Chemistry 40S Acid-Base Equilibrium (This unit has been adapted from https://bblearn.merlin.mb.ca) Name: 1
Lesson 1: Defining Acids and Bases Goals: Outline the historical development of acid base theories. Identify conjugate acid-base pairs. Define amphoteric substances. Write balanced acid/base chemical equations. Early Theories of Acids and Bases Antoinne Lavoisier 2
Sir Humphry Davy Justus Liebig Svante Arrhenius Svante Arrhenius was a chemist studying the conductivity of solutions. At that time, very little was known about the structure of atoms and molecules. He proposed that electrolytes break up into charged particles in water. Scientists of the time did not accept Arrhenius' theory. They believed that atoms and molecules were indestructible and could not break apart. In 1887, Arrhenius discovered that all acidic and basic solutions he tested were electrolytes. He determined that acids and bases must ionize or dissociate in water. According to Arrhenius, an acid is defined as a substance, when dissolved in water, which releases hydrogen ions or increases the hydrogen ion concentration. Example 1: Hydrochloric acid: HCl (aq) H + (aq) + Cl (aq) From a particle view, HCl forms ions in water: 3
Example 2: Acetic acid (vinegar): HC2H3O2 (aq) H + (aq) + C2H3O2 (aq) According to Arrhenius, a base is defined as a substance, when dissolved in water, releases hydroxide ions. Example 3: Sodium hydroxide: NaOH (s) Na + (aq) + OH (aq) From a particle level, NaOH dissociates like an ionic compound. Example 4: Ammonia is a little troublesome to explain with Arrhenius' theory. It must first be shown to react with water then dissociate. NH3 (g) + H2O (l) NH4OH (aq) NH4 + (aq) + OH (aq) Until Arrhenius there was no satisfactory explanation for why acids and bases neutralize each other. He proposed that hydrogen ions produced by acids react with hydroxide ions released by bases to form water. The Arrhenius definition says that acids and bases can only occur in water solutions. However, we notice that hydrogen chloride gas and ammonia gas will react together. That is, as a gas hydrogen chloride is an acid and ammonia is a base. The Arrhenius definition does not account for this. There are many substances which are acidic or basic but do not have a hydrogen ion or a hydroxide ion. For example: Baking soda, sodium bicarbonate NaHCO3, in water turns litmus blue, but has no apparent hydroxide ion. Metal ions, such as iron (III) and aluminum, turn litmus red, but have no hydrogen ions. The Arrhenius definition does not account for the acidity and basicity of the above examples. However, Arrhenius theory was important in establishing the concept of dissociation and explaining the process of neutralization. 4
Bronsted-Lowry Theory In 1923, Johannes Brønsted and Thomas Lowry independently developed a more general definition of acids and bases. This definition accounted for the acidity of metal ions AND the basicity, alkalinity, of baking soda. In the example below, ammonia accepts a proton from water, making ammonia a base and water the acid. The result is the ammonium ion and the hydroxide ion. 5
Bases have a strong desire for protons. So much so that, if they are strong enough, they will steal the proton from an acid and combine it with itself. The stronger the bond between the hydrogen and the rest of the molecule, the more difficult it is for the base to take the proton. Another example of an amphoteric substance is the hydrogen carbonate ion, HCO3, found in baking soda. The hydrogen carbonate can act as an acid by losing its hydrogen in water, or as a base by stealing a proton from water: HCO3 + H2O CO3 2 + H3O + HCO3 + H2O H2CO3 + OH 6
Lewis Gilbert 7
Conjugate Pairs 8
Practice: The Nature of Light 1. Identify the acid, base, conjugate acid and conjugate base for each of the following: a) HClO4 (aq) + H2O (l) H3O + (aq) + ClO4 (aq) b) H2SO3 (aq) + H2O (l) H3O + (aq) + HSO3 (aq) c) HC2H3O2 (aq) + H2O (l) H3O + (aq) + C2H3O2 (aq) d) H2S (g) + H2O (l) H3O + (aq) + HS (aq) 2. Write the equation for the reaction of each of the following with water. Indicate whether the ion or molecule is an acid or base. Which of these are Arrhenius acids/bases, Bronsted-Lowry acids/bases, and/or Lewis acids/bases? For example, HBr (aq) + H2O (l) H3O + (aq) + Br (aq) a) HI (aq) + H2O (l) b) HF (aq) + H2O (l) c) C2H3O2 (aq) + H2O(l) 9
