Chapter 16: Acid-Base Equilibria

Transcription

1 Previous Chapter Table of Contents Next Chapter Chapter 16: Acid-Base Equilibria Section 16.1: The Common Ion Effect Here, we will discuss the acid-base properties of a solution with two solutes containing the same (i.e. a common) ion. This ion can be a cation or an anion. Consider two solutes, NaF and HF (for HF, K a = 7.2 x 10-4 ). NaF is a salt and it dissociates completely into ions when mixed with water. This is expressed by the reaction: NaF (s) Na (aq) F - (aq) In the same solution, we have HF (aq), a weak acid. The ionization of the weak acid is expressed as: HF (aq) H (aq) F - (aq) K a = 7.2 x 10-4 Note: The fact that K a > K w indicates that in this solution, H is produced predominantly by the dissociation of HF. In this solution, F - is the common ion produced from NaF and HF. The presence of the common ion, F -, may have a significant effect on the equilibrium dissociation of the weak HF acid. Let us focus on the ionization reaction of the weak acid, HF. HF (aq) H (aq) F - (aq) What is the consequence of adding NaF to an equilibrium solution of HF? Adding NaF to the HF solution increases [F - ] and disrupts the equilibrium. Recall Le Châtelier s Principle: When a disturbance is imposed on a reaction system at equilibrium, the equilibrium shifts in the direction that reduces the effects of the disturbance. Hence, increasing [F - ] by addition of NaF results in a shift of the equilibrium to the left. In turn, this shift reduces [H ], and the ph. This is called the common ion effect. In summary, the common ion effect is the shift of an equilibrium caused by addition of a solute having an ion in common with the equilibrium system.

2 The common ion effect has two important roles: 1) It allows one to change the ph of a solution. 2) It allows one to change the solubility of salts that are slightly soluble in water. In this chapter, we will focus on 1) the common ion effect for controlling the ph of a solution. The common ion effect in solubility problems is discussed in Chapter 17. Sections : Equilibrium Calculations for Acids with the Common Ion Effect Consider a general solution of a weak acid, HA, and a salt, NaA. The salt dissociates completely into ions as expressed by the reaction: NaA (s) Na (aq) A - (aq) Note: The complete dissociation of the salt into its ions means that if we are given 0.1 M NaA, then [Na ] = 0.1 M, and [A - ] = 0.1 M. The ionization reaction of the weak acid, HA, is expressed as: HA (aq) H (aq) A - (aq) Recall that A - is the weak conjugate base of the weak acid HA. The equilibrium constant for the ionization reaction is expressed as: K a = [ H ] [ A ] [ HA] Rearrange the equation: K [ HA] [ H a ] = [ A ] Take the negative log of both sides: log [ ] = log log [ HA H K ] a [ A ] Note: - log [H ] = ph and log K a = pk a

3 HA Thus, ph = pk a log [ ] [ A ] A Hence, pk a log [ ph = ] [ HA] Note: Switching the numerator and the denominator reverses the sign. A - is the weak conjugate base of the weak acid HA. conjugate base Therefore, ph = pk a log [ ] [ acid] This equation is called the Henderson-Hasselbalch equation. One can use this equation to calculate the ph for solutions of compounds having a common ion. In problems that involve the common ion effect, the initial concentrations of the salt and the weak acid are usually given. As long as the initial concentrations of the salt and weak acid are greater than 0.1 M, we can neglect the ionization of the acid and the hydrolysis of the salt (reaction of the salt with water). Under such conditions, we can use the initial concentrations for the acid and the conjugate base in the Henderson-Hasselbalch equation. Example: (a) Calculate the ph of a 0.15 M CH 3 COOH solution (for CH 3 COOH, K a = 1.8 x 10-5 ). CH 3 COOH is a weak acid which partially dissociates in water. Let us call x the amount of acetic acid that dissociates. Its ionization reaction is expressed as: CH 3 COOH (aq) H (aq) CH 3 COOH - (aq) Initial 0.15 M 0 M 0 M Change - x x x Equilibrium (0.15 x) x x The equilibrium constant for the above reaction is expressed as: K [ H ] [CH COO - 3 ] [CH COOH] a = 3 Substituting the concentrations by their respective values, we get:

