Qualitative Chemical Analysis PDF

 Qualitative Chemical Analysis Learning outcomes  To understand the purpose of this kind of chemical analysis.  To understand the theoretical aspects behind this type of chemical analysis.  To know how to determine the anions and cations in a test sample practically....

 Qualitative Chemical Analysis Learning outcomes  To understand the purpose of this kind of chemical analysis.  To understand the theoretical aspects behind this type of chemical analysis.  To know how to determine the anions and cations in a test sample practically.  To know the applications of this kind of chemical analysis. 1  First lecture: inorganic qualitative chemical analysis. Its classification and its applications  Aims of this lecture  To understand the meaning of qualitative chemical analysis.  To know the classifications of this kind of chemical analysis.  To know some applications that related to this kind of chemical analysis. General types of Chemical analysis  Qualitative chemical analysis. (Quality) concerns with the sample identity. In other words, by this analysis you can determine what chemical elements present in a sample.  Quantitative chemical analysis. (Quantity) concerns with the amount of a substance in a sample. In other words, by this analysis you can determine the amount of the chemical elements that present in a sample.  Any element or compound under the examination of any of the above analysis is termed “an analyte”. These analysis can be done either handily or instrumently. 4 Qualitative Chemical analysis concerns about studying the methods that are used for identifying basic elements or radicals that are present in a compound or a mixture of compounds. This can be done to liquid or solid samples.  It can be classified to qualitative inorganic analysis or qualitative organic analysis. 5  Qualitative inorganic analysis aims at determining the elemental composition of any inorganic compound. Mainly at ions (positive and negative radicals) in aqueous solutions.  Qualitative organic analysis aims at determining the functional group of any organic compound. For example, by using IR spectroscopy, or the separation and identification of amino acids by paper chromatography. 6 Applications of Inorganic qualitative Chemical analysis  The presence of some chemical elements in water, air or food can badly affect the consumers, humans and animals or plants, thus by using qualitative analysis one can determine the presence of these harmful chemicals in a given sample.  The base of this type of analysis is that every element or compound posses distinguishable properties, such as color, boiling point, melting point, density and can react with certain chemicals in a special way. “Lead (Pb), found in house paints, can react with hydrogen sulfide (H2S) to give black precipitate. 7  Chemists have developed the field of qualitative analysis, this by coming up with a variety of unique chemicals, ions or molecules, that can react selectively with specific materials to produce distinctive properties, gas or color.  Chemists made qualitative chemical analysis instrumentally performed. This is by using UV-Vis spectroscopy or mass spectrometer or chromatography, by which one can identify the components of a mixture. These methods have found a wide range of daily applications in different fields. 8  Qualitative analysis can help the police, this by solve crimes’ problems. A chemist can analyze the human DNA samples taken from a crime scene to reveal whether a person was present in a crime scene or not.  Qualitative analysis can be used to detect D9-THC, the active ingredient in marijuana, by analyzing a urine sample. 