Chapter 5 Reactions in Aqueous Solution PDF

Chapter 5 Reactions in Aqueous Solution Prof. Nasser Y. Mostafa Chemistry Department – Faculty of Science Suez Canal University  Ionization of Water Water is a weak electrolyte as it partially ionized according to the following equation: - H2O + H2O ⇄ H3O+(aq) + OH-(aq)...

Chapter 5 Reactions in Aqueous Solution Prof. Nasser Y. Mostafa Chemistry Department – Faculty of Science Suez Canal University  Ionization of Water Water is a weak electrolyte as it partially ionized according to the following equation: - H2O + H2O ⇄ H3O+(aq) + OH-(aq) This is called the auto-ionization Kw = [H3O+] [OH-] = 1.0 x 10-14 at 25oC. (Kw is called the ion product constant for water) Aqueous solutions contain both H3O+ and OH- such that [H3O+] x [OH-] = 1.0 x 10-14 at 25oC. ▪ If [H3O+] increases (> 1.0 x 10-7 M ), [OH-] will decreases (< 1.0 x 10-7 M), and vice versa. Thus, a solution with: [H3O+] = [OH-] = 1.0 x 10-7 M, ➔ the solution is neutral, (such as in pure water). [H3O+] > 1.0 x 10-7 M, or [H+] > [OH-], ➔ the solution is acidic. 2 [H3O ] < 1.0 x 10 M, or [H ] < [OH ], ➔ the solution is basic. + -7 + - pH Scale 3 Below, you can see where several common items range on the pH scale. 4  The pH Scale by Sorensen (1909) pH is a scale that measures the acidity of an aqueous solution where [H+] is very small (i.e. in diluted solutions). pH = - log [H3O+] ▪ If [H3O+] = 1.0 x 10-2 M, pH = -log (1.0 x 10-2) = -(-2.00) = 2.00 (➔ acidic) We can also express basicity using the log scale for [OH-], such that: pOH = -log [OH-] - Since, at 25oC, Kw = [H3O+][OH-] = 1.0 x 10-14 pKw = -log(Kw) = -log[H3O+] + (-log[OH-]) = -log(1.0 x 10-14) = - (-14.00) = 14.00 pKw = pH + pOH = 14.00; and pOH = 14.00 – pH Thus, in aqueous solutions, pH = 2 ➔ pOH = 12, and pOH = 2 ➔ pH = 12. In pure water or neutral solutions, [H3O+] = [OH-] = 1.0 x 10-7 M, and pH = pOH = 7.00; 5 [H+] > 1 x 10-7 M, ➔ pH < 7.0; the solution is acidic.  H+ Conc. pH Value Pure Water Example EASY to remember ? 0 10 000 000 battery acid 1 1 000 000 gastric acid 2 100 000 lemon juice pH = - log [H3O+] 3 10 000 orange juice, soda 4 1 000 tomato juice pOH = -log [OH-] 5 100 bananas 6 10 urine, milk pKw = pH + pOH = 14 7 1 pure water 8 0.1 sea water, eggs 9 0.01 baking soda 10 0.001 Great Salt 11 0.000 1 ammonia solution 12 0.000 01 soapy water 13 0.000 001 bleach, cleaner 14 0.000 000 1 drain cleaner Living Systems Molecules that make up or are produced by living organisms usually function within a narrow pH range (near neutral) and a narrow temperature range (body temperature). pH influences the structure and the function of many enzymes (protein catalysts) in living systems. Many of these enzymes have narrow ranges of pH activity. Cellular pH is so important that death may occur within hours if a person becomes acidotic (having increased acidity in the blood). As one can see pH is critical to life, biochemistry, and important chemical reactions. Common examples of how pH plays a very important role in our daily lives are given below: Whenever we get a heartburn, more acid build up in the stomach and causes pain. We needs to take antacid tablets (a base) to neutralize excess acid in the stomach. The pH of blood is slightly basic. A fluctuation in the pH of the blood can cause in serious harm to vital organs in the body. Certain diseases are diagnosed only by checking the pH of blood and urine. Certain crops thrive better at certain pH range. Enzymes activate at a certain pH in our body. 7  Buffer Solution A buffer solution is that one which resists the change in pH value when SMALL AMOUNTS of acid or base is added to it. It is a mixture of a weak acid and its salt or a weak base and its salt. Buffer solutions are used as a means of keeping pH at a nearly constant value in a wide variety of chemical applications. Buffers can be divided in two main groups: 1) An acidic buffer solution is simply one which has a pH less than 7. Acidic buffer solutions are commonly made from a weak acid and one of its salts - often a sodium salt. e. g. (CH3COOH + CH3COONa). 