Equilibria. Unit Outline

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1 Acid Base Equilibria 17Advanced Unit Outline 17.1 Acid Base Reactions 17.2 Buffers 17.3 Acid Base Titrations 17. Some Important Acid Base Systems In This Unit We will now expand the introductory coverage of acid base equilibria and explore the chemistry of more complicated aqueous solutions containing acids and bases. First, we will address the different types of acid base reactions, and then we will move on to study buffer solutions, acid base titrations, and polyprotic acids. One of the more important types of acid base solutions in terms of commercial and biological applications is buffers because they allow us to control the p of a solution. So that you can get a feeling for the importance of buffers in your world, we will also briefly discuss the chemistry of two important buffers in biological systems. In Precipitation and Lewis Acid Base Equilibria (Unit 18), we will conclude our coverage of chemical equilibria with Lewis acids and bases and the equilibria of sparingly soluble compounds. Vasilyev/Shutterstock.com

2 17.1 Acid Base Reactions 17.1a Strong Acid/Strong Base Reactions In Chemical Reactions and Solution Stoichiometry (Unit ) you learned that acids and bases react to form water and a salt and that these reactions are called neutralization reactions because, on completion of the reaction, the solution is neutral. owever, acid base reactions do not always result in the formation of a solution with a neutral p, and not all acid-base reactions proceed to 100% completion (Interactive Figure ). Interactive Figure Investigate the extent of acid-base reactions. Cl(aq) 1 NaO(aq) 2 O(,) 1 NaCl(aq) 0% 100% Cl(aq) 1 N 3 (aq) N +(aq) 1 Cl (aq) 0% 100% F(aq) 1 C 3 CO 2 (aq) C 3 CO 2 (aq) 1 F (aq) 0% 100% CN(aq) 1 N 3 (aq) N +(aq) 1 CN (aq) 0% 100% Figure Not all acid-base reactions go to 100% completion 2013 Cengage Learning There are four classes of acid base reactions: strong acid + strong base, strong acid + weak base, weak acid + strong base, and weak acid + weak base. For each, we will investigate the extent of the reaction and the p of the resulting solution when equimolar amounts of reactants are combined. Unit 17 Advanced Acid Base Equilibria 55

3 Reaction Example p at Equilibrium Strong acid + strong base Strong acid + weak base Strong base + weak acid Cl(aq) + NaO(aq) S 2 O(l) + NaCl(aq) 7 Cl(aq) + N 3 (aq) S N Cl(aq) < 7 NaO(aq) + ClO(aq) S 2 O(l) + NaClO(aq) > 7 Weak acid + weak base ClO(aq) + N 3 (aq) N ClO(aq) Depends on K a and K b values The reaction between a strong acid and a strong base produces water and a salt (an ionic compound consisting of the cation from the strong base and the anion from the strong acid): Cl(aq) + NaO(aq) S 2 O(l) + NaCl(aq) acid base water salt The net ionic equation for any reaction between a strong acid and a strong base is the reverse of the K w reaction. Complete ionic equation: 3 O + (aq) + Cl (aq) + Na + (aq) + O (aq) S 2 2 O(l) + Na + (aq) + Cl (aq) Net ionic equation: 3 O + (aq) + O (aq) S 2 2 O(l) K = 1/K w = The large value of K for this reaction indicates that in a strong acid + strong base reaction, the reactants are completely consumed to form products. The resulting solution is p neutral (p = 7). Notice that a single arrow (S) is used to indicate that the reaction goes essentially to completion. 17.1b Strong Acid/Weak Base and Strong Base/ Weak Acid Reactions A strong acid/weak base or strong base/weak acid reaction has a large equilibrium constant (K >> 1) and goes essentially to completion. That is, when combined in stoichiometric amounts, all reactants are consumed to form products. For example, consider the reaction between the strong acid Cl and the weak base N 3. Cl(aq) + N 3 (aq) S N Cl(aq) Unit 17 Advanced Acid Base Equilibria 555

