DOCSLIB.ORG

Equilibrium Constant Equilibrium Constant

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Equilibrium Constant Equilibrium Constant Equilibrium Constant Chapters 6 and 8 Most chemical systems are governed by equilibria Chemical Equilibrium such that if: aA+ bB cC+ dD, then & Activity c d = ()()ac a d = γ γ K a b where ax x[X], x is the activity coefficient ()()aa a b In dilute solutions, γx →1, if we assume γx =1, [][]CDc d K = [][]ABa b EquilibriumEquilibrium constants constants may may be be written written for for dissociations, dissociations, associations, associations, reactions, reactions, oror distributions. distributions. Equilibrium Constant aA+ bB cC+ dD [][]CDc d K = [][]ABa b • If K is very large, that the equilibrium lies far to the right (or towards products). If K is small, the reaction lies towards reactants. • Knowledge of reaction stoichiometry and the equilibrium ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley) constant allows us to make some predictions about the system. Important items to regarding K expressions Equilibrium and Thermodynamics 1. All solute concentrations should be in mol/L (M). • Gibbs free energy: 2. All gas concentrations should be in atmospheres. 3. By convention, all K’s are calculated relative to 1 M ΔG o = -RT ln K solutions or 1 atm gas, so the resulting constants are o dimensionless. K= e−ΔG / RT 4. Concentrations of pure solids, pure liquids and solvents are omitted from the equilibrium constant expression. • If K >1 Æ ΔGo<0 Æ Spontaneous Equilibrium constant expressions are Æ o Æ thermodynamic relations. • If K<1 ΔG NOT Spontaneous 1 Using equilibria to characterize Reaction Quotient (Q) systems The equilibrium constant expression with non- equilibrium concentrations plugged in. Solubility: solubility product (Ksp) Complexation: formation constants (Kn), • If Q>K, the reaction must proceed to the left cumulative formation constants (βn) (Ch.12) • If Q<K, the reaction must proceed to the right Acids and Bases: acid dissociation constant • If Q=K the reaction is at equilibrium (Ka), base hydrolysis constant (Kb) Solubility Solubility - Separation by The solubility product (Ksp) describes the concentrations of precipitation species present when ions are in equilibrium with undissolved salt. It is possible to quantitatively separate two or more species based on their solubility. 2+ 2- Ability to do so is related to the magnitudes of the Ksp for each Example: CaCO3 = Ca + CO3 ion. 2+ 2- 2+ 2- -9 K= [Ca ][CO3 ]/[CaCO3]= [Ca ][CO3 ] = Ksp = 6.0 x 10 Example: What is the concentration of calcium in a saturated solution Is it possible to precipitate 99% of 0.010M Ce3+ by adding 2- 2+ of calcium carbonate? oxalate (C2O4 ) without precipitate 0.010M Ca -8 This presence of carbonate results in a common ion effect. CaC2O4 Ksp = 1.3 x 10 What should happen? -29 Ce2(C2O4)3 Ksp = 3.0 x 10 Acids and Bases Solubility – Common Ion Effects • Definitions: Lewis – Electrons (acid: electron pair acceptor); BrØnsted-Lowry (acid: proton donor) A salt will be less soluble is one of its constituent • Conjugate Acid-Base Pairs: related by the gain or loss of one ions is already present in the solution. proton (ex. Acetic acid & acetate ion). • Neutralization Reactions: reactions of acid and base to form salts and water • Solvent Autoprotolysis or self-ionization: water is the most common, in which it acts both an acid and a base: + - Example: What is the solubility of ferric hydroxide 2H2O H 3O +OH - in pH 8.0 buffered aqueous solution OH +H+ H 2O Autoprotolysis constant (equilibrium constants) + + − = − = − = 14 Kw [ 3OH ][OH [] H ][OH ] 10 2 pH Acid and Base Strength - H O OH +H+ 2 Based on percent