9  In the western countries, police use “breathalyzer” to see whether a vehicle driver has drunken alcohol, in huge amount, or not. A “breathalyzer” has a chemical called Potassium dichromate (K2Cr2O7) which change color, from orange to black, if it contacts alcohol vapors. 10 References list The contents of this lecture is taken from 1. A textbook of macro and semimicro qualitative inorganic analysis. Journal of chemical education. H. Rogers.1956. 2. (https://www.cdc.gov/mmwr/preview/mmwrhtml/0 0000138.htm). 3. (http://www.forensicsciencesimplified.org/drugs/pri nciples.html) 11  Second lecture: Solutions The aims of this lecture are to increase students knowledge about the following: 1. Type of solutions. 2. Solubility and the factors affecting it. 3. Solubility in aqueous solutions for ionic and non- ionic compounds. 4. Units of solution concentrations used in qualitative analysis lab. 1  Aims of this lecture  To understand the meaning of solubility and what factors affect it.  To know the difference between ionic and non- ionic compounds.  To understand different types of concentration units. Types of solutions A solution consists of a solute and a solvent. Normally a solute is smaller in quantity than a solvent. A solution can be of homogeneous or heterogeneous nature. A homogeneous solution is composed of a matter that all exist in the same state. A heterogeneous solution is composed of a matter that exist in different states. 3 What does solubility mean?  If table salt (NaCl) and sugar are placed into two separate cups of water, water would dissolve them. This is as they are highly soluble in water. But, the solution of NaCl can conduct the electricity whereas the sugar solution can not.  NaCl is ionic compound, if is placed into a cup of water, the bonds Na-Cl would break, and ions would be surrounded by water molecules. This resulted in hydrated ions. NaCl(s) → Na+(aq) + Cl-(aq) Aq stands for aqueous and means the ions are attached by H2O molecules. Na+ cation and Cl- anion. These ions can transfer the electrical charges. This is called “electrolyte” solution. 4  Sugar in a non-ionic compound, if it is placed into a cup of water, the sugar molecules would swim in water as a bulk without dissolution. C12H22O11(s) + H2O → C12H22O11(aq) So, there is no ions to transfer the electrical charges. This is called “non-electrolyte” solution. 5 Solubility and factors affecting it. A. Solubility. Consider the reaction BaSO4(s) = Ba2+(aq) + SO42-(aq) The solubility product (Ksp) Ksp = ([Ba2+].[ SO42-])/[ BaSO4] As BaSO4 is solid, its concentration is constant as it is in its standard state, thus Ksp = [Ba2+].[ SO42-]  The solubility product, Ksp, is used for systems at equilibrium like saturated solutions, it also can be used for solid formation (precipitate). Systems that are not at equilibrium like unsaturated or supersaturated solutions, their equilibrium constant, ion product, Qsp, can be used. Both Ksp and Qsp have the same expression. 6  Solubility product is “the equilibrium constant for the reaction in which a solid salt dissolves to give its constituent ions in solution”. Consider the dissolution of mercurous chloride Hg2Cl2(s) = Hg22+ + 2 Cl- Ksp = [Hg22+][Cl-]2 = 1.2 × 10-18 If the aqueous solution is left in contact with excess solid, Hg2Cl2, the solid will dissolve until the condition [Hg22+][Cl-]2 = Ksp is satisfied. Thereafter, the amount of undissolved solid remains constant. Unless excess solid remains, there is no guarantee that [Hg22+][Cl-]2 = Ksp. If Hg22+ and Cl- are mixed together (with appropriate counterions) such that the product [Hg22+][Cl-]2 exceeds Ksp, then Hg2Cl2 will precipitate. 7  In general: the solubility product is the molar concentration of ions raised to their stoichiometric powers.  Solubility is the amount (grams) of substance that dissolves to form a saturated solution.  