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Acids and Bases Amphiprotic Substances – acid or base depending on reaction Example – Water HCl + H 2O  Cl- + H3O+ + NH 3 + H 2O ⇌ NH4 + OH

H2O = proton donor H2O = proton acceptor

Aqueous Solutions Autoionisation – 2 water molecules combine H2O + H2O ⇌ H+ + OHKw – Ion product of water

= [H+] [OH-] = (1.0 x 10-7) x (1.0 x 10-7) = 1.0 x 10-14 pH (OH) Scale (Strong) •

pH change of 1 = ten-fold change in [H+]

pH = -log[H+] pOH = -log[OH-] pH + pOH = 14 -pH 10 = [H+] 10-pOH = [OH-] Classifications STRONG Dissociation

WEAK

100%

Partial and Reversible

⇌

 Equilibrium Formula

Right

Left

HA  H + A

HA ⇌ H+ + A-

[H+] = conc. Of solution

[H+] < conc. Of solution

+

-

Acids Acid – Proton (H+) donor, H+ in aqueous solutions (water) 6 Strong • HCl • HBr • HI • HNO3 • H2SO4 • HClO4 (Perchloric)

Rest Weak • Inorganic: HF, HNO2, + H2CO3, NH4 • Organic: RCOOH, phenols • Acetic (CH3COOH) • Most cations

Ka & pK a (weak) • Ka = equilibrium constant  PORK  >Ka = stronger acid •

pKa = –log10(Ka)  Kb = stronger •

pKa = –log10(Kb)  Ka2 (pKa1 < pKa2) Triprotic – 3 protons donated (3 pKa values) Example – H3PO4

pH Solutions (Weak) • •

Both dissociations are weak – discard other dissociations (doesn’t affect [H+] much) Exception – Sulphuric Acid (H2SO4), 1st dissociation is strong

Example – pH of 0.0020 M solution of Carbonic Acid (pKa1 = 6.35, pKa2 = 10.33) Ka1 = 10-6.35 = 4.5 x 10-7 + 2 [H ][HCO3 ] X = 4.5 x 10-7 = K =

H2CO3 ⇌ H+ + HCO 3-

I C E

H2CO3 0.002 -x 0.002-x

H+

HCO3-

0 +x x

0 +x x

a

[H2CO3]

0.002-X

X2 = 0.002 x 4.5 x 10-7 X = √0.002x4.5x10-7

= 3.0 X 10-5 = [H+]

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pH = -log10(3.0 x 10-5) = 4.5

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Hydrolysis of Ions Hydrolysis – with water Cations • Most cations are weak acids = hydrolysis  Except G1 metal ions (Na+…) & some G2, don’t hydrolyse (pH-neutral ions) • HSO 4- = weak acid Example – Fe3+ + 3H2O  Fe(OH)3 (s) + 3H+ Anions • Anion (A-) of weak acid (HA) are weak base, hydrolyses to produce OHA- + H2O ⇌ HA + OH• Most anions are weak bases  Except – OH-, HSO 4-, Cl-, Br-, I-, NO -3& ClO - 4(anions of 6 strong acids) Example – Solution of sodium acetate (NaCH3COO) is basic CH3COO- + H2O ⇌ CH3COOH + OH-

pH of Salt Solutions Formula of salts indicates acidic, basic or neutral solutions in water  Formula & look at cation & anion separately  Cation is acidic & anion is basic, look at pKa & pkB – smallest wins

•

Examples 1. NaF • Na+ - neutral • F-– basic (anion of HA) • Overall – basic

2. NH 4Cl • NH4 +- acidic • •

Cl- - neutral (anion of strong acid) Overall – Acidic

3. NaNO 3 • Na+ - neutral • NO3- neutral •

4. pH of 0.0100M solution of sodium acetate (NaCH3COO) pKa of acetic acid is 4.76 at 25oC Na+ = neutral, pKb (CH3COO-) = 14 – 4.76 = 9.24 Kb = 10-9.24 = 5.75 x 10-10