dissociation in solution P-function: pX = - log10[X] • Strong acids/bases dissociate essentially completely pH = -log [H+] in water (Table 6-2, p.108) pH > 7 is basic, pH<7 is acidic • Weak acids/bases only partially dissociate, results in - pOH=-log[OH ] equilibrium concentrations of both the acid and its pH + pOH = pKw = 14.00 at 25oC conjugate base Ka - + Ka - + Ex. Concentration of H+ and OH- in pure water at 25oC HA +H2O A +H3O HA A + H -7 ANS: 1.0x10 M Kb + - B + H2O BH +OH Weak Acid/Base Equilibria Diprotic acids/Dibasic bases Polyprotic acids/Polybasic bases • Weak Acid: acid dissociation constant (Ka) + − Ka [H ][HA ] HA +H O A- +H O+ − + − + = 2 3 [A ][H O ] [A ][H ] + - Ka1 = 3 = H A H + HA Ka 2 []HA2 Ka - + HA A + H []HA []HA - + 2- + − HA H + A [H ][A2 ] Ka = • Weak Base, hydrolysis reaction: base hydrolysis 2 []HA− constant (Kb) + − Ka1, Ka2 … [BH ][OH ] Kb + - = + - -3 B + H2O BH +OH Kb H PO H + H PO Ka1 = 7.11 x 10 []B 3 4 2 4 - + 2- -8 H2PO4 H + HPO4 Ka2 = 6.32 x 10 • Relationship between Ka and Kb for conjugate HPO 2- H+ + PO 3- Ka3 = 7.1 x 10-13 acid/base pairs: 4 4 + + − = × = − = − = 14 Kw Ka Kb [HO3 ][OH ][H ][OH ] 10 Complex Formation Activity Formation of coordinate bonds between Lewis Acids/Bases + + K1 + If aA+ bB cC+ dD, then Ag ()aq NH 3 (aq) Ag ()()NH 3 aq K2 + Ag ()()NH + aq + NH (aq) Ag ()()NH aq 3 3 3 2 c d c d ()()ac a d ()()γc[C] γ d[D] + + = = [()]Ag NH [()]Ag NH K = 3 = 3 2 a b γa γ b K1 + K 2 + ()()aa a b ()()a[A] b[B] [Ag ][NH 3 ] [()Ag NH3 ][ NH 3 ] where ax is the activity of x Formation constants (Kf) are the equilibrium constants for complex ion formation. The overall, or cumulative, formation constants are ax = γx[X], γ x is the activity coefficient denoted βi c d c d + + γ γ Ag (aq )+ 2NH (aq) Kf Ag()()NH aq ()()ac a d ()()c[C] d[D] 3 3 2 K = = + a b γa γ b [()]Ag NH ()()aa a b ()()a[A] b[B] K = 3 2 =β =KK ⋅ f + 2 2 1 2 If γx →1, ax →[X ] [Ag ][NH 3 ] 3 Activity Coefficient Ionic Strength • Related to the size of the hydrated species • Calculate γ using the extended Debye-Hückel equation, μ =1 2 +2 + = 1 2 which relates activity coefficients to the ability of ions in (c1 z 1 c2 z 2 ...) ∑ci z i solution to interact with one another. 2 2 i where c is concentration and z is charge of each ion .0 51z 2 μ log r = − α μ Ex. What is the ionic strength of a 0.010 M Na2SO4? 1+ ½*{(0.020*1)+(0.010*(-2)2)}=0.030 M 305 If add 0.020 M KBr? z = charge of the ion ½*{(0.020*1) + (0.020*1) + (0.020*1) + (0.010*4)}=0.050 M α = effective diameter "hydrated" of the ion in nanometers μ = ionic strength of the solution Ionic Strength and Ionic Atmospheres Hydrated Radius • Ions with small ionic radii and large charge tend to more strongly bind to solvent molecules (Ion-dipole interactions). • The result of this binding is a larger hydrated radius, causing diminished interaction The greater the ionic strength of a solution, the higher the charge in the with other ions. ionic atmosphere. Each ion-plus-atmosphere contains less net charge and there is less attraction between any particular cation and anion. Ionic Strength, Ion Charge, and Ion Size effect pH Revisited .0 51z 2 μ • Increased ionic conc. Æ log r = − α μ • Concentration is replaced with activity decreased activity coefficient 1+ 305 + γ • Increased ion charge (±) Æ pH = -log aH+ = -log [H ] H+ increased departure of activity coefficient from unity (Multiply charged ions are generally more likely to interact with other ions than Examples (P.147): singly charged.) • Calculate the pH of pure water using activity coefficients • Smaller hydrated radius Æ correctly. increased importance of activity effects • Calculate the pH of water containing 0.10 M KCl at 25oC. 4.
