INORGANIC CHEMISTRY Acids and Bases

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INORGANIC CHEMISTRY Acids and Bases

Solubility Equilibria

Period Trends

Metal Complexes

Intermolecular Forces

Transitional Metal Properties

Physical States and Phase Diagrams

Metals in Biology

Entropy

Chemical Kinetics

Crystal Structures

Acids and Bases Common Types of Acids & Bases   



Many common acids have a proton attached to an O atom: Carboxylic acids Many bases have a nitrogen atom with a lone pair Arrhenius: H+(aq) + OH-(aq)  H2O (l)  Acid = H+ produced in aqueous solution e.g. HCl  Base = OH- producer e.g. NaOH Bronsted – Lowry: H+ + A-  HA  Acid = proton donor (H+) e.g. HCl  Base = proton acceptor e.g. NH3

Conjugate Acid-Base Pairs  

A conjugate base has one less proton than its conjugate acid HSO4 Conjugate bas is SO42 Conjugate acid is H2SO4

Acid Base Reactions

The pH Scale pH = -log10[H3O+] pOH = -log10[OH-] pH + pOH = 14

Strong Acids and Bases 

 

Strong acids and bases completely ionise in water: E.g. HCl (aq) + H20 (l)  H3O+ (aq) + Cl- (aq)  Strong acids: H2SO4, HCl, HBr, HI, HNO3, HClO4  Strong bases: All hydroxides of Groups 1 and 2 (except Be): NaOH, Ca(OH)2 Equilibrium lies completely to the right Examples  What is the pH of a 0.1M HCl solution? HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)

[H3O+] = 0.1M pH = -log10[H3O+] = 1.0  What is the pH of a 0.002M NaOH solution? Completely ionised, so [OH-] = 0.002M pOH = -log10[OH-] = -log10[0.002)] = 2.7 pH = 14 – 2.7 = 11.3

Weak Acids and Equilibrium         

HA(aq) + H2O(l)  H3O+(aq) + A-(aq) A Strong acid has equilibrium to the right (HA completely ionised) A Weak acid has equilibrium to the right (HA partly/most intact) Equilibrium Equation: Ka (Acid Ionisation Constant)= [H3O+][A-] [HA] pKa = -log10Ka The larger the value of Ka the stronger the acid and the lower the value of pKa The concentration of water [H2O] is assumed to be constant Most acids or bases are weak – they do not completely ionise in water Example  Find the pH of 0.1M accetic acid (CH3COOH (HAc)), when pKa = 4.7, Ka = 10-4.7 Conc Initial Change Equil.

HAc(aq) + H2O(l)  H3O+(aq) + Ac-(aq) 0.1 large 0 0 -x +x +x 0.1 – x x x

Ka = 10-4.7 = x2/0.1 – x (Assume that the equilibrium constant is very small so that 0.1 – x = 0.1) 10-4.7 = x2/0.1 x = √(10-4.7 x 0.1) = 10-2.85 pH = -log10[H3O+] = -log10[10-2.85] = 2.9 #

Autoionisation of Water H2O(l) + H2O(l)  H3O+(aq) + OH-(aq) 

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Equilibrium constant (Kw) = [H3O+][OH-] = 1.0 x 10-14 pKw = -log10Kw = 14.00 0 At 25 C: Kw = 1.0 x 10-14  reaction is endothermic: more favourable at higher temperature Neutral solution: [H3O+] = [OH-] = 1.0 x 10-7Acidic solution: [H3O+] > 1.0 x 10-7 Basic solution: [H3O+] < 1.0 x 10-7

Temperature Dependence of pH

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Kw = 1.0 x 10-14 only at 25oC Reaction is endothermic: more favourable at higher temperature  For T > 25oC, Kw > 10-14  For T < 25oC, Kw < 10-14  If T was not 25oC, then pH + pOH does not = 14 and neutral pH is not 7

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