-

[CH3COOH][OH ]

=

CH3COO- + H2O ⇌ CH3COOH + OHX = √0.01 x 5.75 x 10-10 -6 = = 2.4 [OHx-] 10 pOH = -log10(2.4 x 10-6)

-

[CH3COO ] X2

2

X

= 5.62 pH = 14 – 5.62 = 8.38

= 0.01-X = 0.01 x 5.75 x 10 -10

Buffer Solutions Buffer – Mixture of HA and conjugate A- in similar concentrations • Resists change in pH when small acid/base are added • Conjugate base consumes added H+ & conjugate acid consumes added OH• Maintain safe body fluid pH • Different conjugate pairs = different pH values = Buffers of any pH ] [A • Best buffer solutions = ratio of =1 [HA]

Overall - neutral

Page 5 of 32sdfdsfg Examples 1. Blood Plasma = pH of 7.35-7.45 by H2CO3/HCO 3- buffer 2. Need buffer near pH 4/5 Choose Acetate buffer (CH3COOH/CH3COO-) & pKa of CH3COOH is 4.76 CH3COO- + H+  CH3COOH or CH3COOH + OH-  CH3COO- + H2O Buffer Equations

HA ⇌ H+ + A+

Ka =

For Weak Acid (H+ Ions) H+ + A-  HA

-

[H ][A ] [HA] [HA]

For weak base (OH- ions) OH- + HA  H2O + X-

[H+ ] = Ka

[A ]

Calculate pH [HA] -Log10[H+ ] = -log10Ka – log10 [A -] Henderson-Hasselbalch Equation -

pH = pKa + log10

[A ]

[HA]

Examples 1. Need pH 9 buffer, use NH 4+/NH3 (NH4+ pKa = 9.25) Concentrations of Each Species Approximation used pH = pKa + log10 NH 3 • Conc in HH eqn are meant to be NH4+ equilibrium concs (after hydrolysis) NH 3 • Usually put initial concs (prior to = pH – pK a log10 + dissolving) = 9 – 9.25 NH4 = –0.25 NH 3 = 10-0.25 NH4 + = 0.562 Concentration ratio of 0.562 or higher (better buffer capacity) 2. Calculate pH of buffer which is 0.05M in acetic acid & 0.07M in sodium acetate pKa(CH3COOH) = 4.76 pH = pKa + log10 [CH3COO ] [CH3COOH] 0.07

= 4.76 + log 10(

)

0.05

= 4.91 Buffer Capacity – amount of added acid/base buffer can neutralise before pH changes • Depends on acid/base used & ka

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Acid/base Titrations • • •

Unknown A- is added to known HA until desired pH is obtained Acid-base indicators – signal equivalence point pH meter – produce titration curve

Titration SASB • Flask – Acid • Burette – Base

Initial pH Low (1/2)

Equivalence point 7

WASB • Flask – Acid • Burette – Base

>2

>7

SAWB • Flask – Base • Burette – Acid

KSP, precipitate forms

= [Ag+][Cl-] = [5.0 x 10-5] [ 5.0 x 10-3] = 2.5 x 10-7

Common Ion Effect Common Ion Effect – ion from 2 different sources reduce solubility of ionic solutes Examples 1. Saturated solution of AgCl (low [Ag+] & [Cl-]) in pure water + AgNO3 (High [Ag+]) • Added Ag+ - shifts equilibrium left, decreasing solubility of AgCl • Ag+ = common ion 2. Solubility of AgCl in pure water

AgCl(s) ⇌ Ag+ + Cl-

K SP = [Ag+] [Cl-] = s2 = 1.8 x 10 -10M S = [Ag+] = [Cl-] = 1.34 x 10 -5M 3. Solubility of AgCl in 0.1 M NaCl (common ion = Cl-)

