GENERAL CHEMISTRY CHAPTER 13 04/07/08 1CHAPTER THIRTEEN AQUEOUS SOLUTION EQUILIBRIA. 1 INTRODUCTION. A substantial fraction of the chemical reactions that are considered to be of major importance take place in aqueous solution. In this chapter, we shall consider the nature of several types of equilibria for chemical processes that take place in aqueous solution. In the previous chapter, we found that it is appropriate to express the reaction quotients and equilibrium constants for gas reactions in terms of the partial pressures of the reactant and product species. Specifically, the pressures should be given in atmospheres since 1 atm. pressure is that of the standard state. For a solute species, the standard state has to be some type of unit concentration. We have two choices. The most widely adopted choice of standard state for a solute species is unit molarity: 1 mole of solute per liter of solution. The alternative, which is generally looked upon as more correct is unit molality: 1 mole of solute per kilogram of solvent. In this chapter, we shall work with molarities. As in the previous chapter, we might usefully look at a general equation for a solution reaction: a A(aq) + b B(aq) Å Æ c C(aq) + d D(aq) The reactant and product species might be molecular or ionic. 1GENERAL CHEMISTRY CHAPTER 13 04/07/08 2 The equilibrium constant expression is: KC = { [C]c [D]d } / { [A]a [B]b } The square brackets indicate equilibrium molarities. A reaction quotient can be written with exactly the same form but no longer equilibrium concentrations. 2 Acid – Base Equilibria. 2.1 Weak and Strong Acids. We recall from our previous discussions of acid – base chemistry that a strong acid is one which dissociates completely in aqueous solution while a weak acid is only partially dissociated. Thus we write for the strong nitric acid: HNO3(aq) Î H+(aq) + NO3-(aq) We may also recall that many authors prefer to write equations of this type in the form: HNO3(aq) + H2O(l) Î H3O+(aq) + NO3-(aq) This has two effects. It recognizes that protons are unlikely to run around the solution free but attach to themselves one or more water molecules. It also tends to make the equation more complicated than it need be. We shall stick with the first version. 2GENERAL CHEMISTRY CHAPTER 13 04/07/08 3 We shall use our old friend acetic acid as the prime example of a weak acid. Instead of using its full formula CH3COOH, we shall use the abbreviation HAc. Thus the partial dissociation of acetic acid is represented by: HAc(aq) Í Î H+(aq) + Ac-(aq) The equilibrium constant for this reaction at 25o C is 1.8 x 10-5. The equilibrium constant expression is: Ka = [H+] [Ac-] / [HAc] We add a subscript ‘a’ to the equilibrium constant symbol to identify it as an acid dissociation constant or simply acidity constant. For solutions of acetic acid, we can use the equilibrium constant to determine the concentration of hydrogen ions. We use the following approach. We first show the concentrations of the reactant and product species before any dissociation has occurred and then at equilibrium. [HAc] [H+] [Ac-] initial [HAc]o 0 0 equilibrium [HAc]o - x x x Thus we can write: 3GENERAL CHEMISTRY CHAPTER 13 04/07/08 4 Ka = 1.8 x 10-5 = x2 / ([HAc]o - x) or [H+]2 = x2 = { 1.8 x 10-5 ([HAc]o - x) } [H+] = x = { 1.8 x 10-5 ([HAc]o - x) }1/2 We are faced with the modest task of solving a quadratic equation in x. Even that task can be made much simpler by recognizing that there are many occasions when x is so much smaller than [HAc]o that we can simplify the equation by writing: [H+] = x = { 1.8 x 10-5 [HAc]o }1/2 with very little loss of accuracy. There is a rule of thumb that tells us whether or not the approximation was justified and that is that the value obtained for x must be no more than 5% of [HAc]o. It is instructive to look at values of [H+] derived for various values of [HAc]o [HAc]o [H+] [H+] / [HAc]o 1.0 M 4.2 x 10-3 M 4.2 x 10-3 0.1 1.34 x 10-3 1.34 x 10-2 0.01 4.2 x 10-4 4.2 x 10-2 0.001 1.34 x 14-4 1.34 x 10-1 We observe two things. The fractions or percentages of the dissociation of acetic acid increase as we decrease the initial concentrations. By the time we have got down to .001 M, the percentage dissociation has risen above 10 %. We 4GENERAL CHEMISTRY CHAPTER 13 04/07/08 5should no longer ignore the x {= [H+]} above. What we can do is substitute the preliminary value in the full equation giving us: [H+] = x = { 1.8 x 10-5 ( .001 - .000134) }1/2 = 1.25 x 10-4 M On this occasion there was only a change of 9 x 10-6 going the assumed value to the solution value of x, which is less than 1% of the acetic acid concentration. The value of 1.25 x 10-4 is then acceptable. 2.2 Weak and Strong Bases. We recall that the only strong bases are the metal hydroxides of which only a limited number are soluble in water. Weak bases are molecules or ions that generate hydroxide ions when dissolved in water. For the strong bases sodium hydroxide, we write: NaOH(aq) Æ Na+(aq) + (OH)- For the weak base ammonia, we write: NH3(aq) + H2O(l) Í Î NH4+(aq) + (OH)- The equilibrium constant expression for this reaction is: Kb = [NH4+] [OH-] / [NH3] 5GENERAL CHEMISTRY CHAPTER 13 04/07/08 6 The solvent water is being treated as though a pure liquid and makes no contribution to the equilibrium constant expression. The subscript ‘b’ attached to the equilibrium constant symbol identifies this as a base hydrolysis(or basicity) constant. Some authors prefer to use the term base dissociation constant but that seems inappropriate since the ammonia does not dissociate. We may use base hydrolysis constant values and expressions to determine hydroxide ion concentrations in weak base solutions. By chance, the Kb value for ammonia is 1.8 x 10-5, the same as Ka for acetic acid. We can perform the same calculations of [OH-] for different concentrations of aqueous ammonia as we did of [H+] for different concentrations of acetic acid. [NH3]o [OH-]