4 The net ionic equation for this reaction is the reverse of the hydrolysis reaction for the weak acid N +. 3 O + (aq) + N 3 (aq) S 2 O(l) + N + (aq) K = 1/K a (N + ) = 1/( ) = We can predict the large magnitude of the equilibrium constant for this reaction by recognizing that of the two Brønsted acids in this reaction ( 3 O + and N + ), the hydronium ion is a much better proton donor (a much stronger acid) than the ammonium ion. Therefore, the acid base reaction favors the formation of the weaker acid, N +. In general, all acid base reactions favor the direction where a stronger acid and base react to form a weaker acid and base. When the reaction between hydrochloric acid and ammonia is complete (all reactants are consumed), the solution contains the chloride ion (a p-neutral anion) and the ammonium ion (an acidic cation). As a result, the solution is acidic (p < 7). N + (aq) + 2 O(l) N 3 (aq) + 3 O + (aq) K a = In general, the p of a solution resulting from the reaction of a strong acid with weak base is less than 7 (acidic) because of the presence of the conjugate acid of the weak base. Similarly, the reaction between a weak acid and a strong base has a large equilibrium constant (K >> 1) and therefore goes essentially to completion. For example, consider the reaction of the strong base NaO with the weak acid ClO. NaO(aq) + ClO(aq) S 2 O(l) + NaClO(aq) The net ionic equation for this reaction is the reverse of the hydrolysis reaction for the weak base ClO. O (aq) + ClO(aq) S 2 O(l) + ClO (aq) K = 1/K b (ClO ) = 1/( ) = In this reaction, ClO is a stronger acid than 2 O (and O is a stronger base than ClO ); therefore, the reaction favors the formation of products, the weaker acid ( 2 O) and base (ClO ). When the reaction is complete (all reactants are consumed), the solution contains the sodium ion (a p-neutral cation) and the hypochlorite ion (a basic anion). As a result, the solution is basic (p > 7). OCl (aq) + 2 O(l) OCl(aq) + O (aq) K b = In general, the p of a solution resulting from the reaction of a strong base with a weak acid is greater than 7 (basic) because of the presence of the conjugate base of the weak acid. Unit 17 Advanced Acid Base Equilibria 556

5 Example Problem Describe strong acid/weak base and strong base/ weak acid reactions. Write the net ionic equation for the reaction between nitrous acid and potassium hydroxide. Is the reaction reactant- or product-favored? Is the solution acidic or basic after all reactants are consumed? Solution: You are asked to write a net ionic equation for an acid base reaction and to predict whether the reaction is reactant- or product-favored and whether the solution is acidic or basic after all reactants are consumed. You are given the identity of the acid and the base. NO 2 (aq) + O (aq) S 2 O(l) + NO 2 (aq) NO 2 is a much stronger acid than 2 O (and O is a much stronger base than NO 2 ). The reaction will favor the formation of the weaker acid and base, so the reaction is productfavored. Reactions between weak acids and strong bases are assumed to be 100% complete. The p of the solution is greater than 7 (basic) after all reactants are consumed because of the presence of the weak base NO 2. Video Solution Tutored Practice Problem c Weak Acid/Weak Base Reactions In the reaction between a weak acid and a weak base, the magnitude of the equilibrium constant and therefore the extent of reaction depend on the relative strength of the acids and bases in the reaction. For example, consider the reaction between the weak acid ClO and the weak base N 3. ClO(aq) + N 3 (aq) N + (aq) + ClO (aq) In this example, ClO (K a = ) is a stronger acid than N + (K a = ), so the reaction is product-favored. owever, the equilibrium constant for the reaction is not large, so the reaction does not go essentially to completion. A significant concentration of all four species (along with 3 O + and O ) is found at equilibrium. ClO(aq) + 2 O(l) 3 O + (aq) + ClO (aq) N 3 (aq) + 2 O(l) N + (aq) + O (aq) K a = K b = O + (aq) + O (aq) 2 2 O(l) K = 1/K w = 1/( ) ClO(aq) + N 3 (aq) N + (aq) + ClO (aq) K net = K a K b /K w = 63 Unit 17 Advanced Acid Base Equilibria 557

