Chemistry 132 NT. Acid-Base Equilibria. The face of a child can say it all. Especially the mouth part of the face.

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1 Chemistry 1 NT The face of a child can say it all. Especially the mouth part of the face. Jack Handey 1 Chem 1 NT AcidBase Equilibria Module 1 Solutions of a Weak Acid AcidIonization Equilibria Polyprotic Acids Reaction of zinc metal with hydrochloric acid. 1

2 Review Identifying acid and base species Identifying Lewis acid and base species Deciding whether reactants or products are favored in an acidbase reaction Calculating the concentration of H and OH in solutions of strong acid or base Calculating the ph from the hydrogenion concentration, or vice versa. 4 Solutions of a Weak Acid or Base The simplest acidbase equilibria are those in which a single acid or base solute reacts with water. In this chapter, we will first look at solutions of weak acids and bases. We must also consider solutions of salts, which can have acidic or basic properties as a result of the reactions of their ions with water. 5 AcidIonization Equilibria Acid ionization (or acid dissociation) is the reaction of an acid with water to produce hydroniumion (hydrogen ion) and the conjugate base anion. When acetic acid is added to water it reacts as follows. HC H O (aq) HO(l) H O (aq) CHO (aq) Because acetic acid is a weak electrolyte, it ionizes to a small extent in water. 6

3 AcidIonization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acidionization constant (also called the acid dissociation constant). Consider the generic monoprotic acid, HA. HA(aq) HO(l) HO (aq) A (aq) 7 AcidIonization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acidionization constant (also called the acid dissociation constant). The corresponding equilibrium expression is: K c [HO ][A ] = [HA][H O] 8 AcidIonization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acidionization constant (also called the acid dissociation constant). Since the concentration of water remains relatively constant, we rearrange the equation to get: [HO ][A ] Ka = [HO]Kc = [HA] 9

4 AcidIonization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acidionization constant (also called the acid dissociation constant). Thus, K a, the acidionization constant, equals the constant [H O]K c. [HO ][A ] Ka = [HA] AcidIonization Equilibria For a weak acid, the equilibrium concentrations of ions in solution are determined by the acidionization constant (also called the acid dissociation constant). A table in your text lists acidionization constants for various weak acids. Here are a couple of examples. Substance Acetic acid Hydrocyanic acid Formula HC H O HCN K a 1.7 x x 11 Experimental Determination of K a The degree of ionization of a weak electrolyte is the fraction of molecules that react with water to give ions. Electrical conductivity or some other colligative property can be measured to determine the degree of ionization. With weak acids, the ph can be used to determine the equilibrium composition of ions in the solution. 1 4

5 Nicotinic acid is a weak monoprotic acid with the formula H NO. A 0.01 M solution of nicotinic acid has a ph of.9 at 5 o C. Calculate the acidionization constant, K a, for this acid at 5 o C. It is important to realize that the solution was made 0.01 M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than 0.01 M. We will abbreviate the formula for nicotinic acid as HNic. 1 Nicotinic acid is a weak monoprotic acid with the formula H NO. A 0.01 M solution of nicotinic acid has a ph of.9 at 5 o C. Calculate the acidionization constant, K a, for this acid at 5 o C. Let x be the moles per liter of product formed. HNic(aq) HO(l) H O (aq) Nic (aq) Starting 0.01 Change x Equilibrium 0.01x 0 0 x x x x 14 Nicotinic acid is a weak monoprotic acid with the formula H NO. A 0.01 M solution of nicotinic acid has a ph of.9 at 5 o C. Calculate the acidionization constant, K a, for this acid at 5 o C. The equilibriumconstant expression is: [H O ][Nic ] = [HNic ] Ka 15 5

6 Nicotinic acid is a weak monoprotic acid with the formula H NO. A 0.01 M solution of nicotinic acid has a ph of.9 at 5 o C. Calculate the acidionization constant, K a, for this acid at 5 o C. Substituting the expressions for the equilibrium concentrations, we get K a x = (0.01 x) 16 Nicotinic acid is a weak monoprotic acid with the formula H NO. A 0.01 M solution of nicotinic acid has a ph of.9 at 5 o C. Calculate the acidionization constant, K a, for this acid at 5 o C. We can obtain the value of x from the given ph. x = [HO ] = antilog( ph) x = antilog(.9) x = = Nicotinic acid is a weak monoprotic acid with the formula H NO. A 0.01 M solution of nicotinic acid has a ph of.9 at 5 o C. Calculate the acidionization constant, K a, for this acid at 5 o C. Substitute this value of x in our equilibrium expression. Note first, however, that ( 0.01 x) = ( ) = @ 0.01 the concentration of unionized acid remains virtually unchanged. 18 6