4 1.8 x 10 5 (x)(x) = (0.15 x) If x is assumed to be negligible in comparison to 0.15 in (0.15 x), then 2 5 x x 10 = x = 1.5 x Check the assumption that x is negligible in comparison to The % dissociation for the acetic acid is expressed as: x 10 % = 0.15 x 100 = 1.1% Hence, the assumption is valid. x = [H ] = 1.6 x 10-3 ph = - log(1.6 x 10-3 ) ph = 2.80 Example: (b) Calculate the ph of a solution which contains 0.25 M CH 3 COONa and 0.15 M CH 3 COOH (K a = 1.8 x 10-5 for CH 3 COOH). The salt, CH 3 COONa, dissociates completely in water. CH 3 COONa (s) Na (aq) CH 3 COO - (aq) Therefore, [Na ] = 0.25 M and [CH 3 COO - ] = 0.25 M. The ionization reaction of the weak acid is written as: CH 3 COOH (aq) H (aq) CH 3 COO - (aq) CH 3 COO - is the weak conjugate base of the weak acid, CH 3 COOH. Note: Since K a is very small, the fraction of the acid that is dissociated is very small. Hence, [CH 3 COO - ] produced by the salt (0.25 M) is much greater than that produced by the acid.

5 Hence, [conjugate base] = [CH 3 COO - ] = 0.25 M and [acid] = [CH 3 COOH] = 0.15 M K a = 1.8 x 10-5 pk a = - log(1.8 x 10-5 ) pk a = 4.74 Use the Henderson-Hasselbalch equation to calculate the ph. conjugate base ph = pk a log [ ] [ acid] 025 = 474. log (. ) ( 015. ) In conclusion, when a salt containing the common ion is added to the weak acid solution, the ph increases from 2.80 to In Section 16.3, practice the Interactive Problems. Sections : Equilibrium Calculations for Bases with the Common Ion Effect The common ion effect can also be applied to a solution containing a weak base and a salt of the base. Consider a solution of a weak base, NH 3, and a salt, NH 4 Cl. The salt, NH 4 Cl, dissociates completely in H 2 O according to the reaction: NH 4 Cl (s) NH 4 (aq) Cl - (aq) We know that NH 4 is an acid, and that it can donate a proton to water. Thus, NH 4 (aq) NH 3 (aq) H (aq) The equilibrium constant expression for this reaction is: K a = [ NH ] [ H ] 3 4 [ NH ]

6 Rearrange this expression to solve for [H ]. [ H ] Take the negative log of both sides. = Ka [ NH4 ] [ NH ] 3 log [H ] = log K - log a 4 [ NH ] [ NH ] 3 ph = pk a log 4 [ NH ] [ NH ] 3 NH ph = pk a log [ 3 ] [ NH ] Recall that NH 4 is the weak conjugate acid of the weak base, NH 3. 4 ph = pk log a [ base] [ conjugate acid] In Summary: For a weak acid-weak conjugate base pair (such as HF/F - ), the Henderson- Hasselbalch equation is written as: conjugate base ph = pk a log [ ] [ acid] For a weak base-weak conjugate acid pair (such as NH 3 /NH 4 ), the Henderson- Hasselbalch equation is written as: ph = pk log a [ base] [ conjugate acid] From here onwards, whether we consider a weak acid and its conjugate base or a weak base and its conjugate acid, the Henderson-Hasselbalch equation will be written as: base ph = pk a log [ ] [ acid]

7 Example: Calculate the ph of a solution that is 0.25 M NH 3 (K b = 1.8 x 10-5 for NH 3 ) and 0.30 M NH 4 Cl. NH 4 Cl is a salt that dissociates completely in water. NH 4 Cl (s) NH 4 (aq) Cl - (aq) Since [NH 4 Cl] = 0.30 M initially, then [NH 4 ] = 0.30 M, and [Cl - ] = 0.30 M at the end of the dissolution process. Remember that NH 4 is a weak acid which can donate one of its protons to H 2 O according to the reaction: Ignoring water: NH 4 (aq) H 2 O (l) NH 3 (aq) H 3 O (aq) NH 4 (aq) NH 3 (aq) H (aq) NH 3 is the base and we are given K b = 1.8 x However, the above reaction is the deprotonation of an acid. Therefore, we need to calculate K a. Recall K a x K b = K w K a x (1.8 x 10-5 ) = K w K a = 5.6 x Thus, we know that: [acid] = [NH 4 ] = 0.30 M [base] = [NH 3 ] = 0.25 M K a = 5.6 x pk a = 9.25 Now, use the Henderson-Hasselbalch equation to calculate ph. base ph = pk a log [ ] [ acid] 025 = 925. log (. ) ( 030. )