6 The equilibrium p of the solution depends on the acid base strength of the predominant species in solution, N + and ClO. In this example, because OCl is a weaker acid than N 3 is a base [K a (OCl) < K b (N 3 )], the hypochlorite ion is a stronger base than the ammonium ion is an acid [K b (ClO ) > K a (N + )] and the solution is basic (p > 7). Consider the reaction between the weak acid ClO and the weak base CO 2. ClO(aq) + CO 2 (aq) CO 2 (aq) + ClO (aq) We can predict the extent of reaction by comparing the relative strength of the bases in this reaction. ere, ClO (K b = ) is a stronger base than CO 2 (K b = ), so the reaction is reactant-favored. The p of the solution at equilibrium is predicted by comparing the relative acid and base strength of the two predominant species at equilibrium. ypochlorous acid is a stronger acid than the formate ion is a base [K a (ClO) > K b (CO 2 )], so the solution is acidic (p < 7). Example Problem Describe weak acid/weak base reactions. When equimolar amounts of hydrocyanic acid and the acetate ion react, is the reaction reactant- or product-favored? Predict the p of the solution after all reactants are consumed. Solution: You are asked to predict whether an acid base reaction is reactant- or product-favored and what the p of the solution is after the reactants are consumed. You are given the identity of the acid and the base. CN(aq) + C 3 CO 2 (aq) C 3 CO 2 (aq) + CN (aq) Acetic acid is a stronger acid than hydrocyanic acid (and the cyanide ion is a stronger base than the acetate ion). The equilibrium will favor the weaker acid and base (reactant-favored). The p of the solution at equilibrium is controlled by the predominant species in solution, CN and C 3 CO 2. The solution is basic (p > 7) because K b (C 3 CO 2 ) > K a (CN). Video Solution Tutored Practice Problem Section 17.1 Mastery 17.2 Buffers 17.2a Identifying Buffers Buffers are one of the more important types of acid base solutions in terms of commercial and biological applications because they allow us to control the p of a solution. For example, humans are composed of molecules that depend on hydrogen bonding for their structure and function and are therefore highly sensitive to p. Most of the reactions in a Unit 17 Advanced Acid Base Equilibria 558

7 human body occur in aqueous solutions containing buffering agents. It is not surprising that human blood is highly buffered, for if blood is not maintained at a p near 7., death can occur. A buffer solution contains a mixture of a weak acid and a weak base, typically the conjugate base of the weak acid. For example, a buffer solution commonly used in chemistry laboratories contains both acetic acid (C 3 CO 2, a weak acid) and sodium acetate (NaC 3 CO 2, the sodium salt of the conjugate base of acetic acid). Some other examples of buffers are K 2 PO /K 2 PO ( 2 PO is a weak acid and PO 2 is a weak base) and N Cl/ N 3 (N + is a weak acid and N 3 is a weak base). The principle property of a buffer solution is that it experiences a relatively small change in p when a strong acid or a strong base is added. It is the weak acid and weak base components of a buffer that make it possible for buffer solutions to absorb strong acid or strong base without a significant p change. When a strong acid is added to a buffer, the acid reacts with the conjugate base and is completely consumed. Despite the addition of a strong acid, the p of the buffer solution decreases only slightly. Example: When 3 O + is added to a C 3 CO 2 /NaC 3 CO 2 buffer, it consumes some of the conjugate base, forming additional acetic acid. 3 O + (aq) + C 3 CO 2 (aq) S 2 O(l) + C 3 CO 2 (aq) When a strong base is added to a buffer, the base reacts with the weak acid and is completely consumed. Despite the addition of a strong base, the p of the buffer solution increases only slightly. Example: When O is added to a C 3 CO 2 /NaC 3 CO 2 buffer, it consumes some of the weak acid, forming additional acetate ion. O (aq) + C 3 CO 2 (aq) S 2 O(l) + C 3 CO 2 (aq) Example Problem Identify buffer solutions. Identify buffer solutions from the following list. a M sodium hydroxide M sodium bromide b M nitrous acid M sodium nitrite c. 0.2 M nitric acid M sodium nitrate d M calcium chloride M calcium bromide e. 0.3 M ammonia M ammonium bromide c Unit 17 Advanced Acid Base Equilibria 559

8 b Example Problem (continued) Solution: You are asked to determine whether a given combination of species results in a buffer solution. You are given the identify of the species in the solution. a. This is not a buffer. Sodium hydroxide is a strong base, and sodium bromide is a neutral salt. b. This is a buffer solution. Nitrous acid is a weak acid, and sodium nitrate is a source of its conjugate base, the nitrite ion. c. This is not a buffer. Nitric acid is a strong acid, and sodium nitrate is a neutral salt. d. This is not a buffer. Both are neutral salts. e. This is a buffer solution. The ammonium ion (present as ammonium bromide) is a weak acid, and ammonia is its conjugate base. Video Solution Tutored Practice Problem b Buffer p The common ion effect is the shift in an equilibrium that results from adding to a solution a chemical species that is common to an existing equilibrium; we can use this to help us understand the p of buffer solutions. Consider a solution containing acetic acid, a weak acid: C 3 CO 2 (aq) + 2 O(l) 3 O + (aq) + C 3 CO 2 (aq) The addition of sodium acetate, a source of the weak base C 3 CO 2, shifts the equilibrium to the left, suppressing the acid hydrolysis (the forward reaction) and increasing the p. The following example problem demonstrates the common ion effect in an acetic acid/ sodium acetate buffer solution. Example Problem Calculate p of a weak acid/conjugate base buffer solution. Calculate the p of 125 ml of a 0.15-M solution of acetic acid before and after the addition of mol of sodium acetate. Solution: You are asked to calculate the p of a weak acid solution before and after the addition of its conjugate base. You are given the identity of the weak acid and its conjugate base, the volume and concentration of the weak acid, and the amount of conjugate base. c Unit 17 Advanced Acid Base Equilibria 560

