Honors Chemistry

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Honors Chemistry Chapter 4: Reactions in Aqueous Solutions

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Honors Chemistry. Chapter 4: Reactions in Aqueous Solutions. 4.1 Aqueous Solutions. Solution – homogeneous mixture Solute – gets dissolved, often smaller quantity Solvent – dissolves solute, often greater quantity Aqueous solution – solvent is water! - PowerPoint PPT Presentation

Transcript of Honors Chemistry

Page 1: Honors Chemistry

Honors Chemistry

Chapter 4: Reactions in Aqueous Solutions

Page 2: Honors Chemistry

4.1 Aqueous Solutions

• Solution – homogeneous mixture• Solute – gets dissolved, often smaller quantity• Solvent – dissolves solute, often greater quantity• Aqueous solution – solvent is water!

• Electrolyte – conducts electricity when dissolved in water

• Strong electrolyte – good conductor• Weak electrolyte – poor conductor• Nonelectrolyte – does not conduct when dissolved

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4.1 Aqueous Solutions

• Dissolution process• Solvation – solute surrounded by solvent

molecules• Hydration – solvation with water molecules

• Strong• NaCl (s) Na+ (aq) + Cl- (aq)

• Weak• HF (g) H+ (aq) + F- (aq)

• Equilibrium

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4.2 Precipitation Reactions

• Formation of an insoluble product• Solubility Rules

• Always soluble– Alkalai metal salts and ammonium salts– Nitrates, bicarbonates, and chlorates

• Usually soluble– Halides (except Ag+, Hg2+, Pb2+)– Sulfates (except Ag+, Ca2+, Sr2+, Ba2+, Hg2+, Pb2+)

• Usually insoluble– Carbonates, phosphates, chromates, sulfides,

hydroxides

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4.2 Ionic Equations

• Molecular Equation• Formulas written out as normal• Pb(NO3)2 (aq) + 2 KI (aq) PbI2 (s) + 2 KNO3 (aq)

• Ionic Equation• Dissolved substances shown as free ions• Pb2+ (aq) + 2 NO3

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

- (aq)

• Net Ionic Equation• Remove “spectator ions”• Pb2+ (aq) + 2 I- (aq) PbI2 (s)

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4.2 Ionic Equations

• Try this…• Write a molecular equation, ionic equation, and

net ionic equation for the reaction between silver nitrate and iron (III) chloride

• 3 AgNO3 (aq) + FeCl3 (aq) 3 AgCl (s) + Fe(NO3)3 (aq)

• 3 Ag+ (aq) + 3 NO3- (aq) + Fe3+ (aq) + 3 Cl- (aq)

3 AgCl (s) + Fe3+ (aq) + 3 NO3- (aq)

• Ag+ (aq) + Cl- (aq) AgCl (s)

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4.3 Acid-Base Reactions

• Arrhenius definition

• Acid• produces H+ ion in aqueous solution• HCl (g) H+ (aq) + Cl- (aq)

• CH3COOH (l) H+ (aq) + CH3COO- (aq)

• Base• produces OH- ion in aqueous solution• NaOH (s) Na+ (aq) + OH- (aq)

• NH3 (g) + H2O (l) NH4+ (aq) + OH- (aq)

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4.3 Acid-Base Reactions

• Properties of Acids• Sour taste (vinegar, citrus fruits)• React with metals to liberate hydrogen gas

• Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)

• React with carbonates and bicarbonates

• 2 HCl (aq) + CaCO3 (s) CaCl2 (aq) + H2O (l) + CO2 (g)

• Properties of Bases• Bitter taste• Feel slippery (saponification)

• Indicators – change color in acid / base solutions

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4.3 Brønsted Theory

• Define based on movement of H+ ion• Acid = proton donor• Base = proton acceptor

• HCl (g) + H2O (l) H3O+ (aq) + Cl- (aq)• acid base conjugate acid conjugate base

• H3O+ is the hydronium ion• Same as H+ (aq)• Shows a water of hydration

• NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq)

• base acid CA CB

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4.3 Brønsted Theory

• Strong acid• HNO3 + H2O H3O+ + NO3

-

• Weak acid• HBr + H2O H3O+ + Br-

• Diprotic acid• H2SO4 + H2O H3O+ + HSO4

-

• HSO4- + H2O H3O+ + SO4

2-

• Triprotic acid• 3 ionizations, e. g., H3PO4

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4.3 Autoprotolysis of Water

• HOH (l) H+ (aq) + OH- (aq)• Water is both a weak acid and a weak base• Using Brønsted theory,• H2O + H2O H3O+ (aq) + OH- (aq)• B A CA CB• [H3O+] = 1.0 x 10-7 M in pure water• Define pH = -log [H3O+] • pH = 7 in pure water• Acid = lower pH; Base = higher pH

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4.3 Acid-Base Neutralization

• Acid + base salt + water• HCl (aq) + NaOH (aq) NaCl (aq) + H2O• Ionic equation

• H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) Na+ (aq) + Cl- (aq) + H2O (l)