AgCl(s) ⇌ Ag+ + ClLet [Ag+] = s At equilibrium [Ag+] = s [Cl-] = s + 0.1 = [Ag+] [Cl-] = s(s + 0.1) Neglect (s) = 0.1s = 1.8 x 10-10M S = 1.8 x 10 -9 M KSP

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Dependence of Solubility on pH Metal Hydroxides Sparingly Soluble – Most metal hydroxides e.g. Cu(OH)2, Cr(OH)3, Fe(OH)3 • Solubility of metal ion (cation) depends on pH (OH-)  Example – pH = 9 [H+] = 10-9 M & [OH-] = 10-5 M Use KSP to determine if cation will be soluble at given pH • More soluble in acidic conditions – lowering pH reduces equilibrium (OH-), drives reaction right Example – Max pH required to prevent Cu(OH)2 precipitation from 0.1 M Cu(NO3)2 Given – KSP (Cu(OH)2) = 2.2 x 10-20 Cu(OH)2 (s) ⇌ Cu2+ + 2OHKSP = [Cu2+][OH-]2 = 2.2 x 10-20 Equilibrium [OH ] = √ KSP2+ [Cu ] =√

2.2 x 10

-20

0.1

= 4.69 x 10-10M pOH = -log10(4.69 x 10-10) = 9.33 pH = 14 – 9.33 = 4.67 **above pH (basic) = precipitate **below pH (acidic) = soluble

Salts with Basic Anions •

Sparingly soluble salt with weak base anion, adding H+ = increase solubility

Example – CaF2 ⇌ Ca2+ + 2F+ Acid = H+ + F- ⇌ HF Shifts equilibrium right, so more CaF2 dissolves Solubility increases as solution becomes acidic (low pH)

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Thermochemistry Thermodynamics – study of energy transformation Thermochemistry – examines heat released (or absorbed) by chemical reactions

Enthalpy of Reactions Enthalpy Change (ΔH) – heat transferred (gained/lost) to a system at constant pressure • State Function – only initial & final values

ΔH = Hfinal – HInitial Exothermic

Endothermic

–ΔH

+ΔH

Example

H2(g) + Cl 2(g)  2HCl(g)

ΔH = -185 kJ

Calculate ΔH: a) 1 mole of HCl is formed

ΔH = –185kJ 2 = –92.5kJ b) 1.00g of Cl2 reacts Mass

Moles = Molar Mass 1.00

= 70.9 = 0.0141 mol ΔH = 0.0141 mol x -185 kJ mol-1 = –2.61 kJ

Standard Enthalpy of Reaction ΔHo – enthalpy change when all substances are in their standard states Standard States • Temperature – 25oC (298K) • Liquids – pure • Gases – 1 atm pressure • Concentration of solution – 1M (molar)

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Calorimetry Calorimetry – measure of heat flow (heat released or absorbed)

Heat Capacity & Specific Heat Capacity Heat Capacity (C)

Heat required to raise its temp by 1oC (1 K)

JK-1 or JoC-1

Molar Heat Capacity (Cmolar or Cm)

C of 1 mole of pure substance

JK-1 mol-1 or JoC-1 mol-1

Specific Heat Capacity (C s)

C of 1 gram of pure substance

JK-1g-1 or JoC-1 g-1

q

Cs = m x ΔT q = mCs ΔT q = nCmΔT

q = quantity of heat transferred m = grams of substance ΔT = temperature Change = Tfinal – Tinitial n = moles of substance

q = CΔT (for objects e.g. calorimeter) Example 1. 466g sample of water is heated from 8.50oC to 74.60oC. Calculate the amount of heat absorbed by the water. Cs of water = 4.184 JoC-1g-1 Q = mC s ΔT = mCs(Tfinal – Tinitial) = 466 x 4.184 x (74.60 – 8.50) = 1.29 x 105 J = 129 kJ

Constant-Pressure Calorimetry “Coffee Cup” No barrier between system (R & P) and surroundings (water)