Load more

Recommended publications
  • Characteristics of Chemical Equilibrium
    Characteristics of Chemical Equilibrium Chapter 14: Chemical Equilibrium © 2008 Brooks/Cole 1 © 2008 Brooks/Cole 2 Equilibrium is Dynamic Equilibrium is Independent of Direction of Approach Reactants convert to products N2(g) + 3 H2(g) 2 NH3(g) a A + b B c C + d D Species do not stop forming OR being destroyed Rate of formation = rate of removal Concentrations are constant. © 2008 Brooks/Cole 3 © 2008 Brooks/Cole 4 Equilibrium and Catalysts The Equilibrium Constant For the 2-butene isomerization: H3C CH3 H3C H C=C C=C H H H CH3 At equilibrium: rate forward = rate in reverse An elementary reaction, so: kforward[cis] = kreverse[trans] © 2008 Brooks/Cole 5 © 2008 Brooks/Cole 6 1 The Equilibrium Constant The Equilibrium Constant At equilibrium the concentrations become constant. We had: kforward[cis] = kreverse[trans] kforward [trans] or = kreverse [cis] kforward [trans] Kc = = = 1.65 (at 500 K) kreverse [cis] “c” for concentration based © 2008 Brooks/Cole 7 © 2008 Brooks/Cole 8 The Equilibrium Constant The Equilibrium Constant For a general reaction: a A + b B c C + d D [NO]2 N2(g) + O2(g) 2 NO(g) Kc = Products raised to [N2] [O2] stoichiometric powers… k [C]c [D]d forward …divided by reactants Kc = = a b kreverse [A] [B] raised to their stoichiometric [SO ] 1 2 powers 8 S8(s) + O2(g) SO2(g) Kc = [O2] © 2008 Brooks/Cole 9 © 2008 Brooks/Cole 10 Equilibria Involving Pure Liquids and Solids Equilibria in Dilute Solutions [Solid] is constant throughout a reaction. density g / L • pure solid concentration = mol.
    [Show full text]
  • Laboratory 1: Chemical Equilibrium 1
    1 Laboratory 1: Chemical Equilibrium 1 Reading: Olmstead and Williams, Chemistry , Chapter 14 (all sections) Purpose: The shift in equilibrium position of a chemical reaction with applied stress is determined. Introduction Chemical Equilibrium No chemical reaction goes to completion. When a reaction stops, some amount of reactants remain. For example, although we write → ← 2 CO 2 (g) 2 CO (g) + O 2 (g) (1) as though it goes entirely to products, at 2000K only 2% of the CO 2 decomposes. A chemical reaction reaches equilibrium when the concentrations of the reactants and products no longer change with time. The position of equilibrium describes the relative amounts of reactants and products that remain at the end of a chemical reaction. The position of equilibrium for reaction (1) is said to lie with the reactants, or to the left, because at equilibrium very little of the carbon dioxide has reacted. On the other hand, in the reaction → ← H2 (g) + ½ O2 (g) H2O (g) (2) the equilibrium position lies very far to the right since only very small amounts of H 2 and O 2 remain after the reaction reaches equilibrium. Since chemists often wish to maximize the yield of a reaction, it is vital to determine how to control the position of the equilibrium. The equilibrium position of a reaction may shift if an external stress is applied. The stress may be in the form of a change in temperature, pressure, or the concentration of one of the reactants or products. For example, consider a flask with an equilibrium mixture of CO 2, CO, and O 2, as in reaction (1).