• Net ionic equation• H+ (aq) + OH- (aq) H2O (l)• All neutralizations reduce to this

• Try this: H2SO4 (aq) + Mg(OH)2 (aq)

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4.4 Redox Reactions

• Oxidation-reduction reaction

• Electron transfer reactions

• Oxidation – loss of e- (higher charge)

• Reduction – gain of e- (lower charge)

• 2 Ca (s) + O2 (g) 2 CaO (s)

• Ca goes from neutral to 2+ ..... Oxidation

• O goes from neutral to 2- ..... Reduction

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4.4 Half-Reactions

• Separate oxidation from reduction

• 2 Ca (s) + O2 (g) 2 CaO (s)

• 2 Ca 2 Ca2+ + 4 e- oxidation

• O2 + 4 e- 2 O2- reduction

• Ca is reducing agent• Donates e-, gets oxidized

• O is oxidizing agent• Accepts e-, gets reduced

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4.4 Half-Reactions

• Try this...• Write half reactions for

Zn + 2 HCl ZnCl2 + H2

• Zn + 2 H+ + 2 Cl- Zn2+ + 2 Cl- + H2

• Cl- ion is a spectator• Ox: Zn Zn2+ + 2 e-

• Red: 2 H+ + 2 e- H2

• Zn is reducing agent, H+ is oxidizing agent

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4.4 Oxidation Numbers

• Charge an atom has within a molecule• Free element: ox # = 0• Monatomic ions: ox # = charge• Oxygen: always 2- (except peroxides)• Hydrogen: +1 (except in alkalai hydrides)• Fluorine: always -1• Neutral molecule: all ox #’s add up to 0• Ox #’s can occasionally be fractions

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4.4 Oxidation Numbers

• Try this…

• Find oxidation numbers for all elements:

• Al2O3

• H2SO4

• PO43-

• Fe3O4

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4.4 Types of Redox Reactions

• Combination Reactions (synthesis)• Two or more substances combine, single product

• S + O2 SO2

• Decomposition Reactions• Compound breaks down into components

• 2 KClO3 2 KCl + 3 O2

• Displacement Reactions• One ion displaces another in a compound

• Zn + 2 HCl ZnCl2 + H2

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4.4 Displacement Reactions

• Hydrogen Displacement• IA and IIA metals displace H+ from water

• 2 Na + 2 HOH 2 NaOH + H2

• Most metals displace H+ from acid

• Metal Displacement• Cu + AgNO3 Cu(NO3)2 + Ag

• Activity series shows who can displace whom• p. 134• Cu is higher in series and so can displace Ag

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4.4 Redox Reactions

• Halogen Displacement• F > Cl > Br > I

• 2 NaCl + F2 2 NaF + Cl2• NaCl + Br2 NR

• Disproportionation Reactions• Same element is oxidized and reduced

• 2 H2O2 2 H2O + O2

• O starts with 1- charge

• Ends with 2- charge in H2O and 0 charge in O2

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4.5 Concentration of Solutions

• Molarity (M)• Number of moles solute in 1 L of solution• n

M = ---- V

• Find the Molarity of 15 g NaCl in 250 mL.• 15 g 1 mol

------ x --------- = 0.256 mol 1 58.5 g

• M = 0.256 mol / 0.250 L = 1.03 M

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4.5 Concentration of Solutions

• How would you make 175 mL of a 0.500 M solution of CaCl2?

• n0.500 M = ----------- 0.175 L

• n = 0.0875 mol• 0.0875 mol 111 g

--------------- x -------- = 9.71 g 1 1 mol

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4.5 Dilution of Solutions

• Reduce the concentration by adding water

• Moles of solute stay constant

• From M = n/V, we get n = MV

• Since n for soln 1 equals n for soln 2,

• M1V1 = M2V2

• Try this...• How would you prepare 500 mL of a 0.75 M HCl

solution from a 11.7 M stock HCl solution?

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4.6 Gravimetric Analysis

• Lab technique based on measuring mass• Used to find percent composition• E.g., find % Cl in NaCl• AgNO3 + NaCl NaNO3 + AgCl• Collect AgCl precipitate and mass• Determine mass of Cl in precipitate• Use it to find percent Cl in original sample• We’ll do this in lab

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4.7 Acid-Base Titration

• Titration• solution of known concentration is reacted with

solution of unknown concentration• Reaction must go to completion• Calculate the unknown concentration

• Standard Solution• Solution whose concentration is accurately known• KHP (Potassium Hydrogen Phthalate)• Used to standardize base solution• Base solution then used to titrate an unknown acid

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4.7 Acid-Base Titration

• Equivalence Point• Point where acid and base have neutralized• Added in exact stoichiometric ratio• Need an indicator to show end point

• Titration Equation• mol acid = mol base at equivalence point

• MaVa = MbVb

• If we know the Molarity and volume of base, we can measure the volume of acid used and calculate its Molarity

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4.8 Redox Titration

• Titration can be done with redox reactions

• Often need to use a reactant whose color changes during the reaction

• Takes the place of an indicator

• Works just like acid-base titration

• End of chapter 4 – Finally!