•

Exothermic Reaction • +ΔT (Temp increases)

• •

+qsolution -qreaction

Endothermic Reaction • –ΔT (temp decreases)

• •

-qsolution +qreaction

qrxn + qsolution = 0 (No Energy Lost) Example 1. 50.0 ml of 0.100 M AgNO3 & 50.0 ml of 0.100 M HCl are mixed in a constant-pressure calorimeter, temperature of mixture increases from 22.30°C to 23.11°C

AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(aq) Calculate ΔH in kJ mol-1 of HCl, assuming that the combined solution has a mass of 100.0 g and a specific heat capacity of 4.18 J°C-1g-1 Heat Transferred Q

= mCs ΔT = 100 x 4.18 x (23.11 – 22.30) = 388 J

Page 13 of 32sdfdsfg Heat gained Q

Moles of HCl n

= heat released in reaction = -338 J

= molarity x volume = 0.100 M x 50.0 x 10-3 L = 5.00 x 10-3 mol -338J

ΔH = 5.00 x 10−3 = -6.76 x 104 J mol-1 = -67.6 kJ mol-1

Constant-Volume (Bomb) Calorimetry Bomb Calorimeter – measures heat released in a combustion reaction • Fuel Value – energy released when 1g of substance is combusted  Food – CHO (17 kJg-1) & Fat (38kJg-1) Example 0.5865 g of lactic acid (C3H6O3) is burned in a bomb calorimeter, the heat capacity of which is 4.812 kJ°C-1. The temperature increases from 23.10°C to 24.95°C. Calculate the enthalpy of combustion of lactic acid per gram and per mole. Heat transferred q = C ΔT = 4.812 kJ°C-1 x (24.95°C – 23.10°C) = 8.902 kJ Heat liberated (given off: exothermic) during combustion q = -8.902 kJ ΔHc = -8.902kJ 0.5865g

= -15.18 kJ g-1 moles = =

mass molar mass 0.5865 g -1

90.08 g mol

= 6.511 x 10-3 mol

ΔH c =

–8.902 kJ -3

6.511 x 10 mol

= -1.367 x 103 kJ mol-1

Hess’s Law Hess’s Law – overall sum of all enthalpy change stages

ΔHOverall = ΔH1 + ΔH2 + ΔH3 Calculate unknown ΔH: 1. Rearrange equations of known ΔH • Change ΔH sign if equation is reversed • Adjust moles & ΔH by same factor

2. Add altered equations, cancel substances on both sides

Page 14 of 32sdfdsfg Example Given: (1) C2 H2 (g) + 5/2 O2 (g)  2CO2 (2) C (s) + O2 (g)  CO 2 (g) (3) H2 (g) + ½O2 (g)  H2O (l)

(g)

+ H2 O (l)

ΔH = -1299.6 kJ ΔH = -393.5 kJ ΔH = -285.8 kJ

Calculate ΔH: 2C (s) + H2 (g)  C2H2 (g) (acetylene) (1) – C2 H2 (g) as product, reverse equation 2CO2 (g) + H2O (l)  C2H2 (g) + 5/2 O2 (g)

ΔH = 1299.6 kJ

(2) – x2, correct moles of C (s) 2C (s) + 2O2 (g)  2CO2 (g)

ΔH = -787.0 kJ

(3) H2 (g) + ½O2 (g)  H2O (l)

ΔH = -285.8 kJ

= 2C (s) + H2 (g)  C2H2 (g)

ΔH = 226.8 kJ

Standard Enthalpy of Formation Standard Enthalpy of formation (ΔHo f) – forms 1 mole of compound from its elements in their standard states (=0) or most stable form (at 25oC)

ΔH°f = ΣnΔH°f(products) - ΣmΔH°f(reactants) Example 1. Calculate standard enthalpy of reaction: 2C2H6 (g) + 7O2 (g)  4CO2 (g) + 6H2O (g) (ethane) Data Given: • CO2 = -393.5 kJ mol-1 • H2O = -241.8 kJ mol-1 • C2H6 = -84.7 kJ mol-1 • O2 = elemental state = 0 ΔH° ΔHo