Lesson 2: Product of Water Goals: Describe the equilibrium between hydronium and hydroxide ions in water and the effect of changing concentration of these ions using Le Chatelier's Principle. Define the ion product of water (Kw). Describe the relative concentrations of hydronium and hydroxide ions in acidic, basic and neutral solutions. Given the concentration of hydronium or hydroxide ions, calculate the other. Ionization of Water Ion Product of Water If equilibrium is established between hydronium ions, hydroxide ions and water molecules, an equilibrium law can be written: 10
Since water is a liquid, the product of Keq and water results in the ion product for water, Kw. The equilibrium law for water becomes Kw = [H3O + ][OH ] At 25 C, the concentration of the hydronium and hydroxide ions are equal at 1.0 10-7 mol/l. Therefore, at 25 C, the value of Kw is 1.0 10-14. Effect of Adding Acid and Base to Water If the ionization of water occurs by the equation H2O (l) + H2O (l) H3O + (aq) + OH - (aq). We can predict the effect of dissolving an acid or base on hydronium and hydroxide ion concentrations by using Le Chatelier's Principle. Adding Acid When an acid is dissolved in water, the acid produces a large amount of H3O + ions. If the H3O + ion concentration increases, the equilibrium will shift to the left to use up some of the added hydronium and maintain Kw at 1.0 10-14. Since equilibrium shifts left, the hydroxide ion concentration is reduced. Therefore, adding a strong acid to water increases the hydronium ion concentration and reduces the hydroxide ion concentration. Adding Base When a base is dissolved in water, the hydroxide ion concentration increases. According to Le Chatelier's Principle, the equilibrium shifts left to use up some of the added hydroxide and maintain Kw at 1.0 10-14. Since equilibrium shifts left, the hydronium ion concentration is reduced. Therefore, adding a strong base to water increases the hydroxide ion concentration and reduces the hydronium ion concentration. This means that hydronium (hydrogen) ions AND hydroxide ions are BOTH present in any solution - whether they are acidic or basic. 11
Example 1: If 2.5 moles of hydrochloric acid is dissolved in 5.0 L of water, what is the concentration of the hydroxide ions? Assume the volume remains unchanged. Example 2: 0.40 g of NaOH is dissolved in water to make a solution with a volume of 1.0 L. What is the hydronium ion concentration in this solution? 12
Practice: Product of Water 1. What is the hydroxide ion concentration in a solution with a hydronium concentration of 6.80 10-10 mol/l? 2. What is the [H3O + ] in a solution with [OH - ] of 5.67 10-3? 3. If the [OH - ] in a sodium hydroxide solution is 0.050 mol/l, what is [H3O + ]? 4. 0.25 moles of hydrogen chloride gas is dissolved in 2.0 L of water. Write the dissociation equation for this gas and calculate both [H3O + ] and [OH - ]. 13
5. 10.0 g of lithium hydroxide is dissolved in 750 ml of water. Write the dissociation equation and calculate both [H3O + ] and [OH - ]. 6. 10.0 g of calcium hydroxide is dissolved in 400.0 ml of solution. Write the dissociation equation and calculate both [H3O + ] and [OH - ]. 14
Lesson 3: ph and poh Goals: Define ph and poh. Given any one of the values ph, poh, [H3O + ] or [OH ], calculate the remaining values. Calculate the ph or poh, given the concentration of a strong acid or base. Describe how an acid-base indicator works in terms of the colour shifts and Le Chatelier's Principle. Defining ph and poh 15
Example 1: Find the log of each of the following values: a) 1.3 10-5 b) 7.2 10-11 c) 0.0054 The ph Values for ph in most solutions range from 0.0 to 14.0. Pure water is considered to be neutral, or a ph of 7.0. The lower the ph, the more acidic the solution. The higher the ph the more alkaline or basic the solution. Example 2: Calculate the ph of an HCl solution whose concentration is 5.0 10-6 mol/l. 16