    [Show full text]
  • Solubility Equilibria (Sec 6-4)
    Solubility Equilibria (Sec 6-4) Ksp = solubility product AgCl(s) = Ag+(aq) + Cl-(aq) Ksp = 2+ - CaF2(s) = Ca (aq) + 2F (aq) Ksp = n+ m- in general AmBn = mA + nB n+ m m- n Ksp = [A ] [B ] We use Ksp to calculate the equilibrium solubility of a compound. Calculating the Solubility of an Ionic Compound (p.131) 2+ - PbI2 = Pb + 2I 1 n+ m- in general for AmBn = mA + nB The Common Ion Effect (p. 132) What happens to the solubility of PbI2 if we add a - second source of I (e.g. the PbI2 is being dissolved in a solution of 0.030 M NaI)? The common ion = 2+ - PbI2 = Pb + 2I 2 Ch 12: A Deeper Look at Chemical Equilibrium Up to now we've ignored two points- 1. 2. 2+ - PbI2(s) = Pb (aq) + 2I (aq) -9 + - Ksp = 7.9 x 10 (ignoring PbOH , PbI3 , etc) K'sp = 3 Activity Coefficients - concentrations are replaced by "activities" aA + bB = cC + dD [C]c [D]d K old definition [A]a [B]b γc [C]c γd [D]d K C D new definition using activity γa [A]a γb [B]b A B We can calculate the activity coefficients if we know what the ionic strength of the solution is. Charge Effects - an ion with a +2 charge affects activity of a given electrolyte more than an ion with a +1 charge = ionic strength, a measure of the magnitude of the electrostatic environment 1 2 μ C Z Ci = concentration 2 i i Zi = charge e.g. calculate the ionic strength of an aqueous soln of 0.50M NaCl and 0.75M MgCl2 4 The Extended Huckel-Debye Equation 0.51z 2 log A 1 / 305 A = activity coefficient Z = ion charge = ionic strength (M) = hydrated radius (pm) works well for 0.10M 5 6 Example (p.
    [Show full text]
  • Answer Key Chapter 16: Standard Review Worksheet – + 1
    Answer Key Chapter 16: Standard Review Worksheet – + 1. H3PO4(aq) + H2O(l) H2PO4 (aq) + H3O (aq) + + NH4 (aq) + H2O(l) NH3(aq) + H3O (aq) 2. When we say that acetic acid is a weak acid, we can take either of two points of view. Usually we say that acetic acid is a weak acid because it doesn’t ionize very much when dissolved in water. We say that not very many acetic acid molecules dissociate. However, we can describe this situation from another point of view. We could say that the reason acetic acid doesn’t dissociate much when we dissolve it in water is because the acetate ion (the conjugate base of acetic acid) is extremely effective at holding onto protons and specifically is better at holding onto protons than water is in attracting them. + – HC2H3O2 + H2O H3O + C2H3O2 Now what would happen if we had a source of free acetate ions (for example, sodium acetate) and placed them into water? Since acetate ion is better at attracting protons than is water, the acetate ions would pull protons out of water molecules, leaving hydroxide ions. That is, – – C2H3O2 + H2O HC2H3O2 + OH Since an increase in hydroxide ion concentration would take place in the solution, the solution would be basic. Because acetic acid is a weak acid, the acetate ion is a base in aqueous solution. – – 3. Since HCl, HNO3, and HClO4 are all strong acids, we know that their anions (Cl , NO3 , – and ClO4 ) must be very weak bases and that solutions of the sodium salts of these anions would not be basic.
    [Show full text]
  • Drugs and Acid Dissociation Constants Ionisation of Drug Molecules Most Drugs Ionise in Aqueous Solution.1 They Are Weak Acids Or Weak Bases
    Drugs and acid dissociation constants Ionisation of drug molecules Most drugs ionise in aqueous solution.1 They are weak acids or weak bases. Those that are weak acids ionise in water to give acidic solutions while those that are weak bases ionise to give basic solutions. Drug molecules that are weak acids Drug molecules that are weak bases where, HA = acid (the drug molecule) where, B = base (the drug molecule) H2O = base H2O = acid A− = conjugate base (the drug anion) OH− = conjugate base (the drug anion) + + H3O = conjugate acid BH = conjugate acid Acid dissociation constant, Ka For a drug molecule that is a weak acid The equilibrium constant for this ionisation is given by the equation + − where [H3O ], [A ], [HA] and [H2O] are the concentrations at equilibrium. In a dilute solution the concentration of water is to all intents and purposes constant. So the equation is simplified to: where Ka is the acid dissociation constant for the weak acid + + Also, H3O is often written simply as H and the equation for Ka is usually written as: Values for Ka are extremely small and, therefore, pKa values are given (similar to the reason pH is used rather than [H+]. The relationship between pKa and pH is given by the Henderson–Hasselbalch equation: or This relationship is important when determining pKa values from pH measurements. Base dissociation constant, Kb For a drug molecule that is a weak base: 1 Ionisation of drug molecules. 1 Following the same logic as for deriving Ka, base dissociation constant, Kb, is given by: and Ionisation of water Water ionises very slightly.