9 b Example Problem (continued) Step 1. Write the balanced equation for the acid hydrolysis reaction. C 3 CO 2 (aq) + 2 O(l) 3 O + (aq) + C 3 CO 2 (aq) When solving problems involving buffers, it is important to first write the weak acid hydrolysis reaction before considering the effect of added conjugate base. Step 2. Set up an ICE table and calculate the p of the weak acid solution. C 3 CO 2 (aq) + 2 O(l) 3 O + (aq) + C 3 CO 2 (aq) Initial (M) Change (M) 2x 1x 1x Equilibrium (M) x x x K a C 3CO O 1 3C 3 CO 2 5 x = [ 3 O + ] = M p = x2 1x x < x Step 3. Set up a new ICE table for the buffer that now includes the concentration of the common ion, the acetate ion. 3C 3 CO 22 initial mol C 3CO L M C 3 CO 2 (aq) + 2 O(l) 3 O + (aq) + C 3 CO 2 (aq) Initial (M) Change (M) 2x 1x 1x Equilibrium (M) x x x Step. Substitute these equilibrium concentrations into the K a expression. K a C 3CO O 1 3C 3 CO x2 1x x2 < 0.152x 0.15 We can make the assumption that x is small compared with the initial acid and conjugate base concentrations because (1) K a is small and (2) the presence of a common ion suppresses the weak acid hydrolysis. Step 5. Solve the expression for [ 3 O + ] and calculate p. x O 1 5 K a 3C 3 CO 2 3C 3 CO a b M p =.65 Is your answer reasonable? The p of the solution has increased (the solution is more basic) because a weak base, the acetate ion, was added to the solution to form a buffer. Video Solution Tutored Practice Problem Unit 17 Advanced Acid Base Equilibria 561

10 A buffer solution can consist of either a weak acid and its conjugate base or a weak base and its conjugate acid. To be consistent, we will treat all buffers as weak acid systems as shown in the following example. Example Problem Calculate p of a weak base/conjugate acid buffer solution. A 0.30-M aqueous solution of N 3 has a p of Calculate the p of a buffer solution that is 0.30 M in N 3 and 0.23 M in ammonium bromide. Solution: You are asked to calculate the p of a buffer solution. You are given the identity of the weak acid and weak base in the buffer and the concentration of the species in the buffer solution. Step 1. Write the balanced equation for the acid hydrolysis reaction. In this example the weak acid is the ammonium ion. N + (aq) + 2 O(l) 3 O + (aq) + N 3 (aq) Step 2. Set up an ICE table for the buffer solution. N + (aq) 1 2 O(l) 3 O 1 (aq) 1 N 3 (aq) Initial (M) Change (M) 2x 1x 1x Equilibrium (M) x x x Step 3. Substitute these equilibrium concentrations into the K a expression. K a N 3 3 3O x2 1x x2 5 < 3N x 0.23 Once again, we can make the assumption that x is small compared with the initial acid and conjugate base concentrations. Step. Solve the expression for [ 3 O + ] and calculate p. x O 1 5 K a 3N 1 3N a b M p = 9.37 Is your answer reasonable? Addition of a weak acid, ammonium ion, decreases the p of the solution (the solution is more acidic than the solution containing only ammonia). Video Solution Tutored Practice Problem Unit 17 Advanced Acid Base Equilibria 562