8 In Section 16.5, practice the Interactive Problems. Sections : Buffer Solutions Buffer solutions have the ability to resist changes in ph upon addition of small amounts of either acid or base. Buffers are very important in chemical and physiological processes. The ph in the human body varies from one fluid to another. For example, the ph of blood is 7.4, whereas the ph of gastric juices in the stomach is about 1.5. Blood can absorb the acids and bases produced in biological reactions without changing its ph. Thus, blood is a buffer solution. A buffer solution consists of a weak acid and its salt or a weak base and its salt. In other words, a buffer solution follows the concepts of common ion solutions. An example of an acidic buffer solution is a mixed solution of the weak acid, HF, and its salt, NaF. Similarly, an example of a basic buffer solution is a mixed solution of the weak base, NH 3, and its salt, NH 4 Cl. By choosing appropriate solutes and concentrations, a buffer solution can be created for any desired ph. In Section 16.7, practice the Interactive Problems. Sections : The Working of a Buffer Solution Consider a buffer solution of a weak acid, HA, and its salt, NaA. The salt dissociates as: and the weak acid as: NaA (s) Na (aq) A - (aq) HA (aq) H (aq) A - (aq) When the buffer solution is at equilibrium, we can calculate the ph using the Henderson-Hasselbalch equation. ph = pk a A log [ ] [ HA]

9 Now, let s disturb this buffer solution by adding some NaOH. NaOH dissociates completely in solution: NaOH (s) Na (aq) OH - (aq) OH - is a strong base and accepts a proton from the acid. In solution, we have the acid, HA. Thus, OH - reacts with HA as: OH - (aq) HA (aq) H 2 O (l) A - (aq) Hence, adding a strong base to an acid buffer leads to the formation of more A -. Now, look at the equilibrium dissociation of the weak acid. HA (aq) H (aq) A - (aq) The equilibrium constant for this reaction is expressed as: According to Le Châtelier s principle, increasing [A - ] leads to a shift of the equilibrium to the left in the direction that consumes A -. Thus, the shift of the equilibrium results in an increase in [HA]. [ HA] Thus, the ratio changes very little. [ A ] This implies that [H ] changes very little. Hence, the ph of an acid buffer solution does not change much when a strong base is added. Example: (a) Calculate the ph of a buffer solution that contains 0.40 M HC 2 H 3 O 2 (K a = 1.8 x 10-5 ) and 0.40 M NaC 2 H 3 O 2. NaC 2 H 3 O 2 dissociates completely in H 2 O. NaC 2 H 3 O 2 (aq) Na (aq) C 2 H 3 O 2 - (aq) Given the information provided and the stoichiometry: [Na ] = [C 2 H 3 O 2 - ] = 0.40 M HC 2 H 3 O 2 is a weak acid whose concentration is given as 0.40 M: [HC 2 H 3 O 2 ] = 0.40 M

10 K a = 1.8 x Hence, pk a = 4.74 Use the Henderson-Hasselbalch equation to calculate ph. base ph = pk a log [ ] [ acid] CHO ph = pk a log [ ] [ HC H O ] = 474. log (. ) ( 040. ) Example: (b) Calculate the ph when mol solid NaOH is added to one liter of the buffer solution discussed in part (a). NaOH dissociates completely in H 2 O: NaOH (s) Na (aq) OH - (aq) From the information given, [OH - ] = M. OH - is a strong base; thus, it accepts a proton from the acid. The only acid in the solution is HC 2 H 3 O 2. Hence, the reaction is: OH - (aq) HC 2 H 3 O 2(aq) H 2 O (l) C 2 H 3 O 2 - (aq) Initial: M 0.40 M 0.40 M Change M M M Final 0 M 0.39 M 0.41 M The Henderson-Hasselbalch equation is written as: CHO ph = 474. log [ ] [ HC H O ] 2 3 2