    [Show full text]
  • Dissociation Constants and Ph-Titration Curves at Constant Ionic Strength from Electrometric Titrations in Cells Without Liquid
    U. S. DEPARTMENT OF COMMERCE NATIONAL BUREAU OF STANDARDS RESEARCH PAPER RP1537 Part of Journal of Research of the N.ational Bureau of Standards, Volume 30, May 1943 DISSOCIATION CONSTANTS AND pH-TITRATION CURVES AT CONSTANT IONIC STRENGTH FROM ELECTRO­ METRIC TITRATIONS IN CELLS WITHOUT LIQUID JUNCTION : TITRATIONS OF FORMIC ACID AND ACETIC ACID By Roger G. Bates, Gerda L. Siegel, and S. F. Acree ABSTRACT An improved method for obtaining the titration curves of monobasic acids is outlined. The sample, 0.005 mole of the sodium salt of the weak acid, is dissolver! in 100 ml of a 0.05-m solution of sodium chloride and titrated electrometrically with an acid-salt mixture in a hydrogen-silver-chloride cell without liquid junction. The acid-salt mixture has the composition: nitric acid, 0.1 m; pot assium nitrate, 0.05 m; sodium chloride, 0.05 m. The titration therefore is performed in a. medium of constant chloride concentration and of practically unchanging ionic­ strength (1'=0.1) . The calculations of pH values and of dissociation constants from the emf values are outlined. The tit ration curves and dissociation constants of formic acid and of acetic acid at 25 0 C were obtained by this method. The pK values (negative logarithms of the dissociation constants) were found to be 3.742 and 4. 754, respectively. CONTENTS Page I . Tntroduction __ _____ ~ __ _______ . ______ __ ______ ____ ________________ 347 II. Discussion of the titrat ion metbod __ __ ___ ______ _______ ______ ______ _ 348 1. Ti t;at~on. clU,:es at constant ionic strength from cells without ltqUld JunctlOlL - - - _ - __ _ - __ __ ____ ____ _____ __ _____ ____ __ _ 349 2.
    [Show full text]
  • Electrochemistry –An Oxidizing Agent Is a Species That Oxidizes Another Species; It Is Itself Reduced
    Oxidation-Reduction Reactions Chapter 17 • Describing Oxidation-Reduction Reactions Electrochemistry –An oxidizing agent is a species that oxidizes another species; it is itself reduced. –A reducing agent is a species that reduces another species; it is itself oxidized. Loss of 2 e-1 oxidation reducing agent +2 +2 Fe( s) + Cu (aq) → Fe (aq) + Cu( s) oxidizing agent Gain of 2 e-1 reduction Skeleton Oxidation-Reduction Equations Electrochemistry ! Identify what species is being oxidized (this will be the “reducing agent”) ! Identify what species is being •The study of the interchange of reduced (this will be the “oxidizing agent”) chemical and electrical energy. ! What species result from the oxidation and reduction? ! Does the reaction occur in acidic or basic solution? 2+ - 3+ 2+ Fe (aq) + MnO4 (aq) 6 Fe (aq) + Mn (aq) Steps in Balancing Oxidation-Reduction Review of Terms Equations in Acidic solutions 1. Assign oxidation numbers to • oxidation-reduction (redox) each atom so that you know reaction: involves a transfer of what is oxidized and what is electrons from the reducing agent to reduced 2. Split the skeleton equation into the oxidizing agent. two half-reactions-one for the oxidation reaction (element • oxidation: loss of electrons increases in oxidation number) and one for the reduction (element decreases in oxidation • reduction: gain of electrons number) 2+ 3+ - 2+ Fe (aq) º Fe (aq) MnO4 (aq) º Mn (aq) 1 3. Complete and balance each half reaction Galvanic Cell a. Balance all atoms except O and H 2+ 3+ - 2+ (Voltaic Cell) Fe (aq) º Fe (aq) MnO4 (aq) º Mn (aq) b.
