2O= 0.291 L - 0.205 L = 0.086 L = 86 mL

• Determination of the amount or

concentration of a substance in a sample

• Two Quantitative Analysis Techniques 1. Gravimetric Analysis – based on the

measurement of mass (Section 4.6)

2. Titration - solution of known concentration is added to solution of unknown concentration until chemical reaction is complete (endpoint)

a. Acid-Base Titrations (Section 4.7) b. Redox Titrations (Section 4.8)

Quantitative Analysis

How many grams of NaCl are required to precipitate all the Ag⁺ions from 125 mL of 0.0081 M AgNO3solution?

NaCl (aq) + AgNO3(aq) d AgCl (s) + NaNO3(aq)

[work on blackboard]

0.059 g

4.7 Acid-Base Titration

• Equivalence point – point at which acid is completely neutralized by added base (or vice versa)

• Equivalence point (or end point) is signaled by an indicator – substance that is different color in acid and base

H₂SO₄+ 2 KOH d K2SO₄+ 2 H₂O M KOH = (0.0225 L H₂SO₄)(0.383 mol

H₂SO₄/L H₂SO₄)(2 mol KOH/ 1 mol H₂SO₄)/ (0.0200 mL KOH) = 0.862 M KOH

22.5 mL of 0.383 M H2SO4are required to neutralize 20.0 mL of a KOH solution.

Calculate the molarity of the KOH solution.

4.8 Redox titrations

• Read on your own

• Won’t be on exam