11 The enderson asselbalch Equation The rearranged K a expression used in previous example problems can be used for calculating the p of any buffer solution. In general, for a weak acid/conjugate base buffer, 3weak acid 3 3 O 1 5 K a 3conjugate base We can rewrite this equation in terms of p and pk a by taking the negative logarithm of both sides. Because a2log 3weak acid 2log3 3 O 1 5 2logaK a 3conjugate base b 3weak acid 2log3 3 O 1 5 2log1K a 2 1 a2log 3conjugate base b 3weak acid p 5 pk a 2log 3conjugate base 3weak acid b 5 a1 log 3conjugate base 3conjugate base b, 3weak acid 3conjugate base p 5 pk a 1 log 3weak acid (17.1) Equation 17.1 is the enderson asselbalch equation, a very useful form of the K a expression that is often used to calculate the p of buffer solutions. It is important to note that the enderson asselbalch equation is used only for calculations involving buffer solutions. It is not used if a solution contains only a weak acid or only a weak base. The enderson asselbalch equation shows the mathematical relationship between the weak acid pk a and the p of a buffer, and that buffer p can be manipulated by changing the ratio of [conjugate base] to [weak acid]. Notice that in the special case where [weak acid] = [conjugate base], the ratio of concentrations is equal to 1 and the p of the buffer solution is equal to the weak acid pk a. When [weak acid] = [conjugate base], p = pk a + log(1) = pk a + 0 = pk a Buffer components are chosen based on the relationship between weak acid pk a and the target p for the buffer. For the buffer to be effective, it should contain significant amounts of both weak acid and conjugate base. Effective buffers, those that can best resist p change upon addition of a strong acid or base, have a [conjugate base] to [weak acid] ratio between 1:10 and 10:1. Because log(10/1) = 1, this results in a buffer p that is approximately equal to the weak acid pk a ± 1. This type of buffer solution has a high Unit 17 Advanced Acid Base Equilibria 563

12 buffer capacity, which is defined as the amount of strong acid or base that can be added to a buffer without a drastic change in p. [conjugate base]/ [weak acid] p of buffer solution Example Problem Use the enderson asselbalch equation to calculate p of a buffer solution. Using the enderson asselbalch equation, calculate the p of a buffer solution that is 0.18 M in 2 PO and 0.21 M in PO 2. 1 p = pk a 10/1 p = pk a + 1 1/10 p = pk a 1 Solution: You are asked to calculate the p of a buffer solution using the enderson asselbalch equation. You are given the concentration and identity of the species in the buffer solution. Step 1. Write the balanced equation for the acid hydrolysis reaction. 2 PO (aq) + 2 O(l) 3 O + 2 (aq) + PO (aq) Step 2. Set up an ICE table for the buffer solution. 2 PO (aq) 1 2 O(l) 3 O 1 (aq) 1 2 PO (aq) Initial (M) Change (M) 2x 1x 1x Equilibrium (M) x x x Step 3. Substitute these equilibrium concentrations into the enderson asselbalch equation and calculate p. p 5 pk a 1 log 3PO PO log x2 2 1 log < log a x b Once again, we can make the assumption that x is small compared with the initial acid and conjugate base concentrations. Is your answer reasonable? Notice that the p of the buffer solution is greater than pk a for the weak acid because [conjugate base]/[weak acid] > 1. If [conjugate base]/[weak acid] < 1, the p is less than pk a. Video Solution Tutored Practice Problem Adding Acid or Base to a Buffer It is a common misconception that buffer p remains constant when some strong acid or base is added. This is not the case. As shown in the following example problem, a buffer minimizes the p change upon addition of a strong acid or base because only the weak acid/conjugate base ratio of the buffer is affected. The p changes, but only by a small amount. Unit 17 Advanced Acid Base Equilibria 56

13 Example Problem Calculate buffer p after adding strong acid or strong base. Determine the p change when mol Cl is added to 1.00 L of a buffer solution that is 0.10 M in C 3 CO 2 and 0.25 M in C 3 CO 2. Solution: You are asked to calculate the p change when a strong acid is added to a buffer solution. You are given the amount of strong acid, the concentration of the species in the buffer solution, and the volume of the buffer solution. Step 1. Write the balanced equation for the acid hydrolysis reaction. C 3 CO 2 (aq) + 2 O(l) 3 O + (aq) + C 3 CO 2 (aq) Step 2. Use the enderson asselbalch equation to calculate the p of the buffer solution before the addition of Cl. p 5 pk a 1 log 3C 3CO 22 3C 3 CO 2 5 2log loga b Step 3. Assume that the strong acid reacts completely with the conjugate base. Set up a table that shows the amount (in moles) of species initially in the solution, the change in amounts of reactants and products (based on the amount of limiting reactant), and the amounts of reactants and products present after the acid base reaction is complete. 3 O + (aq) 1 C 3 CO 2 (aq) S 2 O(l) 1 C 3 CO 2 (aq) Initial (mol) Change (mol) After reaction (mol) Step. Use the new weak acid and conjugate base concentrations to calculate the buffer p after adding strong acid mol 3C 3 CO L M 3C 0.23 mol 3CO L M p 5 pk a 1 log 3C 3CO 22 3C 3 CO 2 5 2log loga b p = = 0.11 Is your answer reasonable? Addition of mol Cl to the buffer decreases the p only slightly, by 0.11 p units. If the same amount of Cl is added to 1.00 L of water, the p decreases by 5.30 p units, from a p of 7.00 to a p of Video Solution Tutored Practice Problem Unit 17 Advanced Acid Base Equilibria 565