    [Show full text]
  • Chapter 15 Chemical Equilibrium
    Chapter 15 Chemical Equilibrium Learning goals and key skills: Explain what is meant by chemical equilibrium and how it relates to reaction rates Write the equilibrium-constant expression for any reaction Convert Kc to Kp and vice versa Relate the magnitude of an equilibrium constant to the relative amounts of reactants and products present in an equilibrium mixture. Manipulate the equilibrium constant to reflect changes in the chemical equation Write the equilibrium-constant expression for a heterogeneous reaction Calculate an equilibrium constant from concentration measurements Predict the direction of a reaction given the equilibrium constant and the concentrations of reactants and products Calculate equilibrium concentrations given the equilibrium constant and all but one equilibrium concentration Calculate equilibrium concentrations given the equilibrium constant and the starting concentrations Use Le Chatelier’s principle to predict how changing the concentrations, volume, or temperature of a system at equilibrium affects the equilibrium position. The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. 1 Concept of Equilibrium • As a system approaches equilibrium, both the forward and reverse reactions are occurring. • At equilibrium, the forward and reverse reactions are proceeding at the same rate. • Once equilibrium is achieved, the amount of each reactant and product remains constant. The same equilibrium is reached whether we start with only reactants (N2 and H2) or with only product (NH3). Equilibrium is reached from either direction. 2 The Equilibrium Constant • Consider the generalized reaction aA + bB cC + dD The equilibrium expression for this reaction would be [C]c[D]d K = c [A]a[B]b Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written c d (PC) (PD) Kp = a b (PA) (PB) Chemical equilibrium occurs when opposing reactions are proceeding at equal rates.
    [Show full text]
  • Simple Mixtures
    Simple mixtures Before dealing with chemical reactions, here we consider mixtures of substances that do not react together. At this stage we deal mainly with binary mixtures (mixtures of two components, A and B). We therefore often be able to simplify equations using the relation xA + xB = 1. A restriction in this chapter – we consider mainly non-electrolyte solutions – the solute is not present as ions. The thermodynamic description of mixtures We already considered the partial pressure – the contribution of one component to the total pressure, when we were dealing with the properties of gas mixtures. Here we introduce other analogous partial properties. Partial molar quantities: describe the contribution (per mole) that a substance makes to an overall property of mixture. The partial molar volume, VJ – the contribution J makes to the total volume of a mixture. Although 1 mol of a substance has a characteristic volume when it is pure, 1 mol of a substance can make different contributions to the total volume of a mixture because molecules pack together in different ways in the pure substances and in mixtures. Imagine a huge volume of pure water. When a further 1 3 mol H2O is added, the volume increases by 18 cm . When we add 1 mol H2O to a huge volume of pure ethanol, the volume increases by only 14 cm3. 18 cm3 – the volume occupied per mole of water molecules in pure water. 14 cm3 – the volume occupied per mole of water molecules in virtually pure ethanol. The partial molar volume of water in pure water is 18 cm3 and the partial molar volume of water in pure ethanol is 14 cm3 – there is so much ethanol present that each H2O molecule is surrounded by ethanol molecules and the packing of the molecules results in the water molecules occupying only 14 cm3.
