Acid Base Equilibria

Transcription

1 Acid Base Equilibria Acid Ionization, also known as acid dissociation, is the process in where an acid reacts with water to produce a hydrogen ion and the conjugate base ion. HC 2 H 3 O 2(aq) H + (aq) + C 2 H 3 O 2 - (aq) For a weak acid the concentration of ions in solution is determined by the acid ionization constant which is the equilibrium constant for the ionization of a weak acid. HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) K a = [H 3 O + ] [A - ] [HA] There are two methods to determine the ionization constant for a weak acid are: 1. Determination of K a from the degree of ionization - is the fraction of molecules that react with water to give ions. 2. Determination of K a from the solution ph - finding the ph of a solution will give you the hydronium concentration. Example: Nicotinic acid is a monoprotic acid with the formula, HC 6 H 4 NO 2. A solution that is 0.012M has a ph of 3.39 at 25ºC. What is the acid ionization constant, K a, for this acid at 25ºC? What is the degree of ionization of nicotinic acid in this solution?

2 Calculations with K a Give the K a for a weak acid you can determine the concentrations of HA, H + and A -. Using the method that we used to determine the equilibrium constant, K c, we can apply that theory in a more simplified form to determining the different molarities of our weak acid. Example: What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in the solution of a 0.10 M nicotinic acid? What is the ph of the solution? What is the degree of ionization?

3 To determine whether you can use the simplified assumption check your answer. It can be shown that this simplifying assumption gives an error of less than 5% if the concentration of the acid, c a, divided by K a equal 100 or more. Example: an acid concentration of 10-2 M and the K a of 10-5 C a / K a = 10-2 / 10-5 = 10-3 so the assumption is valid. If the simplifying assumption is not valid, you can solve the equilibrium equation with the quadratic formula. What is the ph at 25ºC of the solution obtained by dissolving a 5.00 grain tablet of aspirin in 0.500L of water? The tablet contains 5.00 grains, or g of acetylsalicylic acid, HC 9 H 7 O 4. The acid is monoprotic and the K a = 3.3 x 10-4 at 25ºC. Polyprotic Acids are acids that have more than one ionizable hydrogen Example: Carbonic acid H 2 CO 3(aq) H + (aq) + HCO - 3(aq) K a1 = 4.3 x HCO 3 (aq) H + (aq) + CO -2 3(aq) K a2 = 4.8 x The K a1 is much larger than the K a2 which indicates that the carbonic acid loses the first proton more easily than the second one.

4 Ascorbic acid is a diprotic acid, H 2 C 6 H 6 O 6. what is the ph of a 0.10 M solution? What is the concentration of ascorbate ion? The acid ionization constants are K a1 = 7.9 x 10-5 and K a2 = 1.6 x Base Ionization Equilibria involves weak bases and are treated similarly to those of weak acids. NH 3(aq) + H 2 O (l) K b = [NH 4 + ][ OH - ] [NH 3 ] NH 4 + (aq) + OH - (aq) Morphine, C 17 H 19 NO 3, is a naturally occurring base, or alkaloid. What is the ph of a M solution of morphine at 25C? The base ionization constant, K b, is 1.6 x 10-6.

5 A salt can be regarded as an ionic compound from a neutralization reaction in aqueous solution. Often it is acidic or basic because it reacts with the water. The hydrolysis of an ion is the reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and the hydrogen ion. Generally : The anions of a weak acid are basic and the cations of a weak base are acidic. To predict the acidity or basicity of a salt solution you need to examine the acidity or basicity of the ions composing the salt. 1. A salt of a strong base and a strong acid gives a neutral aqueous solution 2. A salt of a strong base and a weak acid gives a basic solution. 3. A salt of a strong acid and a weak base gives an acidic solution. 4. A salt of a weak acid and a weak base is dependent on the relative acid-base strengths of the two ions. Calculating the K a from K b or K b from K a K a K b = K w Determine the K b for CN - and the K a for NH 4 + Common Ion Effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion that takes part in the equilibrium. CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq)

6 Buffers A buffer is a solution characterized by the ability to resist changes in ph when limited amounts of acid and base are added to it. Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. A buffer solution resists change in ph through the ability to combine with both H 3 O + ions and OH - ions. Two important characteristics of a buffer are : ph buffer capacity Given instructions on how to make up a buffer we can calculate the ph of the final solution Instructions for making a buffer say to mix 60 ml of M NH 3 with 40 ml of M NH 4 Cl. What is the ph of the buffer?

7 Henderson-Hasselbalch Equation is an equation relating the ph of a buffer for different concentrations of conjugate acid and base. ph = pk a + log [ ] [ ] You add 1.5 ml of a 1.0M HCl to each of the following solutions. Which one will show the least change of ph? a. 15 ml of 0.1 M NaOH b. 15 ml of 0.1 M CH 3 COOH c. 30 ml of 0.1 M NaOH and 30 ml of 0.1 M CH 3 COOH d. 30 ml of 0.1 M NaOH and 60 ml of 0.1 M CH 3 COOH Acid Base Titration Curves is a plot of the ph of a solution of acid (or Base) against a volume of added base (or acid). Curve for the titration of a strong acid by a strong base Equivalence point - the point in a titration when a stoichiometric amount of reactant has been added.

8 Calculating the ph of a solution of a strong acid and a strong base. Calculate the ph of a solution in which 15 ml of a 0.10 M NaOH has been added to 25 ml of a 0.10 M HCl. Curve for the titration of a weak acid by a strong base Calculate the ph of the solution at the equivalence point when 25 ml of 0.10 M nicotinic acid is titrated by 0.10 M NaOH. K a for nicotinic acid is 1.4 x 10-5

9 Curve for the titration of a weak base by a strong acid Calculate the ph of the NH 3 solution at the following points during the titration: a) Prior to the addition of any HCl b) After the addition of 12 ml of M HCl c) At the equivalence point d) After the addition of 31 ml of HCl