= ΣnΔH°f(products) - ΣmΔH°f(reactants) = [4ΔHof(CO )2 + 6ΔHo (H O)] – [2ΔHo (C H2 ) 6+ 7ΔHo (Of )] 2 f 2 f = [4(-393.5) + 6(-241.8)] – [2(-84.7) + 7(0)] kJ = -2855.4 kJ

Entropy Entropy (S) – measure of disorder (randomness) • Spontaneous Process – occurs on its own

ΔS = SFinal – SInitial Increases (+ΔS) when: • Solids/liquids form gas • Solids form liquids (solutions) • Number of gas molecules increases

Page 15 of 32sdfdsfg Examples 1. Predict if ΔS is positive or negative

H2O (l)  H2O (g) Gas formed from liquid = +ΔS 2. Predict if ΔS is positive or negative

C2H2 (g) + 2H2 (g)  C2H6 (g) 3 moles on left and 1 on right = decreased = –ΔS

Standard Molar Entropy (ΔSo) •

Substances in standard states

ΔSo = ΣnSo(products) – ΣmSo(reactants) •

N & m = coefficients in equation

Observations of ΔSo:  S° of elements is at 298K  S° values of gases> liquids & solids  S° values increase with molar mass  S° values increase with more atoms in a substance

Example – Determine standard entropy change ΔS°, at 298K: Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g) Data Given, So: • Al (s) = 28.32 kJ-1 mol-1 • H2O (g) = 188.83 kJ-1 mol-1 • Al2O3 (s) = 51.00 kJ-1 mol-1 • H2 (g) = 130.58 kJ-1 mol-1 ΔS° ΔSo

= ΣnSo(products) – ΣmSo(reactants) = [2 x 28.32 + 3 x 188.83] – [51.00 + 3 x 130.58] = 180.39 JK-1

Gibbs Free Energy Gibbs Free Energy – Energy used to do work (constant temp and pressure) G = Free Energy H = Enthalpy Gibbs-Helmholtz T = Temperature (kelvin – K = oC + 273) ΔG = ΔH – TΔS S = Entropy Standard-State Free Energy Changes

ΔG° = ΣnΔGf°(products) – ΣmΔGf°(reactants) Example – Determine ΔGo:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Data Given, ΔGfo: • CO2 (g) = -394.4 kJ mol-1 • H2O (g) = -228.6 kJ mol-1 • CH4 (g) = -50.8 kJ mol-1

ΔGfo = standard free energy of formation n,m = coefficients in equation ΔGfo of an element in its most stable form = 0

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Free Energy and Temperature • •

Estimate ΔGo at temps other than 25oC, Use – Gibbs-Helmholtz  Assume ΔH° and ΔS° don’t change with temp Spontaneity of reaction depends on sign & magnitude of ΔHo and TΔSo

ΔH ΔS – +

-TΔS –

ΔG = ΔH – T ΔS –

+

–

+

+

– +

– +

+ –

+/– +/–

Characteristics • Exergonic (releases heat) • Favourable, Spontaneous at all temps • Endergonic (absorbs heat) • Unfavourable, non-spontaneous at all temps • Spontaneous at Low T • Spontaneous at High T

Example 1. Production of Ammonia via the Haber process involves the equilibrium:

N2(g) + 3H2(g) ⇌ 2NH3(g)

Determine ΔG° at 500oC & comment on effect of temp on equilibrium **Note - ΔG° = –33.3 kJ for the Haber process at 25°C ΔH°

= ΣnΔH f(products) – ΣmΔH f(reactants) = [2 x -46.2] – [0 + 3 x 0] = -92.4 kJ

ΔSo

= ΣnS°(products) - ΣmS°(reactants) = [2 x 192.5] – [191.5 + 3 x 130.7] = -198.6 JK-1