2) An alkaline buffer solution has a pH greater than 7. Alkaline buffer solutions are commonly made from a weak base and one of its salts. A frequently used example is a mixture of ammonia 8 solution and ammonium chloride solution (NH4OH + NH4Cl). How do buffer solutions work? Acidic buffer solutions We'll take a mixture of acetic acid and sodium acetate as typical. Acetic acid is a weak acid, and the position of this equilibrium will be well to the left: CH3COOH (aq)  CH3COO-(aq) + H+(aq) CH3COONa → CH3COO- + Na+ Adding sodium acetate to this adds lots of extra acetate ions. According to Le Chatelier 's principle, that will shift the position of the equilibrium even further to the left. Adding an acid to this buffer solution Hydrogen ions combine with the acetate ions to make acetic acid. Although the reaction is reversible, since the acetic acid is a weak acid, most of the new hydrogen ions are removed in this way. CH3COO-(aq) + H+(aq)  CH3COOH (aq) 9  Adding an alkali to this buffer solution This time the situation is more complicated because there are two processes which can remove hydroxide ions. Removal by reacting with acetic acid The most likely acidic substance which a hydroxide ion is going to react with is an acetic acid molecule. They will react to form acetate ions and water. CH3COOH(aq) + OH-  CH3COO- (aq) + H2O(aq) Because most of the new hydroxide ions are removed, the pH doesn't increase very much. Removal of the hydroxide ions by reacting with hydrogen ions Remember that there are some hydrogen ions present from the ionization of the acetic acid. CH3COOH(aq) ⇄ CH3COO- (aq) + H+(aq) Hydroxide ions can combine with these to make water. 10 As soon as this happens, the equilibrium shifts to replace them. 11  ❑ Calculating the pH of Strong Acid and Strong Base Solutions Strong acids are assumed to ionize completely in aqueous solution. For monoprotic acids (that is, acids with a single ionized hydrogen atom) such as HCl and HNO3, the concentration of hydronium ion in solution is the same as the molar concentration of the acid. That is: [H3O+] = [Acid] ▪ For example, in 0.10 M HCl (aq), [H3O+] = 0.10 M, and pH = - log(0.10) = 1. Also, in 0.01 M HNO3 (aq), [H3O+] = 0.01 M, and pH = - log(0.01) = 2 A strong base such as NaOH has [OH-] equals to the molar concentration of dissolved NaOH. That is, a solution of 0.10 M NaOH(aq) has [OH-] = 0.10 M and 12 pOH = - log[OH-] = - log(0.10) = 1; pH = 13 Exercises: 1. Calculate the pH of the following solutions: (a) 0.005 M HCl (b) 0.005 M NaOH (a) pH = - log (0.005) = 2.30 (b) pOH = - log (0.005) = 2.30 pH = 14 – 2.3 = 11.7 2. What is [H3O+] and [OH-], respectively, in a solution where: (a) pH = 3.60; (b) pOH = 4.40 (a) pH = 3.60 = - log [H3O+] [H3O+] = shift log (- 3.6) = 2.5 x 10-4 moles / L [OH-] = 1 x 10-14 / 2.5 x 10-4 = 4 x 10-11 moles / L (b) pOH = 4.40 = - log [OH-] [OH-] = shift log (- 4.40) = 3.9 x 10-5 moles / L [H3O+] = 2.5 x 10-10 moles / L 13  Calculating the pH of Weak Acid and Weak base Solutions Weak acids do not ionize completely. At equilibrium, [H+] is much less than the concentration of the acid. The concentration of H+ in a weak acid solution depends on the initial acid concentration and the Ka of the acid. To determine [H+] of a weak acid we can set up the “ICE” table as follows: ▪ Consider a solution of 0.10 M acetic acid and its ionization products: Concentration: CH3COOH(aq) ⇄ H+(aq) + CH3COO- (aq) Initial [ ], M: 0.10 0.00 0.00 Change,[ ], M: -x +x +x Equilibrium [ ], M : (0.10 - x) x x ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ 14 The acid ionization constant, Ka , is given by the expression: Ka = = x2 /(0.10 – x) = 1.8 x 10-5 ➔ x2 = 1.8 x 10-6, and x = (1.8 x 10-6 )0.5 = 1.3 x 10-3 M Note that x = [H3O+] = 1.3 x 10-3 M ; ➔ pH = - log (1.3 x 10-3) = 2.89 In general, for weak acids: [H+] = 𝐀𝐜𝐢𝐝  𝐊𝐚 Ammonia (weak base) is very soluble in water and ionizes as follows: NH3(aq) + H2O ⇄ NH4+(aq) + OH-(aq) The base dissociation constant, Kb, is given by the expression: Kb = = 1.8 x 10-5 ▪ The concentration of OH- in a weak base such as NH3 (aq), depends on its Kb value and the initial concentration of the base. For example, to determine [OH-] and pH of 0.10 M NH3(aq), we set up the following “ICE” table: 15 Concentration: NH3(aq) + H2O ⇄ NH4+(aq) + OH-(aq) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ ⎯⎯⎯⎯⎯⎯⎯⎯ Initial [ ], M: 0.10 0.00 0.00 Change,[ ], M: -x +x +x Equilibrium [ ], M: (0.10 - x) x x ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Kb = = x2 / (0.10 – x) = 1.8 x 10-5 ➔x2 = 1.8 x 10-6; x = (1.8 x 10-6)0.5 = 1.3 x 10-3 M Where, x = [OH-] = 1.3 x 10-3 M; pOH = - log (1.3 x 10-3) = 2.87; ➔ pH = 11.13 𝐁𝐚𝐬𝐞  𝐊𝐛 In general, for weak bases: [OH-] = 16 ❑ Salt Hydrolysis Neutralization→ Acid + Base  Salt + Water Salt Hydrolysis ▪ The acidic or basic nature of a salt solution depends on whether it is a product of: a strong acid-strong base reaction; a weak acid- strong base reaction; a strong acid-weak base reaction, or a weak acid-weak base reaction. 1) Salts of Strong Acid-Strong Base Reactions: such as NaCl, NaNO3, KBr, etc. ▪ Salts of this type form neutral solutions (pH = 7), because neither the cation ion nor the anion reacts with water and offset the equilibrium concentrations of H3O+ and OH- in the solution. 2) Salts of Weak Acid-Strong Base Reactions: such as NaF, NaNO2, CH3COONa, etc. ▪ Salts that are products of reactions between weak acids and strong bases form basic solutions (pH  7) when dissolved in water. The anions 17 of such salts react with water that increases [OH-]. ▪ When sodium acetate is dissolved in water, it dissociates into sodium and acetate ions: CH3COONa (aq) → Na+(aq) + CH3COO-(aq) The acetate ion reacts with water and the following equilibrium is established: CH3COO-(aq) + H2O ⇄ CH3COOH(aq) + OH-(aq) 3) Salts of Strong Acid-Weak Base Reactions: such as NH4Cl, NH4NO3. ▪ Aqueous solutions of salts that are products of strong acidic- weak base reactions are acidic (pH  7). The cations react with water and increases [H3O+] in solutions. For example, NH4Cl is produced when HCl (strong acid) reacts with NH3 (weak base). HCl(aq) + NH3(aq) → NH4Cl(aq) → NH4+(aq) + Cl-(aq) ▪ In aqueous solution, NH4+ establishes the following equilibrium that increases [H3O+], thus creates an acidic solution: 18 NH4+(aq) + H2O ⇄ H3O+(aq) + NH3(aq) 4) Salts of Weak Acid-Weak Base Reactions: such as CH3COONH4, NH4CN, NH4NO2, etc. ▪ Solutions of salts that are products of weak acid-weak base reactions can be neutral, acidic, or basic, depending on the relative magnitude of the Ka of the weak acid and the Kb of the weak base. If Ka  Kb, the salt will form approximately a neutral solution. If Ka > Kb, the salt solution will be acidic. If Ka < Kb, the salt solution will be basic. 19 ❑ Solubility Solubility of a substance is the maximum amount of substance that can be dissolved in a specified amount of solvent at a specified temperature. i.e., the amount of substance required to get a saturated solution. e.g., The solubility of NH4Cl in water is 41 g / 100 g of water at 20oC. e.g., The solubility of NaOH in water is 109 g / 100 g of water at 20oC. e.g., The solubility of NaOH in water is 129 g / 100 g of water at 40oC. ❖ Generally, solubility depends on: 1. Nature of solute and solvent 2. Temperature 3. Pressure. ❑ Soluions of Gases in Liquids ▪ Most of the gases are soluble in liquids such as water and form REAL SOLUTIONS, e. g., HCl, NH3, Cl2, O2 & CO2 gases, all soluble in water. ▪ The solubility of a gas in a liquid depends on: 1. Nature of solute and solvent 2. Temperature 3. Pressure. 