Example 3: The ph of a solution is 3.25. Calculate the hydronium ion concentration in the solution. poh 17
Example 4: The ph of a solution is 10.30, what is the hydroxide ion concentration? Example 5: What is the ph of a 5.0 10-5 mol/l Mg(OH)2 solution? 18
Practice: ph and poh 1. Determine the ph of each of the following values: a) [H + ] = 1.0 10-6 b) [OH - ] = 1.0 10-2 c) [H + ] = 2.0 10-3 d) [OH - ] = 7.50 10-7 19
2. Determine the concentration of [H + ] of each of the following: a) ph = 7.00 b) poh = 5.00 c) ph = 8.90 d) poh = 13.80 3. Determine the poh of the following if the ph is given: a) ph = 5.40 b) ph = 8.60 20
4. Determine the concentration of the OH - ions in the solutions with the following ph values: a) ph = 4.10 b) ph = 5.10 5. Calculate ph of each of the following solutions: a) 4.0 10-4 mol/l Ba(OH)2 b) 0.50 mol/l HNO3 21
Lesson 4: Strength of Acids and Bases Goals: Define electrolyte and non-electrolyte. Distinguish between weak and strong aqueous solutions of acids and bases. Predict the products of an acid-base reaction and the position of the equilibrium. Electrolytes Pure water and a solution of sugar in water are non-electrolytes because they do not contain any moving charges to carry an electric current. In a wire or metal, the moving charges are electrons. In order for a solution to conduct an electric current, charged particles or ions must be present in the solution. The more ions present in a solution the greater the conductivity, up to a certain limit. The conductivity is due to the movement of the ions in solution. The number of ions in pure water is not sufficient to conduct a current. This explains the difference between the dissolving of sugar and sodium chloride in water. Sugar is a molecular substance, which is, held together by covalent bonds, so it dissolves as whole molecules. These molecules are uncharged so there is nothing to carry the charge. Sodium chloride is an ionic compound when it dissolves, it breaks up into ions, or dissociates. These ions are free to move and carry an electric current through the solution. As a solid, sodium chloride does not conduct a current since its ions are held tightly by the ionic bonds in a crystal lattice structure. However, if we melt the solid sodium chloride crystals, the liquid sodium chloride will conduct a current, since it separates into positively charged sodium ions and negatively charged chloride ions. 22
Strong electrolyte Weak electrolyte Strong and Weak Acids Strong acid Name of Acid Formula 23
Weak acid Name of Acid Formula Strong and Weak Bases Strong base Name of Acid Formula 24
Weak acid Name of Acid Formula Use Example 1: Arrange the following in decreasing strength as an acid and decreasing strength as bases. Note some may be used in both groups. HNO2, OH, HCO3, HPO4 2, HTe, C2H3O2 25
Example 2: Complete the reaction below, indicating the acids and bases. Which direction is favoured and why? HCO3 + PO4 3 Practice: Strength of Acids and Bases 1. In each case: I. Complete the acid-base equation with the help of tables if needed. II. Label the two acids and bases involved. III. Explain whether the reactants or products are favored at equilibrium. a) H3PO4 + C2H3O2 b) C2H3O2 + HSO4 c) Al(H2O)63+ + HSe 26
2. List the following in decreasing, strongest to weakest, order of strength as acids and bases. Some may be used in both lists. CO3 2, HCO3, HPO4 2, CH3COO, HSO3 3. List the following substances in increasing order, weakest to strongest, of strength as acids and bases. Some may be used more than once. F, H2PO4, HSO4, H2S, C2H3O2, HNO2, HTe 27
Lesson 5: Acid and Base Equilibria Goals: Write the equilibrium expression, Ka or Kb, from a balanced chemical equation. Use Ka or Kb to solve problems for ph, percent dissociation and concentration. Acid Dissociation Constant Strong acids dissociate completely, and therefore do not establish equilibrium. However, because both the dissociated and molecular species occur in solution, weak acids do establish equilibrium. This means, for weak acids we can make an equilibrium law. In general, for the weak acid HA: HA + H2O H3O + + A The equilibrium law would be: But water is a liquid and its concentration does not change, so we remove it and replace Keq with Ka, the acid dissociation constant or ionization constant. 28