Molar solubility is the number of moles of solute dissolving to form a litre of saturated solution. To convert solubility to Ksp : solubility needs to be converted into molar solubility (via molar mass). molar solubility is converted into the molar concentration of ions at equilibrium (equilibrium calculation), Ksp is the product of equilibrium concentration of ions. 8  Ksp is used to find the concentration of one ion, involved in a chemical reaction, provided that the concentration of the other ion is known. Question: Calculate the solubility product if the solubility of AgCl is 1.5 mg.L-1 and the molecular weight of AgCl is 143.5 g.mol-1. Answer: The number of AgCl moles = (1.5/1000) / 143.5 = 1 × 10- 5 mole per a litre. AgCl would dissociate, in a saturated solution, AgCl(s) = Ag+(aq) + Cl-(aq) 1 mole AgCl would dissociate to give 1 mole of Ag+ and 1 mole of Cl- [= Ag+].[ Cl-] = [1 × 10-5]. [1 × 10-5] = 1 × 10-10 9 Question: What is the concentration of Hg22+ in equilibrium with 0.1 M Cl- in a solution of KCl containing excess, undissolved Hg2Cl2(s)? Answer: The chemical reaction would be Hg2Cl2(s) = Hg22+ + 2 Cl- Thus, Ksp = [Hg22+][Cl-]2 = 1.2 × 10-18 So, [Hg22+] = (Ksp / [Cl-]2) = (1.2 × 10-18/[0.1]2) [Hg22+] = 1.2 × 10-16 M 10 B. Factors affecting the solubility. There are several factors have an impact on the solubility, such as the following: 1. Common ion effect. 2. pH value. 3. Formation of complex ions. 4. Amphoterism. 5. Temperature. 6. Pressure. 7. Polarity. 8. Molecular size. 11 Common ion effect (1). Solubility is decreased when a common ion is added. This is an application of Le Châtelier’s principle: “a salt will be less soluble if one of its constituent ions is already present in the solution”. Consider the ionic solubility reaction CaSO4(s) Ca2+(aq) + SO4-2(aq) The product [Ca2+][SO4-2] is constant at equilibrium. If the concentration of Ca2+ is increased by adding Ca2+ from other source like CaCl2 then the concentration of SO4-2 must decrease. This is to keep the product [Ca2+][SO4-2] constant. The solubility of CaSO4 will decrease in the presence of dissolved CaCl2. 12 Common ion effect (2). Consider the reaction CaF2(s) Ca2+(aq) + 2F-(aq) If F- is added (from any salt like NaF), the equilibrium shifts away from the increase. Therefore, CaF2(s) is formed and precipitation occurs. If NaF is added to the system, the solubility of CaF2 decreases. 13 pH value. Again we apply Le Châtelier’s principle: CaF2(s) = Ca2+(aq) + 2F-(aq) If the F- is removed, then the equilibrium shifts towards the decrease and CaF2 dissolves. F- can be removed by adding a strong acid. As pH decreases, [H+] increases and the solubility increases. The effect of pH on the solubility is dramatic. 14 Question: Calculate the minimum pH at which Cr(OH)3 will precipitate if the solution has [Cr3+] = 0.0670 M? Ksp of Cr(OH)3 is 6.70 x 10-31 Solution: The dissociation equation is Cr(OH)3 Cr3+ + 3OH- Ksp = [Cr3+] [OH-]3 6.70 x 10-31 = (0.0670) (s)3 s = 2.1544 x 10-10 M pH = 14 - pOH = 14 - 9.667 = 4.333 15 Effect of complex ions formation. (1) Consider the formation of Ag(NH3)2+ according to the equation Ag+(aq) + 2NH3(aq) = Ag(NH3)2(aq) Ag(NH3)2+ is the complex ion. NH3 (the attached Lewis base) is called a ligand. The equilibrium constant for the reaction is called the formation constant, Kf Kf = ([Ag(NH3)2])/([Ag+].[NH3]2) 16 Effect of complex ions formation. (2) Consider the addition of ammonia to AgCl (white precipitate): AgCl(s) = Ag+(aq) + Cl-(aq) Ag+(aq) + 2 NH3(aq) = Ag(NH3)2(aq) The overall reaction is : AgCl(s) + 2 NH3(aq) = Ag(NH3)2(aq) + Cl-(aq) Effectively, the Ag+ (aq) has been removed from solution. By Le Châtelier’s principle, the forward reaction (dissolving AgCl) is favoured. 