11 = 474. log The ratio in the Henderson-Hasselbalch equation here is 1.05, whereas the ratio was 1.0 in part (a). This ratio has changed very little. Since the ratio did not change much, there is no significant change in ph. ph = = 4.76 This is how buffer solutions resist a change in ph. In Section 16.9, practice the Interactive Problems. Sections : Preparing a Buffer Solution with a Specific ph In order to calculate the ph of a buffer solution, one uses the Henderson-Hasselbalch equation. base ph = pk a log [ ] [ acid] Here, A - is the base and HA is the acid. HA (aq) H (aq) A - (aq) K [ H ] [ A ] [ HA] a = A Thus, pk a log [ ph = ] [ HA] Recall, it is the ratio [A - ]/[HA] that is the most resistant to change when either H or OH - is added to the buffer solution. The concentrations of A - and HA in the buffer solution are usually chosen in such a way that the ratio [A - ]/[HA] is as close as possible to 1. Thus, ph = pk a log (1) log (1) = 0 Hence, ph = pk a

12 Thus, when preparing a buffer solution of a desired ph, one should choose a weak acid of pk a values as close as possible to the desired ph. Example: Consider the following acids: Acid K a HClO x 10-2 HF 7.2 x 10-4 HClO 3.5 x 10-8 HCN 6.2 x a) Which acid would be the best choice for the preparation of a ph = 7.00 buffer? First, calculate the pk a values for these acids. Recall that pk a = - log K a Acid K a pk a HClO x HF 7.2 x HClO 3.5 x HCN 6.2 x Recall the Henderson-Hasselbalch equation to calculate ph. base ph = pk a log [ ] [ acid] Thus, ph = pk a The only pk a value which is close to ph = 7.00 is that of HClO. Therefore, HClO would be the best weak acid. b) Calculate the ratio [base]/[acid]. needed for a ph = 7.00 buffer given that pk a = 7.46 for HClO. Substitute the ph and pk a values in the Henderson-Hasselbalch equation to solve for the ratio [base]/[acid] log [ base = ] [ acid]

13 046. = log [ base ] [ acid] [ base] [ acid ] = [ base]. [ acid ] = 035 c) How would you prepare the buffer solution? To ensure that the ph = 7.00, the ratio should be [base]/[acid] = 0.35 = 0.35 / 1. Since we selected HClO as the acid, the conjugate base is ClO -. One of the simplest salts for this acid-base pair is NaClO. [NaClO] and [HClO] should be equal to 0.35 M and 1.0 M, respectively. By selecting the concentrations of NaClO and HClO to be 0.35 M and 1.00 M, respectively, we can prepare a buffer solution for ph = In Section 16.11, practice the Interactive Problem. Section 16.12: Acid-Base Titrations and ph Curves Titration is used to determine the amount of acid or base in a solution. The solution being analyzed, called the analyte, is the acid or the base solution in an Erlenmyer flask or a beaker. The titrant is the acid or the base solution in the buret. The progress of an acid-base titration is often shown on a graph of the ph of the solution being analyzed vs. the amount of titrant added. Such a graph is called a ph curve, or a titration curve. In this section, we will consider three types of reactions: a) Strong acid-strong base titrations b) Weak acid-strong base titrations c) Strong acid-weak base titrations Section 16.13: Strong Acid - Strong Base Titrations