= 773 K = ΔH° – TΔS° = -92.4 kJ – (773 K x -198.6 x 10-3 kJ K-1) = 61.1 kJ ΔG° changes from negative to positive value = unfavourable At 500°C ΔG°

Free Energy and the Equilibrium Constant Free energy change for any condition: ΔG = ΔGo + RTlnQ

R = universal constant (8.314 JK-1mol-1) T = temperature in Kelvin Q = reaction quotient

Free energy change at equilibrium (ΔG = 0) 0 = ΔGo + RTlnK ΔGo = –RTlnK

Q = K (equilibrium constant) More negative ΔG° – larger K More positive ΔG° – smaller K

Example – Calculate equilibrium constant at 25°C for the Haber process: ΔG° = -33.3 kJ mol-1 N 2(g) + 3H 2(g) 2NH 3(g) ΔG° = -RT In K –33.3 = -8.314 x 10-3 x 298 x ln K = –2.477 x lnK -33.3 lnK = -2.477 = 13.44 Large K = more negative ΔG° = NH3 is favoured K = e13.44 5 = 6.9 x 10

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Electrochemistry Electrochemistry – study of chemical reactions involving transfer of electrons (e-)

Revision • • •

OXIDATION E donor • Increased oxidation state • Reducing Agent • OIL RIG

REDUCTION E acceptor Decreased oxidation state Oxidising Agent

Galvanic (Voltaic) Cells Galvanic Cells – electrochemical cell, uses spontaneous reaction to produce electrical energy • Electron Flow – anode to cathode Parts • Electrode  Active – part of reaction  Inactive (e.g. C – graphite or Pt – platinum) – conduct electrons but don’t take part in reaction • Anode – negative electrode, where oxidation occurs

Metal Atom  Metal Ion + Electrons •

Cathode – positive electrode, where reduction occurs

Metal Ion + Electrons  Metal Atom • •

Salt (Ion) Bridge – salt solution that doesn’t interfere with redox reactions Electrolyte – ions in solution involved in reaction or carry charge

Line Notation 1. 2. 3. 4. 5.

Anode first | = phase boundaries (e.g. solid & solution) Comma separates components in same phase | | = salt bridge Reactants then products on each side

Example 1. Zinc-copper cell, 1M each

Zn | Zn2+ (1M) | | Cu2+ (1M) | Cu

Example – Voltaic cell is constructed with Ag/Ag= (positive electrode) & Pb/Pb2+ half-cell Write balanced half equations & overall spontaneous reaction Reduction (Ag+(aq) + e-  Ag(s)) x 2 Pb(s)  Pb2+(aq) + 2eOxidation 2Ag+ (aq) + Pb (s)  2Ag (s) + Pb2+ (aq) Overall

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Cell Potentials Standard Cell Potential (Eo cel ) – emf of a cell when all substances are in standard states Electromotive Force (emf) – difference in potential energy (Volts, 1V = 1JC-1) • Driving force of electrons from anode to cathode = drop in potential energy

Standard Reduction (Half-cell) Potential (Eo) Reduction Potential – tendency of a substance to acquire electrons Eo – potential of a reduction half-reaction under standard conditions, relative to SHE

Eocell = E oxidant (reduction process) – E reductant (oxidation process) Standard Hydrogen Electrode (SHE) – electrode that produces this half-reaction (= 0V) • Anode or Cathode depending on other half-cell Anode Cathode

½ H2(g)  H+ aq) + e( Other half-cell = +Eocell

H+(aq) + e - ½ H2(g) Other half-cell = –Eocell

Table of Standard Reduction Potentials • If listed in order of decreasing reduction potential (electrochemical order)  Top, left – strongest oxidants (easily reduced)  Bottom, right – strongest reductants (easily oxidised) • Cathode = more positive Eo & occurs in that direction • Anode = more negative Eo & reaction occurs in opposite direction

Examples 1. Arrange in order of decr...