14 17.2c Making Buffer Solutions The preparation of a buffer solution with a known p is a two-step process. 1. A weak acid/conjugate base pair is chosen. The weak acid pk a must be within about 1 p unit of the desired p. This guarantees that the [weak acid]/ [conjugate base] ratio is between 10:1 and 1:10, ensuring that the solution will contain significant amounts of weak acid and conjugate base and will be able to buffer against the addition of strong acid or base. 2. The desired p and the weak acid pk a are used to determine the relative concentrations of weak acid and conjugate base needed to give the desired p. Once the desired weak acid and conjugate base concentrations are known, the solution is prepared in one of two ways: Direct addition: The correct amounts of the weak acid and conjugate base are added to water. Acid base reaction: For example, a conjugate base is created by reacting a weak acid with enough strong base to produce a solution containing the correct weak acid and conjugate base concentrations. Example Problem Prepare a buffer by direct addition. Describe how to prepare 500. ml of a buffer solution with p = 9.85 using one of the weak acid/conjugate base systems shown here. Weak Acid Conjugate Base K a pk a C 3 CO 2 C 3 CO PO 2 PO CO 3 2 CO Solution: You are asked to describe how to prepare a buffer solution with a known p using the direct addition method. You are given the identity and K a values for three possible weak acid/conjugate base pairs and the volume and target p of the buffer solution. c Unit 17 Advanced Acid Base Equilibria 566

15 b Example Problem (continued) Step 1. Choose a weak acid/conjugate base pair. The bicarbonate ion/carbonate ion buffer system is the best choice here because the desired p is close to the pk a of the weak acid. Write the balanced equation for the acid hydrolysis reaction. CO 3 (aq) + 2 O(l) 3 O + (aq) + CO 3 2 (aq) Step 2. Determine the necessary weak acid/conjugate base ratio using the rearranged K a expression for the weak acid. (The enderson asselbalch equation can also be used to determine the weak acid/conjugate base ratio.) [ 3 O + ] = 10 p = = M 3 3 O K a 3CO 32 3CO CO 32 3CO CO 3 2 3CO mol /L 1.0 mol/l mol CO mol CO 3 Notice that the volume of buffer is cancelled in the ratio. The required amounts of weak acid and conjugate base are independent of the solution volume, so the volume of a buffer has no effect on the buffer p. Step 3. Determine the amount of weak acid and conjugate base that must be combined to 2 produce the buffer solution. Mixing 2.9 mol CO 3 and 1.0 mol CO 3 (or any multiple of this ratio) will result in a buffer with a p of Alternately, assuming that each is present in the form of a sodium salt, combine 20 g NaCO 3 and 110 g Na 2 CO 3 in a flask and add water (to a total solution volume of 500 ml) to make the buffer solution. 8.0 g 2.9 mol NaCO g NaCO 1 mol NaCO mol Na 2 CO g 1 mol Na 2 CO g Na 2 CO 3 Video Solution Tutored Practice Problem When performing a calculation to determine how to make a buffer solution using an acid base reaction, it is helpful to set up a stoichiometry table to assist in keeping track of the initial amounts of reactants and the amounts of reactants and products in the solution after a reaction takes place. Note that a stoichiometry table is different from an ICE table, which is used to keep track of changes that occur in an equilibrium system. A stoichiometry table shows the amount (in moles) of species initially in the solution, the change in amounts Unit 17 Advanced Acid Base Equilibria 567