17 Amphoterism. Amphoteric oxides will dissolve in either a strong acid or a strong base. Examples: hydroxides and oxides of Al3+ , Cr3+, Zn2+ , and Sn2+. The hydroxides generally form complex ions with four hydroxide ligands attached to the metal: Al(OH)3(s) + OH-(aq) = Al(OH)4-(aq) Hydrated metal ions act as weak acids. Thus, the amphoterism is interrupted: Al(H2O)63+(aq) + OH-(aq) = Al(H2O)5(OH)2+(aq) + H2O(l) Al(H2O)5 (OH)2+(aq) + OH-(aq) = Al(H2O)4(OH)2+(aq) + H2O(l) Al(H2O)4 (OH)2+(aq) + OH-(aq) = Al(H2O)3(OH)3(s) + H2O(l) Al(H2O)3(OH)3(s) + OH-(aq) = Al(H2O)2(OH)4-(aq) + H2O(l) 18 Temperature.  Solubility increases with increasing the temperature. However, gases solubility in water decreases with increasing the temperature. Polarity.  “Like dissolves like”. 19 Units of solution concentrations used in qualitative analysis lab. The concentration of a solution is the amount of solute presents in a given amount of a solvent. The relation is proportion and a ratio. The concentration can be determined either qualitatively or quantitatively. There are several types in qualitative concentration, such as concentrated solution, diluted solution, unsaturated solution, saturated solution and supersaturated solution. 21 To differentiate between the types of solutions, one need to compare the ion product (Qs) with the solubility product (Ksp). If Qs/Ksp < 1, the product concentration is too low for equilibrium, so the net reaction shits to right. Qs/Ksp = 1, the system is at equilibrium, no net change will occur. Qs/Ksp > 1, the product concentration is too high for equilibrium, so the net reaction shits to left. 22  Unsaturated solution is when it can still dissolve more solute. Any solution that has not reached yet its limit. For example, if you have a glass of water and you pour some salt, it will dissolve immediately. In simple words it is like if it has “a room” for the solute. 23  Saturated point is when no more solute can dissolve in the solvent. Solubility, in this case, depends on the temperature. Some solutes will dissolve at high temperatures while others will do at low temperatures.  Super saturated solution is when you put excess solute in the solvent. When you put the solute in a large amount, it might not dissolve into the solvent. The solvent will not be able to absorb the whole solute, because there is too much. This case is called “supersaturated solution”. 24 25 Units of solution concentrations used in qualitative analysis lab. The amount of a solute cab be presented by the following units: milli (m) = 10-3 X micro () = 10-6 X nano (n) = 10-9 X pico (p) = 10-12 X femto (f) = 10-15 X atto (a) = 10-18 X X can stand for the mass units, for grams or moles. 26 The concentration of a solution is the amount of solute presents in a given amount of a solvent. It can be expressed by many units like the following. If the solute amount is represented by grams dissolved in a solution. Then, the concentration can be presented by : Part per million (ppm) = mg.L-1 , Part per billion (ppb) = g.L-1 Part per trillion (ppt) = ng.L-1 If the solute amount is presented by its number of moles, then we can use Molarity (M) which represents the moles of a solute dissolved in a liter of a solvent. M = no. of moles of a solute / solution volume in liter X can stand for the mass units, for grams or moles. 27 Weight percent = (mass of solute / mass of solution or mixture) × 100 Like, 20 g of salt in 50 g of water = (20/70) × 100 = 28.57% Volume percent = (volume of solute / volume of total solution or mixture) × 100 28 References list The contents of this lecture is taken from 1. A textbook of macro and semimicro qualitative inorganic analysis. Journal of chemical education. H. Rogers.1956. 2. Quantitative Chemical Analysis (seventh edition). W.H. Freeman and Company. Daniel C. Harris. 2007 29  Third lecture: Solutions 1. Chemical equilibrium. 