1. The nature of solute and solvent ▪ In general, no two gases dissolve with the same amount in the same liquid at the same conditions. Also, there is no gas dissolves in two different liquids with the same degree. ▪ e.g., NH3 & HCl gases dissolve well in water, however, O2 dissolves only to a small extent. CO2 more soluble in benzene than in water. 2. Effect of temperature ▪ Solubility of gases in liquids generally decreases with increasing temperature because the kinetic energy of gas molecules increases and therefore, the ability of gas molecules to escape from the solution increases. ▪ We can get ride of AIR from water by BOILING. Also, we can collect N2O gas over HOT WATER although it is soluble well in COLD WATER. 3. Effect of pressure (Henry’s law) ▪ Henry’s law states that: “At a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas over the liquid” “The pressure of the gas (P) is proportional to amount of a gas (x) in the solution” and is expressed as: P = KH x Where, KH is the Henry’s law constant. ▪ Henry ,s law failed to account for the solubility of HCl & NH3 gases in water because theses gases react chemically with water as shown below: HCl + H2O → H3O+ + Cl- NH3 + H2O ⇄ NH4+ + OH- ❑Henry ,s law applications Henry’s law in carbonated soft drinks – When soft drinks bottle is opened some of the gas escapes giving a specific pop. This is due to the lower pressure above the liquid and carbon dioxide comes out as bubbles. Henry’s law in SCUPA diving – A knowledge of Henry ,s law can protect divers from a fatal condition known as “THE BENDS” (Nitrogen bubbles in body fluids). ❑ The Solubility Product Constant, Ksp AgCl(s) ⇄ AgCl(aq) ⇄ Ag+(aq) + Cl-(aq) − + 𝑪𝒍 [𝑨𝒈 ] K= K [AgCl] = [Ag+] [Cl-] [𝑨𝒈𝑪𝒍] Ksp = [Ag+] [Cl-] = 1 x 10-10 at 25 oC - In any solution saturated with a sparingly soluble ionic salt, the product of the concentrations of its ions is a constant value at a given temperature. This constant is known as solubility product constant, Ksp 26 ❑ How to calculate the Ksp value? ▪ If the dissociation of a salt gives 2 ions Agl(s) ⇄ Ag+(aq) + l-(aq) Ksp = [Ag+] [l-] CaCO3(s) ⇄ Ca2+(aq) + CO32-(aq) Ksp = [Ca2+] [CO32-] ▪ If the dissociation of a salt gives more than 2 ions Ag3PO4(s) ⇄ 3Ag+(aq) + PO43-(aq) Ksp = [Ag+]3 [PO43-] Fe(OH)3(s) ⇄ Fe3+(aq) + 3OH-(aq) Ksp = [Fe3+] [OH-]3 Problem 1 The solubility of AgCl in water at 25 oC is 1.3 x 10-5 moles/L. Calculate the solubility product constant for this compound? 27 Solution AgCl(s) ⇄ Ag+(aq) + Cl-(aq) Solid 1.3 x 10-5 M 1.3 x 10-5 M Ksp = [Ag+] [Cl-] = (1.3 x 10-5 ) ( 1.3 x 10-5 ) = 1.69 x 10-10 Problem 2 The solubility of barium fluoride, BaF2, is 3.15 x 10-3 M/ L at 25 °C. Calculate the solubility product, Ksp, of the compound. Solution: BaF2 (s) ⇄ Ba2+(aq) + 2F -(aq) Solid 3.15 x 10-3 M 6.30 x 10-3 M Ksp = [Ba2+] [F-]2 total = (3.15 x 10-3 ) x (6.30 x 10-3) 2 = 28 1.25 x 10-7 Problem 3 The solubility product constant of ferric hydroxide in water at 25 oC is 1.2 x 10-36. Calculate the solubility of this compound in water at 25 oC ? Solution: Fe(OH)3(s) ⇄ Fe3+(aq) + 3OH-(aq) Solid S 3S Ksp = [Fe3+] [OH-]3total 1.2 x 10-36 = (S) (3S)3 = 27 S4 S4 = 4.4 x 10-38 29 S (Solubility) = 4.6 x 10-10 mole / L ❑ Predicting Precipitate Formation ▪ Every precipitate has a Ksp value. ▪ If on mixing solutions containing the two constituent ions, the Ksp expression is exceeded, precipitation will take place until the product of the ion concentrations equals the Ksp value. If the Ksp expression is not exceeded, no precipitation will take place. [Cl-] [Ag+] 10-8 mole/ L 10-8 mole/ L 10-7 mole/ L 10-7 mole/ L 10-6 mole/ L 10-6 mole/ L 10-5 mole/ L 10-5 mole/ L 10-4 mole/ L 10-4 mole/ L 30