Base Dissociation Constant Just as with acids, the base dissociation constant, Kb, reflects the strength of a base. The higher the value of Kb, the stronger the base. For example, the weak base ammonia ionizes according to the equation NH3 (g) + H2O (l) NH4 + (aq) + OH (aq) The equilibrium law is: The Kb for ammonia at 25 C is 1.8 10-5. In general, the Kb for the weak base, B, whose dissociation is like that of ammonia is B + H2O( l) BH + + OH The strong base, sodium hydroxide, dissociates completely according to the equation NaOH (s) Na + (aq) + OH - (aq) Since no NaOH is present in solution, the Kb is very large. 29
Example 1: Initially a 0.10 mol/l solution of acetic acid, HC2H3O2, is only partly ionized. If at equilibrium, the hydronium ion concentration is 1.3 10-3 mol/l, what is the acid dissociation constant, Ka? 30
Example 2: HA is a weak acid with a Ka of 7.3 10-8. What is the concentration of all species, HA, H3O + and A -, if the initial concentration of HA is 0.50 mol/l? 31
Percent Dissociation Another way to describe the amount of dissociation is by percent dissociation. Percent dissociation can be calculated by using the hydronium ion, or hydroxide ion concentrations. For the acid HA: For the base B: Where [HA] and [B] are the initial concentrations of the acid and base, respectively. Example 3: Calculate the percent dissociation of a 0.100 mol/l solution of formic acid, HCH2O2, if the hydronium ion concentration is 4.21 10-4 mol/l. 32
Example 4: Calculate the Kb of the hydrogen phosphate ion, HPO4 2-, if a 0.25 mol/l solution of sodium hydrogen phosphate is dissociated is 0.080%. 33
Example 5: What is the ph of an imaginary acid, HA, if the initial concentration of the acid is 0.10 mol/l and its Ka is 4.0 10-9? 34
Example 6: The ph of a 0.20 mol/l solution of CO3 2- ions is 10.75. What is the Kb of carbonate ions? 35
Practice: Acid and Base Equilibria 1. Calculate the concentration of all species in a 0.70 mol/l HNO3 solution. 2. Determine the concentration of [H3O + ] in a weak acid, H2S. The initial concentration of H2S is 0.90 mol/l and its Ka is 1.0 10-7. 36
3. Determine the concentration of all species with the initial concentration of 0.65 mol/l solution of a weak base NH3. The base dissociation constant is 1.8 10-5. 4. If [H3O + ] = 4.5 10-6 mol/l in a 0.45 mol/l solution of the weak acid HX, calculate percent dissociation. 37
5. Find the percent dissociation in an initial concentration of 0.87 mol/l solution of the weak base HPO3 2- if Kb is 1.6 10-7. 6. Calculate the [H3O + ] of a 0.38 mol/l weak acid that is dissociated 0.12%. 38
7. Determine the Ka for an acid, HA, if a 0.45 mol/l solution is dissociated 0.025%. 8. Calculate the percent dissociation of an initial concentration of 0.60 mol/l aniline, C6H5NH2, solution. The Kb is 3.8 10-10 if it dissociates according to the equation C6H5NH2 + H2O C6H5NH3 + + OH -. 39
9. If [H3O + ] = 4.5 10-10 mol/l, calculate the Ka for a weak acid in a solution of 0.80 mol/l of an acid HB. 10. Calculate the ph with an initial concentration of 0.10 mol/l solution of hypochlorous acid, HOCl, if its Ka is 3.5 10-8. 40
11. A 0.20 mol/l solution of the weak base HPO4 2 has a ph of 9.00. Find the Kb. 12. Calculate the Ka of a weak acid, HX, if a 0.25 mol/l solution has a ph of 4.40. 41
13. At 25 C, a 0.010 mol/l ammonia solution is 4.3% ionized. Calculate the poh and ph. 14. Calculate the percent dissociation of a 0.20 mol/l solution of the weak acid, HNO2, if the ph of the solution is 4.20. 42
15. Hydrazine, N2H4, is a weak base with a Kb of 3.0 10-6. The reaction of hydrazine in water is H2NNH2 + H2O H2NNH3 + + OH -. Calculate the ph with an initial concentration of 2.0 mol/l solution of hydrazine. 43
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