2. Kinetic chemistry. 3. Velocity and rate of chemical reactions. 1. Chemical equilibrium. What is chemical equilibrium? Equilibrium constant, Kc Calculating Equilibrium Constant, Kc Relationship between Kc and Kp Le Chatelier’s Principle Factors effecting the equilibrium. What is chemical equilibrium? The state where the concentrations of all reactants and products remain constant with time. Consider the following reactions: CaCO3(s) + CO2(aq) + H2O(l) Ca2+(aq) + 2HCO3-(aq)..(A) and Ca2+(aq) + 2HCO3-(aq) CaCO3(s) + CO2(aq) + H2O(l)..(B) Both reactions are reverse to each other. At equilibrium, the two opposing reactions occur at the same rate. Concentrations of chemical species do not change once equilibrium is established. Equilibrium constant, Kc Consider the following equilibrium system: aA + bB cC + dD Kc value can be calculated by using the concentrations of the reactants and products that exist at equilibrium. [C]c [D]d Kc = [A]a [B]d Consider the reaction of producing the ammonia, N2(g) + 3H2(g) 2NH3(g) [NH3]2 Kc = [N2] [H2]3 Calculating Equilibrium Constant Example: 1.000 mole of H2 gas and 1.000 mole of I2 vapor are introduced into a 5.00-liter sealed flask. The mixture is heated to a certain temperature and the following reaction occurs until equilibrium is established. H2(g) + I2(g) 2HI(g) At equilibrium, the mixture is found to contain 1.580 mole of HI. (a) What are the concentrations of H2, I2 and HI at equilibrium? (b) Calculate the equilibrium constant Kc? Calculating Equilibrium Constant ———————————————————————————— H2(g) + I2(g) ⇄ 2 HI(g) ———————————————————————————— Initial [ ], M: 0.200 0.200 0.000 Change in [ ], M: -0.158 -0.158 + 0.316 Equilibrium [ ], M 0.042 0.042 0.316 ———————————————————————————— [HI]2 (0.316) 2 Kc = = = 57 [H 2 ][I 2 ] (0.042) 2 Relationship between Kc and Kp In general, for reactions involving gases such that, aA + bB cC + Dd where A, B, C, and D are all gases, and a, b, c, and d are their respective coefficients, Δn Kp = Kc(RT) and Δn = (c + d) – (a + b) (In heterogeneous systems, only the coefficients of the gaseous species are counted.) Relationship between Kc and Kp Example: 2SO2(g) + O2(g) 2SO3(g) [SO]2 (PSO3 ) 2 Kc = and Kp = [SO 2 ]2 [O 2 ] (PSO2 ) 2 (PO2 ) Assuming ideal behavior, where PV = nRT and P = (n/V)RT = [M]RT and PSO3 = [SO3]RT; PSO2 = [SO2]RT; PO2 = [O2]RT [SO 3 ]2 (RT) 2 [SO 3 ]2 Kp  2 2  2 (RT) -1  K c (RT) -1 [SO 2 ] (RT) [O 2 ]( RT) [SO 2 ] [O 2 ] Relationship between Kc and Kp For other reactions: 1. 2NO2(g) N2O4(g); Kp = Kc(RT)-1 2. H2(g) + I2(g) 2 HI(g); Kp = Kc 3. N2(g) + 3H2(g) 2 NH3(g); Kp = Kc(RT)-2 Le Chatelier’s Principle The Le Châtelier's principle states that: when stress is applied to a system at chemical equilibrium, the equilibrium will shift in a direction that tends to relieve or counteract that stress. Factors that influence equilibrium: Concentration. Temperature. Partial pressure (for gaseous).  2. Chemical Kinetics Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B D[A] D[A] = change in concentration of A over rate = - Dt time period Dt D[B] D[B] = change in concentration of B over rate = Dt time period Dt Because [A] decreases with time, D[A] is negative. Chemical Kinetics A B time D[A] rate = - Dt D[B] rate = Dt  Factors that Affect Reaction Rate 1. Temperature Collision Theory: When two chemicals react, their molecules have to collide with each other with sufficient energy for the reaction to take place. Kinetic Theory: Increasing temperature means the molecules move faster. 2. Concentrations of reactants More reactants mean more collisions if enough energy is present 3. Catalysts Speed up reactions by lowering activation energy 4. Surface area of a solid reactant Bread and Butter theory: more area for reactants to be in contact 5. Pressure of gaseous reactants or products Increased number of collisions  Fourth lecture: Solutions 1. Acid-base equilibrium. 