7 Nicotinic acid is a weak monoprotic acid with the formula H NO. A 0.01 M solution of nicotinic acid has a ph of.9 at 5 o C. Calculate the acidionization constant, K a, for this acid at 5 o C. Substitute this value of x in our equilibrium expression. x ( ) Ka = 1.4 (0.01 x) (0.01) 5 19 Nicotinic acid is a weak monoprotic acid with the formula H NO. A 0.01 M solution of nicotinic acid has a ph of.9 at 5 o C. Calculate the acidionization constant, K a, for this acid at 5 o C. To obtain the degree of dissociation: Degree of dissociation = = The percent ionization is obtained by multiplying by 0, which gives.4%. 0 Calculations With K a Once you know the value of K a, you can calculate the equilibrium concentrations of species HA, A, and H O for solutions of different molarities. The general method for doing this was discussed in our chapter on equilibrium. 1 7

8 Calculations With K a Note that in our previous example, the degree of dissociation was so small that x was negligible compared to the concentration of nicotinic acid. It is the small value of the degree of ionization that allowed us to ignore the subtracted x in the denominator of our equilibrium expression. The degree of ionization of a weak acid depends on both the K a and the concentration of the acid solution (see the figure on the next slide). Percent Ionization vs Acid Concentration Calculations With K a How do you know when you can use this simplifying assumption It can be shown that if the acid concentration, C a, divided by the K a exceeds 0, that is, C if a > 0 Ka then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5%. 4 8

9 Calculations With K a How do you know when you can use this simplifying assumption If the simplifying assumption is valid, you can approximate the equilibrium calculations by ignoring added or subtracted x s in the equilibrium expression. The next example illustrates this with a solution of nicotinic acid, again. 5 What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. As in our previous example, the solution was made 0. M in nicotinic acid, however, some molecules ionize making the equilibrium concentration of nicotinic acid less than 0. M. We will again abbreviate the formula for nicotinic acid as HNic. 6 Starting 0. Change x Equilibrium 0.x What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. Let x be the moles per liter of product formed. HNic(aq) H O(l) H O (aq) Nic (aq) 0 0 x x x x 7 9

10 What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. The equilibriumconstant expression is: [H O ][Nic ] = [HNic] Ka 8 What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. Substituting the expressions for the equilibrium concentrations, we get x Ka = = 1.4 (0. x) 5 9 What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. Here is where we must decide whether we can simplify the equation or if we must solve the quadratic. x 5 Ka = = 1.4 (0. x) 0

11 What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. C Divide the acid concentration by the K a. = 0. M Hnic 1.4 a K 5 = a 714 which is much greater than 0, so we can use our simplifying assumption that x will be negligible. 1 What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. This means that x is so much smaller than the 0. M HNic concentration, that subtracting it from 0. won t change the value. That is, What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. Going back to our equilibrium expression, we see K a x = = 1.4 (0. x) 5 (0.) 11

12 What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. Solving for x ( ) 1.4 = What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. Checking to see if our simplifying assumption was correct, we see that 0. x = ( ) = 0. (to two significant figures) the assumption was indeed valid. 5 What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. We can now find the equilibrium concentrations of all the species by substituting x into the last line of our original table. The concentrations of HNic, H, and Nic are 0. M, M, and M, respectively. 6 1

13 What are the concentrations of nicotinic acid, hydrogen ion, and nicotinate ion in a solution of 0. M nicotinic acid, H NO, at 5 o C What is the ph of the solution The acid ionization constant, K a, is 1.4 x 5. The ph of the solution is ph = log[h ] = log(0.001) =.9 7 Calculations With K a How do you know when you can use this simplifying assumption If the simplifying assumption is not valid, you can solve the equilibrium equation exactly by using the quadratic equation. The next example illustrates this with a solution of aspirin (acetylsalicylic acid), HC 9, a common headache remedy. 8 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. The molar mass of HC 9 is 180. g. From this we find that the sample contained mol of the acid. 9 1

14 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. The molar mass of HC 9 is 180. g. Hence, the concentration of the acetylsalicylic acid is mol/0.500 L = M (Retain two significant figures, the same number of significant figures in K a ). 40 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. Note that C a = K 4 = a 11 which is less than 0, so we must solve the equilibrium equation exactly. 41 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. We will abbreviate the formula for acetylsalicylic acid as HAcs and let x be the amount of H O formed per liter. The amount of acetylsalicylate ion is also x mol; the amount of nonionized acetylsalicylic acid is (0.006x) mol. 4 14