14 Example: Consider the titration of 50.0 ml of M HCl with M NaOH. Calculate the ph during the course of the titration when specified volumes of NaOH have been added. Note: Here, the analyte is the 50.0 ml solution of M HCl. Thus, the HCl solution is placed in the Erlenmyer flask or beaker. The buret is filled with the titrant, the M NaOH solution. The titrant is added to the analyte, or solution being analyzed. Case 1: Calculate the ph of the HCl solution before any NaOH is added. At this stage, the solution is simply 50.0 ml of M HCl. HCl is a strong acid and is fully ionized in water: HCl (aq) H (aq) Cl - (aq) Considering the molarity of the HCl solution and the stoichiometry of the ionization reaction, we conclude that [H ] = M. ph = - log([h ]) ph = - log(0.200) ph = Case 2: Calculate the ph when 15.0 ml of M NaOH has been added. The analyte solution is a solution of the strong acid, HCl. Hence, HCl is fully dissociated as H and Cl -. The titrant solution is a NaOH solution, and NaOH is a strong base. Hence, in the titrant solution, NaOH is fully dissociated: NaOH (aq) Na (aq) OH - (aq) Thus, after adding 15.0 ml of titrant to the analyte solution, the resulting solution contains H, Cl -, Na and OH - ions. Thus, the only reaction occurring in the Erlenmyer flask is: Na and Cl - are spectator ions. H (aq) OH - (aq) H 2 O (l) Since the limiting reactant is OH -, the amount of this reactant is used to calculate the number of moles of H at the end of the reaction.

15 The total volume is: 50.0 ml 15.0 ml = 65.0 ml = L mol [ H ] = = 013. M L Case 3: Calculate the ph when ml of M NaOH is added. The only reaction occurring in the Erlenmyer flask is: H (aq) OH - (aq) H 2 O (l) Initial # moles H Initial # moles OH - mol = L x M = mol mol = L x M = mol In this case, the numbers of moles of H and OH - are equal. Since these reactants are present in the stoichiometric ratio, they fully react with each other and only H 2 O should be left at the end of the reaction. However, in the above discussion, we ignored the autoionization of H 2 O. Since [H ] = [OH - ] = 10-7 in pure H 2 O at equilibrium, the ph is This is the stoichiometric point or the equivalence point of the titration. At this point, enough OH - is added to react exactly with all H. Case 4: Calculate the ph when ml of M NaOH has been added. The only reaction occurring in the Erlenmyer flask is: H (aq) OH - (aq) H 2 O (l) Initial # moles H Initial # moles OH - mol = M x L = mol mol = M x L = mol Since the limiting reactant is H, the amount of this reactant is used to calculate the number of moles of OH - at the end of the reaction.

16 The total volume is 50.0 ml ml = ml = L mol [ OH ] = = L M Sections : ph Curve for Strong Acid - Strong Base Titrations The plot of ph vs. Volume of Titrant Added is called the Titration Curve, or the ph Curve. In the previous example, the titrant was NaOH. Hence, the ph curve for a strong acid-strong base titration is:

17 In this curve, the equivalence point is at ph = Before the equivalence point, [H ] and ph can be calculated by dividing the moles of [H ] remaining by the total volume of the solution (acid base) in Liters. After the equivalence point, [OH - ] and the poh can be calculated by dividing the moles of OH - remaining by the total volume of the solution (acid base) in Liters. Note: In order to calculate ph, use the relation ph poh = If the titrant is HCl, then the ph curve for a strong acid-strong base titration is:

18 In this curve, the equivalence point is at ph = Before the equivalence point, [OH - ] and the poh can be calculated by dividing the moles of OH - remaining by the total volume of the solution (acid base) in Liters. After the equivalence point, [H ] and the ph can be calculated by dividing the moles of H remaining by the total volume of the solution (base acid) in Liters. Note: In order to calculate ph, use the relation ph poh = In Section 16.15, practice the Interactive Problems. Section 16.16: Weak Acid - Strong Base Titrations Calculating the ph during the titration of a weak acid by a strong base is a two-step process. 1) The stoichiometric step: The reaction of OH - with a weak acid is assumed to be complete. The concentrations of acid remaining in solution and conjugate base formed are determined. 2) The equilibrium step: The equilibrium constant for dissociation of the weak acid, the acid and the conjugate base concentrations are used to calculate [H ], then the ph.