2. Dissociation of water. 3. pH and neutralization indicators.  Acid-base equilibrium Definitions Arrehenius Acids – produce H+ Bases - produce OH- only in water Bronsted-Lowry Acids – donate H+ any solvent Bases – accept H+ Lewis Acids – accept e- pair used in organic chemistry, wider range of substances Bases – donate e- pair  Acid-base equilibrium The hydrogen ion in aqueous solution H+ + H2O  H3O+ (hydronium ion) Examples Arrhenius HCl NaOH Bronsted-Lowry HCl HCN NH3 Lewis BF3 :NH3  Acid-base equilibrium Strong and Weak Acids Strong acids ionize completely in aqueous solution: HCl(aq) + H2O H3O+(aq) + Cl-(aq) H2SO4(aq) + H2O H3O+(aq) + HSO4-(aq) Weak acids ionize only partially in aqueous solution: HF(aq) + H2O H3O+(aq) + F-(aq) HClO(aq) + H2O H3O+(aq) + ClO-(aq)  Acid-base equilibrium Let’s examine the behavior of an acid, HA, in aqueous solution. HA What happens to the HA molecules in solution?  Acid-base equilibrium 100% dissociation of HA HA H+ Strong Acid A- Would the solution be conductive?  Acid-base equilibrium Partial dissociation of HA HA H+ Weak Acid A- Would the solution be conductive?  Acid-base equilibrium HA  H+ + A- HA H+ Weak Acid A- At any one time, only a fraction of the molecules are dissociated.  Acid-base equilibrium Strong and Weak Acids/Bases Strong acids/bases – 100% dissociation into ions HCl NaOH HNO3 KOH H2SO4 Weak acids/bases – partial dissociation, both ions and molecules CH3COOH NH3  Dissociation of water When water dissociates, one of the hydrogen nuclei leaves its electron behind with the oxygen atom to become a hydrogen ion, while the oxygen and other hydrogen atoms become a hydroxide ion. Since the hydrogen ion has no electron to neutralize the positive charge on its proton, it has a full unit of positive charge and is symbolized as H+. The hydroxide ion retains the electron left behind and thus has a full unit of negative charge, symbolized by OH-. The hydrogen ion (proton) does not wander long by itself before it attaches to the oxygen atom of a second un-ionized water molecule to form a hydronium ion (H3O +).  Dissociation of water Pure Water is Neutral H2O + H2O H3O+ + OH- hydronium hydroxide ion ion 1 x 10-7 M 1 x 10-7 M H 3O + OH- Ion Product of Water Kw [ ] = Molar concentration Kw = [ H3O+ ] [ OH- ] = [ 1 x 10-7 ][ 1 x 10-7 ] = 1 x 10-14 What is the pH scale? The pH scale measures how acidic or basic a solution is. The pH scale The pH scale is the concentration of hydrogen ions in a given substance. pH   log H   Identifying Acids and Bases Acids have a pH from 0-7 Lower pH value indicates a stronger acid. Bases have a pH from 7-14 Higher pH value indicates a stronger base. Characteristics Of Acids & Base Acids can be characterized by: 1. A sour taste. 2. It turns blue litmus paper red 3. It tastes sour. Try drinking lemon juice (citric acid) A Base is characterized by: 1. A bitter taste. (Milk of Magnesia) 2. It feels slippery. (Soapy Water) 3. It turns Red Litmus Blue.  ACID-BASE INDICATORS Litmus Litmus is a weak acid. It has a seriously complicated molecule which we will simplify to HLit. The "H" is the proton which can be given away to something else. The "Lit" is the rest of the weak acid molecule. There will be an equilibrium established when this acid dissolves in water. Taking the simplified version of this equilibrium: The un-ionised litmus is red, whereas the ion is blue. Now use Le Chatelier's Principle to work out what would happen if you added hydroxide ions or some more hydrogen ions to this equilibrium.  