16 of reactants and products (based on the amount of limiting reactant), and the amounts of reactants and products present after the reaction is complete. Example Problem Prepare a buffer by acid base reactions. Describe how to prepare a buffer solution with p = 5.25 (using one of the weak acid/ conjugate base systems shown here) by combining a 0.50-M solution of weak acid with any necessary amount of 1.00 M NaO. Weak Acid Conjugate Base K a pk a C 3 CO 2 C 3 CO PO 2 PO CO 3 2 CO Solution: You are asked to describe how to prepare a buffer solution with a known p using an acid base reaction. You are given the identity and K a values for three possible weak acid/conjugate base pairs, the concentration of the weak acid and the strong base used to make the buffer, and the target p of the buffer solution. Step 1. Choose a weak acid/conjugate base pair. The acetic acid/acetate ion buffer system is the best choice here because the desired p is close to the pk a of the weak acid. Write the balanced equation for the acid hydrolysis reaction. C 3 CO 2 (aq) + 2 O(l) 3 O + (aq) + C 3 CO 2 (aq) Step 2. Determine the necessary weak acid/conjugate base ratio using the rearranged K a expression for the weak acid. (The enderson asselbalch equation can also be used to determine the weak acid/conjugate base ratio.) [ 3 O + ] = = M 3 3 O K a 3C 3 CO 2 3C 3 CO 22 3C 3 CO 2 3C 3 CO C 3CO 2 3C 3 CO mol /L 1.0 mol/l mol C 3CO mol C 3 CO 2 c Unit 17 Advanced Acid Base Equilibria 568

17 b Example Problem (continued) Notice that volume units (L) are cancelled in the 3C 3CO 2 ratio. The required amounts of 3C 3 CO 22 weak acid and conjugate base are independent of the total solution volume, so the volume of a buffer has no effect on the buffer p. Step 3. Determine the amount of weak acid and strong base that must be combined to produce the buffer. Recall that a strong base will react completely with a weak acid forming water and the conjugate base of the weak acid. In this case, the weak acid and strong base react in a 1:1 stoichiometric ratio, so initial amount of weak acid required = amount of weak acid in buffer + amount of conjugate base in buffer initial amount of weak acid required = 0.31 mol mol = 1.31 mol C 3 CO 2 The amount of strong base required is determined by the reaction stoichiometry. Set up a stoichiometry table that shows the amount (in moles) of species initially in the solution, the change in amounts of reactants and products (based on the amount of limiting reactant), and the amounts of reactants and products present after the acid base reaction is complete. In this case, the stoichiometry table is used to determine the initial amount of reactants needed to produce a buffer containing 0.31 mol acetic acid and 1.00 mol acetate ion. C 3 CO 2 (aq) 1 O (aq) S 2 O(l) 1 C 3 CO 2 (aq) Initial (mol) Change (mol) After reaction (mol) The combination of 1.31 mol C 3 CO 2 with 1.00 mol O (as NaO) will produce the buffer solution. Step. Determine the volume of weak acid and strong base solutions that must be combined to produce the buffer solution. 1.0 L 1.31 mol C 3 CO mol C 3 CO L C 3CO 2 solution 1.0 L 1.00 mol NaO L NaO solution 1.00 mol NaO Mix 2.6 L of 0.50 M C 3 CO 2 with 1.00 L of 1.00 M NaO to produce a buffer with a p of Note that any ratio of these volumes will produce the buffer with a p of For example, combing 1.0 L of 0.50 M C 3 CO 2 with 0.38 L of 1.00 M NaO also produces a buffer with a p of Video Solution Tutored Practice Problem Unit 17 Advanced Acid Base Equilibria 569

18 Alpha Plots As shown in the preceding example problems, the p of a buffer solution is controlled by the relative amounts of weak acid and conjugate base present and by the weak acid pk a. That is, buffer p is independent of solution volume. If a buffer solution is diluted, the p does not change. Regardless of the amounts of weak acid and conjugate base that are present in the solution, however, the relative amounts can change if the p is changed by external agents such as a strong acid or a strong base. Consider the acetic acid/acetate ion buffer system: C 3 CO 2 (aq) + 2 O(l) 3 O + (aq) + C 3 CO 2 (aq) When the solution is highly acidic, the acid form predominates and very little of the conjugate base is present ([C 3 CO 2 ] >> [C 3 CO 2 ]). When the solution is highly basic, the base form predominates and very little of the acid form is present ([C 3 CO 2 ] >> [C 3 CO 2 ]). We quantify this using an alpha (a) value, which defines the mole fraction of the acid base pair present as either the acid or the base. For the acetic acid/acetate ion buffer system, a C3 CO 2 5 a C3 CO mol C 3 CO 2 mol C 3 CO 2 1 mol C 3 CO 2 2 mol C 3 CO 2 2 mol C 3 CO 2 1 mol C 3 CO 2 2 The relationship between p and solution composition is shown graphically in an alpha (a) plot, a plot of solution composition (a) versus p. The alpha plot for the acetic acid/acetate ion buffer system shown in Interactive Figure has the following features: When the solution p is more than about 2 p units below pk a (.7), the solution consists of almost all weak acid and almost no conjugate base. When the solution p is more than about 2 p units above pk a (.7), the solution consists of almost all conjugate base and almost no weak acid. When the solution p is near pk a, the solution contains a significant concentration of both weak acid and conjugate base. When the solution p is equal to pk a, the solution is composed of equal parts weak acid and weak base (a = 0.5). The ammonia/ammonium ion alpha plot (Figure ) shows that equal amounts of weak acid and conjugate base are found when the p is equal to the ammonium ion pk a (9.26). Interactive Figure Explore alpha plots. Alpha Fraction of C 3 CO p The acetic acid/acetate ion alpha plot Alpha Fraction of N + Section 17.2 Mastery Fraction of C 3 CO 2 Fraction of N p Figure The ammonia/ammonium ion alpha plot 2013 Cengage Learning 2013 Cengage Learning Unit 17 Advanced Acid Base Equilibria 570