19 Example: Consider the titration of 50.0 ml of 0.10 M HC 2 H 3 O 2 solution (K a = 1.8 x 10-5 ) with 0.10 M NaOH. Calculate the ph during the course of the titration for specified volumes of NaOH added. Note: In this example, 50.0 ml of 0.10 M HC 2 H 3 O 2 is the analyte (solution being analyzed), and 0.10 M NaOH is the titrant solution placed in the burette. Case 1: Calculate the ph of the solution before NaOH is added. At this stage, the solution is simply 50.0 ml of 0.10 M HC 2 H 3 O 2. HC 2 H 3 O 2 is a weak acid. Let us call x the amount of acid that is dissociated. The equilibrium dissociation is written as: HC 2 H 3 O 2(aq) H (aq) C 2 H 3 O 2 - (aq) Initial 0.10 M 0 M 0 M Change - x x x Equilibrium (0.1 x) x x H C H O K a = [ ] [ ] [HC H O ] x 10 5 (x)(x) = (0.10 x) Assume x to be negligible in comparison to 0.10 in (0.10 x), then 2 5 x -3 x 10 = x 1.3 x = x = [H ] = 1.3 x 10-3 ph = - log([h ]) ph = - log(1.3 x 10-3 ) ph = 2.89 Case 2: Calculate the ph when 10.0 ml of 0.10 M NaOH is added. Now, we follow the two-step approach. 1) The stoichiometric step: The reaction of OH - with a weak acid is assumed to be complete. The concentrations of acid remaining in solution and conjugate base formed are determined. The reaction between OH - and HC 2 H 3 O 2 is written as:

20 OH - (aq) HC 2 H 3 O 2 (aq) H 2 O (l) C 2 H 3 O 2 - (aq) Initial # moles OH - Initial # moles HC 2 H 3 O 2 mol = M x L = 0.10 M x L = mol mol = M x L = 0.10 M x L = mol Since the limiting reactant is OH -, the amount of this reactant is used to calculate the number of moles of HC 2 H 3 O 2 left at the end of the reaction.. [ CHO mol ] = = M L 2) The equilibrium step: The equilibrium constant for dissociation of the weak acid, the acid and the conjugate base concentrations are used to calculate [H ], and ph. The equilibrium dissociation is written as: HC 2 H 3 O 2(aq) H (aq) C 2 H 3 O 2 - (aq) Initial M 0 M M Change - x x x Equilibrium (0.067 x) x (0.017x) H C H O K a = [ ] [ ] [HC H O ] x 10 5 (x)(0.017 x) = (0.067 x) Assume x to be negligible in comparison to (and 0.067), then (x)(0.017) 1.8 = (0.067) 5-5 x 10 = x 7.1 x 10 x = [H ] = 7.1 x 10-5 ph = - log([h ]) ph = - log(7.1 x 10-5 )

21 ph = 4.15 Case 3: Calculate the ph after addition of 50.0 ml of 0.10 M NaOH. 1) The stoichiometric step: The reaction of OH - with a weak acid is assumed to be complete. The concentrations of acid remaining in solution and conjugate base formed are determined. The reaction between OH - and HC 2 H 3 O 2 is written as: OH - (aq) HC 2 H 3 O 2 (aq) H 2 O (l) C 2 H 3 O 2 - (aq) Initial # moles OH - = 0.10 M x L = mol Initial # moles HC 2 H 3 O 2 = 0.10 M x L = mol OH - was added in the amount necessary to fully react with HC 2 H 3 O 2 present. This point is called the stoichiometric point or the equivalence point in a weak acid-strong base titration. As a result of this reaction, the concentrations change by:. [ CHO mol ] = = M L 2) The equilibrium step: The equilibrium constant for the reaction of the weak conjugate base with water, the acid and the conjugate base concentrations are used to calculate [OH - ], then the ph. The equilibrium reaction between C 2 H 3 O 2 - and H 2 O is written as: C 2 H 3 O 2 - (aq) H 2 O (l) HC 2 H 3 O 2 (aq) OH - (aq) Initial M 0 M 0 M Change - x x x Equilibrium (0.050 x) x x For this reaction, the equilibrium constant is expressed as:

22 K b = [ HC H O ] [ OH ] - [C2H3O 2 ] Recall K a x K b = K w K a x (1.8 x 10-5 ) = K w = 1.0 x K a = 5.6 x x (x)(x) = (0.50 x) Assume x to be negligible in comparison to in (0.050 x), then, 2 10 x -6 x 10 = x 5.3 x = x = [OH - ] = 5.3 x 10-6 poh = - log([oh - ]) poh = - log(5.3 x 10-6 ) poh = 5.28 ph poh = ph 5.28 = ph = 8.72 Thus, in the titration of a weak acid by a strong base, the equivalence point is always at a ph greater than Case 4: Calculate the ph after addition of 25.0 ml of 0.10 M NaOH. Note: This is a special point on the titration curve. At this point, the amount of base added is half the amount added at the equivalence point. 1) The stoichiometric step: The reaction of OH - with a weak acid is assumed to be complete. The concentrations of acid remaining in solution and conjugate base formed are determined. The reaction between OH - and HC 2 H 3 O 2 is written as: OH - (aq) HC 2 H 3 O 2 (aq) H 2 O (l) C 2 H 3 O 2 - (aq) Initial # moles OH - = 0.10 M x L = mol Initial # moles HC 2 H 3 O 2 = 0.10 M x L = mol

23 [ HC H O ] = mol L L mol = = 033. M L. [ CHO mol ] = = 033. M L 2) The equilibrium step: The equilibrium constant for dissociation of the weak acid, the acid and the conjugate base concentrations are used to calculate [H ], and then ph. CH 3 COOH (aq) H (aq) CH 3 COO - (aq) Initial 0.33 M 0 M 0 M Change - x x x Equilibrium (0.33 x) x x K a = [ H ] [ C2H3O2 ] [HC2H3O 2] 1.8 x 10 5 (x)(0.33 x) = (0.33 x) Assume x to be negligible in comparison to 0.33 in (0.33 ± x), then, (x)(0.33) 1.8 = (0.33) 5-5 x 10 = x 1.8 x 10 x = [H ] = 1.8 x 10-5 ph = - log([h ]) ph = - log(1.8 x 10-5 ) ph = 4.74 ph = pk a Hence, when the amount of acid or base added is halfway to the equivalence point, ph = pk a.

24 Thus, at this point, one does not need to do these rigorous calculations but simply remember that ph = pk a. Sections : ph Curve for Weak Acid - Strong Base Titrations The plot of ph vs. Volume of Titrant Added is called the Titration Curve, or ph Curve. The ph curve for a weak acid-strong base titration is shown below: The equivalence point in an acid-base titration is defined by the stoichiometry. Why? Because the equivalence point occurs when sufficient titrant has been added to react exactly with all the acid or base present. In the titration curve of a weak acid by a strong base, the equivalence point is located at ph > At the equivalence point, the solution contains only a soluble salt and water. The soluble salt consists of the cation from the strong base and the anion of the weak acid. The former does not affect the ph, while the latter makes the solution basic (see Section in the Chapter on Acids and Bases).

25 The ph value at the equivalence point is affected by the strength of the acid. The weaker the acid, the stronger its conjugate base and the higher the ph value at the equivalence point. In Section 16.18, practice the Interactive Problems. Section 16.19: Strong Acid - Weak Base Titrations Consider the titration of HCl, a strong acid, with NH 3, a weak base. HCl ionizes as: HCl (aq) H (aq) Cl - (aq) NH 3 is a base; thus, it reacts with the acid by accepting a proton. The reaction is: NH 3 (aq) H (aq) NH 4 (aq) At the stoichiometric point, the solution contains H 2 O, NH 4, and Cl -. Since NH 4 is a weak acid and Cl - is an extremely weak base, the solution is acidic at the stoichiometric point. This means that NH 4 donates a proton to H 2 O, a weak base. Therefore, NH 4 (aq) H 2 O (l) NH 3 (aq) H 3 O (aq) Often, H 2 O is ignored and the reaction is written as: NH 4 (aq) NH 3 (aq) H (aq) The important point to remember is that at the stoichiometric point, the solution is acidic and is characterized by K a. However, because we start off with a weak base (NH 3 ) and the K b value, we need to calculate K a from K b. Example: Calculate the ph of a solution formed by adding ml of M NH 3 (K b = 1.8 x 10-5 ) to 10. ml of 0.10 M HNO 3. HNO 3 is a strong acid; thus, it dissociates completely in H 2 O. HNO 3 (aq) H (aq) NO 3 - (aq) From the information given, and according to the stoichiometry, [H ] = 0.10 M