ACID-BASE INDICATORS Adding hydroxide ions: Adding hydrogen ions:  ACID-BASE INDICATORS Methyl orange Methyl orange is one of the indicators commonly used in titrations. In an alkaline solution, methyl orange is yellow and the structure is: In fact, the hydrogen ion attaches to one of the nitrogens in the nitrogen- nitrogen double bond to give a structure which might be drawn like this:  ACID-BASE INDICATORS Phenolphthalein Phenolphthalein is another commonly used indicator for titrations, and is another weak acid. In this case, the weak acid is colourless and its ion is bright pink. Adding extra hydrogen ions shifts the position of equilibrium to the left, and turns the indicator colourless. Adding hydroxide ions removes the hydrogen ions from the equilibrium which tips to the right to replace them - turning the indicator pink. The pH range of indicators indicator pKind pH range litmus 6.5 5-8 methyl 3.7 3.1 – 4.4 orange phenolphth 9.3 8.3 – 10.0 alein  ACID-BASE INDICATORS Choosing indicators for titrations 1. Strong acid v strong base 2. Strong acid v weak base  ACID-BASE INDICATORS Choosing indicators for titrations 3. Weak acid v strong base 4. Weak acid v weak base References: Harris, Daniel C. Quantitative chemical analysis. W.H. Freeman and Company, 2010. Gary D. Christian. Analytical Chemistry. John Wiley and Sons, inc, 2004.  Fifth lecture: Solutions CONTENTS Salt Hydrolysis Types of salts General Rules Question Salt Hydrolysis Salt Hydrolysis: reaction of the anions or cations in a salt with water in order to form an acidic or basic solution. Types of salts There are 3 types of salts (neutral, acidic or basic). Therefore: when salts dissolve in water, the pH of the resulted solution can be neutral, acidic or basic. Salt + water = ? 1. Neutral solution (pH = 7.0) 2. Basic solution (pH > 7.0) 3. Acidic solution (pH < 7.0) Things to consider: o Metal hydroxides of group I and II except Be are strong bases o All other metal hydroxides are weak bases 1. Neutral solution Conjugate of strong Acid + conjugate of strong Base neutral solution Neutralization equation can tell you the salt type. (e.g.) consider Sodium Chloride, NaCl: HCl + NaOH → NaCl + H2O So, NaCl is a salt resulted from the reaction of strong acid with a strong base, hence the salt is of neutral nature. Note: If the acid and the base are both strong, the resulted salt is neutral salt. Neutral salt dissociates in water to give the positive and negative radicals, but the ions do not hydrolyze or react with water further. 2. Basic solution Conj. of weak Acid + conj. of strong Base basic solution (e.g.) Sodium cyanide (NaCN) HCN + NaOH → NaCN + H2O This salt (NaCN) can be the result of the reaction between a weak acid (HCN) and a strong base (NaOH). Thus, the NaCN salt is of basic nature. NaCN dissociates in water: NaCN → Na+ + CN- Because Na+ came from a strong parent (strong base), it does not hydrolyze. (neutral), but as CN- came from a weak parent (weak acid), it will react with water (hydrolyzes) as a Bronsted base CN- + H2O HCN + OH- 3. Acidic solution Conj. of weak Base + conj. of strong Acid acidic solution (e.g.) Ammonium Chloride (NH4Cl) HCl + NH3 → NH4Cl This salt can be the result of the reaction between a strong acid (HCl) and a weak base (NH3), thus, the NH4Cl salt is of acidic nature. NH4Cl dissociates in water: NH4Cl → NH4+ + Cl- Because Cl- came from a strong parent (strong acid), it does not hydrolyze, but as NH4+ came from a weak parent (weak base), it will react with water (hydrolyzes) as a Bronsted acid NH4+ + H2O H3O+ + NH3 General Rules  When the negative ion is from a weak acid then the salt is basic by hydrolysis.  When the positive ion is from a weak base then the salt is acidic by hydrolysis.  If the salt formed from a strong acid and strong base then it is neutral.  