19 17.3 Acid Base Titrations 17.3a Strong Acid/Strong Base Titrations As you learned in Chemical Reactions and Solution Stoichiometry (Unit ) an acid base titration is when one reactant with a known concentration is placed in a buret and is added to the other reactant (of unknown concentration). The progress of the reaction is monitored externally using an acid base indicator or p meter, and the titration end point is used to calculate the unknown concentration. We will now consider how p changes during acid base titrations involving strong and weak acids and bases. We will construct p titration plots (a plot of solution p versus volume of added titrant, the substance being added during the titration) that reveal information about the nature of the acid or base under study, the equivalence point in the reaction, and, for weak acids and bases, the weak acid K a (or weak base K b ). The plot of p versus volume of NaO solution for the titration of 50.0 ml of M Cl with M NaO is shown in Interactive Figure The strong acid/strong base titration plot has four distinct regions: 1. The initial p is less than 7 because of the presence of an acid. 2. The p rises slowly with the addition of strong base as the base is completely consumed and a p-neutral salt is formed. 3. Near the equivalence point, where the acid is completely consumed by added base, the p increases rapidly. The midpoint of this vertical section of the plot (where the p is 7.00) is the equivalence point of the titration.. After the equivalence point, the p is high and increases slowly as excess base is added. The p plot for a strong base/strong acid titration (Figure ) has a similar shape. Again, there are four distinct regions in the p plot when a strong base is titrated with a strong acid: 1. The initial p is greater than 7 because of the presence of a base. 2. The p decreases slowly with the addition of strong acid as the acid is completely consumed and a p-neutral salt is formed. Unit 17 Advanced Acid Base Equilibria 571

20 3. Near the equivalence point, where the base is completely consumed by added acid, the p decreases rapidly. The midpoint of this vertical section of the plot (where the p is 7.00) is the equivalence point of the titration.. After the equivalence point, the p is low and decreases slowly as excess acid is added. Interactive Figure Explore a strong acid/strong base p titration plot. 1 As shown in the following example problem, we can use solution stoichiometry to calculate the p at four different points of a strong acid/strong base titration. In some of these calculations we will again use a stoichiometry table to assist us in keeping track of the initial amounts of reactants and the amounts of reactants and products in the solution after a reaction takes place. Note that a stoichiometry table is different from an ICE table, which is used to keep track of changes that occur in an equilibrium system. p Example Problem Calculate p for a strong acid/strong base titration. Determine the p during the titration of 75.0 ml of M Cl by M NaO at the following points: a. Before the addition of any NaO b. After the addition of 20.0 ml of NaO c. At the equivalence point d. After the addition of ml of NaO Volume of NaO added (ml) p plot for the titration of Cl with NaO Cengage Learning Solution: 12 You are asked to calculate the p at four different points in a strong acid/strong base titration. You are given the volume and concentration of the strong acid and the concentration of the strong base. p Write the balanced net ionic equation for the acid base reaction. 3 O + (aq) + O (aq) S 2 2 O(l) a. Cl is a strong acid and is 100% dissociated in solution, so [ 3 O + ] = [Cl] = M. p = log(0.100) = b. Before the equivalence point, all NaO is consumed by excess Cl. Set up a stoichiometry table to track the amounts of reactants and products present in solution before and after the reaction is complete. c Volume of Cl added (ml) Figure p plot for the titration of NaO with Cl 2013 Cengage Learning Unit 17 Advanced Acid Base Equilibria 572

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