    [Show full text]
  • Acid Dissociation Constant - Wikipedia, the Free Encyclopedia Page 1
    Acid dissociation constant - Wikipedia, the free encyclopedia Page 1 Help us provide free content to the world by donating today ! Acid dissociation constant From Wikipedia, the free encyclopedia An acid dissociation constant (aka acidity constant, acid-ionization constant) is an equilibrium constant for the dissociation of an acid. It is denoted by Ka. For an equilibrium between a generic acid, HA, and − its conjugate base, A , The weak acid acetic acid donates a proton to water in an equilibrium reaction to give the acetate ion and − + HA A + H the hydronium ion. Key: Hydrogen is white, oxygen is red, carbon is gray. Lines are chemical bonds. K is defined, subject to certain conditions, as a where [HA], [A−] and [H+] are equilibrium concentrations of the reactants. The term acid dissociation constant is also used for pKa, which is equal to −log 10 Ka. The term pKb is used in relation to bases, though pKb has faded from modern use due to the easy relationship available between the strength of an acid and the strength of its conjugate base. Though discussions of this topic typically assume water as the solvent, particularly at introductory levels, the Brønsted–Lowry acid-base theory is versatile enough that acidic behavior can now be characterized even in non-aqueous solutions. The value of pK indicates the strength of an acid: the larger the value the weaker the acid. In aqueous a solution, simple acids are partially dissociated to an appreciable extent in in the pH range pK ± 2. The a actual extent of the dissociation can be calculated if the acid concentration and pH are known.
    [Show full text]
  • Exercise 10 SOLUBILITY PRODUCT CONSTANTS
    Exercise 10 SOLUBILITY PRODUCT CONSTANTS Theory Solubility is a physical property referring to the ability for a given substance, the solute to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium. The resulting solution is called a saturated solution. Solubility is commonly expressed as a concentration, either mass concentration (g of solute per kg of solvent, g per 100 ml of solvent), molarity, molality, mole fraction or other similar descrip- tions of concentration. An oversaturated solution becomes a saturated solution by forming a solid to reduce the dissolved material. The crystals formed are called a precipitate. Often, however, a precipitate is formed when two clear solutions are mixed. For example, when a silver nitrate solution and sodium chloride solution are mixed, silver chloride crystals AgCl (a precipitate) are formed. Silver chloride is one of the few chloride that has a limited solubility. Formations of precipitates are considered between heterogeneous chemical equilibria phenomena. A saturated solution of a slightly soluble salt in contact with its undissolved salt involves an equilibrium like the one below: + - AgX(s) Ag (aq) + X (aq) A where X = Cl, Br, I The symbols (aq) indicate that these ions are surrounded by water molecules. These ions are in the solution. AgCl(s) is a precipitate (s – solid). One can express the equilibrium constant: a a K Ag X (1) aAgX + where: aAg – is activity of the silver cations in the solution - aX – is activity of the halogenide anions in the solution aAgCl – is activity AgX in the solution The activities of ions in solution are: a c a c Ag Ag X X (2) where: γ+ and γ- – are activity coefficients of cation and anion According to the convention, the activity of any solid (s) is equal to 1, i.e.
    [Show full text]
  • Chapter 8:Chemical Reaction Equilibria 8.1 Introduction Reaction Chemistry Forms the Essence of Chemical Processes
    Chapter 8:Chemical Reaction Equilibria 8.1 Introduction Reaction chemistry forms the essence of chemical processes. The very distinctiveness of the chemical industry lies in its quest for transforming less useful substances to those which are useful to modern life. The perception of old art of ‘alchemy’ bordered on the magical; perhaps in today’s world its role in the form of modern chemistry is in no sense any less. Almost everything that is of use to humans is manufactured through the route of chemical synthesis. Such reactive processes need to be characterized in terms of the maximum possible yield of the desired product at any given conditions, starting from the raw materials (i.e., reactants). The theory of chemical reactions indicates that rates of reactions are generally enhanced by increase of temperature. However, experience shows that the maximum quantum of conversion of reactants to products does not increase monotonically. Indeed for a vast majority the maximum conversion reaches a maximum with respect to reaction temperature and subsequently diminishes. This is shown schematically in fig. 8.1. Fig. 8.1 Schematic of Equilibrium Reaction vs. Temperature The reason behind this phenomenon lies in the molecular processes that occur during a reaction. Consider a typical reaction of the following form occurring in gas phase: AgBgCgDg()+→+ () () (). The reaction typically begins with the reactants being brought together in a reactor. In the initial phases, molecules of A and B collide and form reactive complexes, which are eventually converted to the products C and D by means of molecular rearrangement. Clearly then the early phase of the reaction process is dominated by the presence and depletion of A and B.
    [Show full text]