15 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. Starting Change Equilibrium These data are summarized below. HAcs(aq) HO(l) HO (aq) Acs (aq) x 0.006x 0 0 x x x x 4 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. The equilibrium constant expression is [H O ][Acs ] [HAcs] = K a 44 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. If we substitute the equilibrium concentrations and the K a into the equilibrium constant expression, we get x 4 =. (0.006 x) 45 15

16 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. You can solve this equation exactly by using the quadratic formula. Rearranging the preceding equation to put it in the form ax bx c = 0, we get x (. 4 )x (1. 6 ) = 0 46 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. Now substitute into the quadratic formula. b x = b a 4ac 47 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. Now substitute into the quadratic formula (. ) (. ) 4(1. ) x = The lower sign in ± gives a negative root which we can ignore

17 What is the ph at 5 o C of a solution obtained by dissolving 0.5 g of acetylsalicylic acid (aspirin), HC 9, in L of water The acid is monoprotic and K a =. x 4 at 5 o C. Taking the upper sign, we get x = [HO ] = 9.4 Now we can calculate the ph. ph = log( ) =.0 49 Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. Sulfuric acid, for example, can lose two protons in aqueous solution. SO (aq) H O(l) fi H O (aq) HSO (aq) H (aq) HO(l) HO (aq) SO4 HSO (aq) The first proton is lost completely followed by a weak ionization of the hydrogen sulfate ion, HS. 50 Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. For a weak diprotic acid like carbonic acid, H CO, two simultaneous equilibria must be considered. H CO(aq) H O(l) H O (aq) HCO (aq) HCO (aq) H O(l) H O (aq) CO (aq) 51 17

18 Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. Each equilibrium has an associated acidionization constant. K For the loss of the first proton [HO ][HCO ] 7 a1 = = 4. [HCO ] 5 Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. Each equilibrium has an associated acidionization constant. K For the loss of the secondproton [HO ][CO ] 11 a = = 4.8 [HCO ] 5 Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. In general, the second ionization constant, K a, for a polyprotic acid is smaller than the first ionization constant, K a1. In the case of a triprotic acid, such as H P, the third ionization constant, K a, is smaller than the second one, K a

19 Polyprotic Acids Some acids have two or more protons (hydrogen ions) to donate in aqueous solution. These are referred to as polyprotic acids. When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions. However, reasonable assumptions can be made that simplify these calculations as we show in the next example. 55 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. For diprotic acids, K a is so much smaller than K a1 that the smaller amount of hydronium ion produced in the second reaction can be neglected. 56 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. The ph can be determined by simply solving the equilibrium problem posed by the first ionization

20 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. If we abbreviate the formula for ascorbic acid as H Asc, then the first ionization is: H Asc(aq) HO(l) H O (aq) HAsc (aq) 58 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. Starting Change Equilibrium H Asc(aq) H O(l) HO (aq) HAsc (aq) 0. x 0.x 0 0 x x x x 59 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. The equilibrium constant expression is [H O ][HAsc = Ka1 [HAsc] ] 60 0

21 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. Substituting into the equilibrium expression x = 7.9 (0. x) Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. Assuming that x is much smaller than 0., you get 5 (7.9 ) (0.) = Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. The hydronium ion concentration is M, so ph = log(0.008) =

22 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. The ascorbateion, Asc, which we will call y, is produced only in the second ionization of H Asc. HAsc (aq) H O(l) H O (aq) Asc (aq) 64 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. Assume the starting concentrations for HAsc and H O to be those from the first equilibrium. HAsc (aq) H O(l) H O (aq) Asc (aq) 65 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. Starting Change Equilibrium HAsc (aq) H O(l) y 0.008y H O (aq) Asc (aq) y y 0.008y y 66

23 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. The equilibrium constant expression is [H O ][Asc [HAsc ] = K a ] 67 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. Substituting into the equilibrium expression ( y)(y) 1 (0.008 y) = Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. Assuming y is much smaller than 0.008, the equation simplifies to ( 0.008)(y) 1.6 (0.008) 69

24 Ascorbic acid (vitamin C) is a diprotic acid, H What is the ph of a 0. M solution What is the = 7.9 x 5 and K a = 1.6 x 1. Hence, ] = The concentration of the ascorbate ion equals K a. 70 Operational Skills Determining K a (or K b ) from the solution ph Calculating the concentration of a species in a weak acid solution using K a Calculating concentrations of species in a solution of a diprotic acid Time for a few review questions