if the salt formed from a weak acid and weak base the its hydrolysis is determined by the relative Ka and Kb values. Question Determine if the following aqueous solutions will be acidic, basic or neutral:  NH4NO3  KCl  NaHCO3 Na2SO3 sixth lecture: Buffer solutions used in qualitative analysis. CONTENTS What is a buffer solution? Buffer capacity Buffer range Henderson-Hasselbalch Equation Calculating the pH of a buffer solution Buffer solutions - ideal concentration What is Buffer ???  A buffer is a solution (or a substance) that has the ability to maintain pH and bring it back to its optimal value by addition or removal of hydrogen ions.  A buffer solution consists of a mixture of a weak acid [HX] and its conjugated base [X-]. HX(aq) H+(aq) + X-(aq) [H+] = Ka [HX] / [X-]  The buffered solution resists changes in pH value when a small amounts of acids (H+) or bases (OH-) are added or when dilution occurs.  If bases (OH-) or acids (H+) is added to the system above in the presence of buffer solution, the OH- or H+ reacts with HX to produce water and X-. Therefore [HX]/[X-] remains almost unchanged, so pH value is unchanged. Buffer Capacity  The effectiveness of any buffer is described in term of its buffer capacity Buffer capacity is the amount of acid or base neutralized by the buffer before there is a significant change in pH.  Buffer capacity depends on the composition of the buffer.  The greater the amounts of conjugate acid-base pair, the greater the buffer capacity (effectiveness). Buffer Range  The buffer capacity of a buffer is maximum when acid to salt or base to salt ratio is equal 1.  All buffer solutions remain effective over a small ph range, and this pH- range is characteristic of the buffer and is termed as buffer range.  Buffer range in pH unit:  Acid buffer: pKa – 1 to pKa + 1  Basic buffer: (pKw - pKb) – 1 to (pKw - pKb) + 1 Henderson-Hasselbalch Equation  The central equation for buffers is the henderson-Hasselbalch equation, which is simply rearranged form of Ka equilibrium. Ka = [H+] [X-] / [HX] log Ka = log [H+] [X-] / [HX] log Ka = log [H+] + log [X-] / [HX] - log [H+] = - log Ka + log [X-] / [HX] pH pKa Therefore pH = pKa + log ([X-] / [HX])  The pH of the buffer depends on Ka.  where Ka is the acid dissociation constant of the weak acid Calculating the pH of an acidic buffer solution Calculate the pH of a buffer whose [HX] is 0.1 mol dm-3 and [X¯] of 0.1 mol dm-3? Ka = [H+(aq)] [X¯(aq)] [HX(aq)] re-arrange [H+(aq)] = [HX(aq)] x Ka [X¯(aq)] from information given [X¯] = 0.1 mol dm-3 [HX] = 0.1 mol dm-3 If the Ka of the weak acid HA is 2 x 10-4 mol dm-3. [H+(aq)] = 0.1 x 2 x 10-4 = 2 x 10-4 mol dm-3 0.1 pH = - log10 [H+(aq)] = 3.699 Calculating the pH of an acidic buffer solution Calculate the pH of the solution formed when 500cm3 of 0.1 mol dm-3 of weak acid HX is mixed with 500cm3 of a 0.2 mol dm-3 solution of its salt NaX. Ka = 4 x 10-5 mol dm-3? Ka = [H+(aq)] [X¯(aq)] [HX(aq)] re-arrange [H+(aq)] = [HX(aq)] Ka [X¯(aq)] The solutions have been mixed; the volume is now 1 dm3 therefore [HX] = 0.05 mol dm-3 and [X¯] = 0.10 mol dm-3 Substituting [H+(aq)] = 0.05 x 4 x 10-5 = 2 x 10-5 mol dm-3 0.1 pH = - log10 [H+(aq)] = 4.699 Buffer solution – uses  Biological Uses In biological systems (saliva, stomach, and blood) it is essential that the pH stays ‘constant’ in order for any processes to work properly. (e.g.)  If the pH of blood varies by 0.5 it can lead to unconsciousness and coma.  Most enzymes work best at particular pH values.  Agriculture Uses The pH of the soil is very important for having proper crop yield  other Uses Many household and cosmetic products need to control their pH values. (e.g.)  Shampoo Buffer solutions counteract the alkalinity of the soap and prevent irritation  Baby lotion Buffer solutions maintain a pH of about 6 to prevent bacteria multiplying Others Washing powder, eye drops, fizzy lemonade Fe+3 Sn+2