26 In solution, the weak base NH 3 reacts with a proton. Assuming the limiting reactant (H ) fully reacts: NH 3 (aq) H (aq) NH 4 (aq) Initial mol mol 0 mol Change mol mol mol Equilibrium mol 0 mol mol. [ NH ] = mol L = M [ NH mol 4 ] = = M L NH 4 releases a proton in water according to the reaction: NH 4 (aq) NH 3 (aq) H (aq) Initial M M 0 M Change - x x x Equilibrium ( x) (0.036 x) x The equilibrium constant for this reaction is expressed as: K a = [ H ] [ NH3 ] [NH 4 ] In this problem, we are given K b for NH 3. Recall: K a x K b = 1.00 x x a = 5.6 x 10-5 x K = x (x)(0.036 x) = ( x) If x is assumed to be negligible in comparison to , then,

27 (x)(0.036) 5.6 = = (0.0091) x 10 x 1.4 x 10 x = [H ] ph = - log (1.4 x ) ph = 9.85 Example: Calculate the ph of a solution formed by adding ml of M NH 3 (K b = 1.8 x 10-5 ) to 50.0 ml of 0.10 M HNO 3. [H ] = 0.10 M The base in this solution is NH 3. It reacts with the proton from HNO 3. NH 3 (aq) H (aq) NH 4 (aq) Initial # moles NH 3 = L x M = mol Initial # moles H = L x 0.10 M = mol Note: The number of moles of acid is the same as the number of moles of base. Hence, this is the equivalence point of the titration.. [ NH mol 4 ] = = L M Ignoring water, and calling x is the amount of NH 4 that dissociates, the dissociation reaction of NH 4 is written as: NH 4 (aq) NH 3 (aq) H (aq) Initial M 0 M 0 M Change - x x x Equilibrium (0.033 x) x x The equilibrium constant for this reaction is expressed as:

28 K a = [ H ] [ NH3 ] [NH 4 ] In this problem, we are given K b for NH 3. Recall: K a x K b = 1.00 x x K a = = 5.6 x x x (x)(x) = (0.033 x) If x is assumed to be negligible in comparison to 0.033, then 2 10 x 5.6 x 10 = x = x M x = [H ] ph = - log (4.3 x 10-6 ) ph = 5.37 Hence, for the titration of a weak base by a strong acid, the ph at the equivalence point is always less than Example: Calculate the ph of a solution prepared by adding ml of M NH 3 (K b = 1.8 x 10-5 ) to 25.0 ml of 0.10 M HNO 3. [H ] = 0.10 M The base in this solution is NH 3. NH 3 (aq) H (aq) NH 4 (aq) Initial # moles NH 3 = L x M = mol Initial # moles H = L x 0.10 M = mol This is a very special point because the concentrations of the acid and the base are half-way to the equivalence point. Remember, during the titration of a strong acid with a weak base, when the acid-base concentrations are half-way to the equivalence point, then,

29 ph = pk a K b = 1.8 x 10-5 K a x K b = 1.0 x K a = 5.6 x pk a = - log K a pk a = - log (5.6 x ) pk a = 9.25 ph = pk a = 9.25 Note: You will reach the same answer if you follow the stoichiometry or the equilibrium calculations. Sections : ph Curve for Strong Acid - Weak Base Titrations The plot of ph vs. Volume of Titrant Added is called the Titration Curve or ph Curve. The ph curve for a strong acid-weak base titration is shown below: In this curve, the equivalence point is located at ph < This is because of the acidic nature of the conjugate acid of a weak base.

30 The equivalence point in an acid-base titration is defined by the stoichiometry. In this case, the ph value at the equivalence point is determined by the strength of the base. The weaker the base, the stronger its conjugate acid and the lower the ph value at the equivalence point. In Section practice the Interactive Problems. Section 16.22: Acid - Base Indicators In acid-base titrations, the end point is observed by the change in color of an indicator. An indicator is an organic dye whose color depends on the ph of the solution. For example: The indicator methyl red is red at ph below 4. This means that if the ph of a solution is 4 or below and a drop or two or methyl red indicator is added, then the solution turns red. The indicator methyl red is yellow at ph above 7. This means if the ph of a solution is 7 or higher and a drop or two of methyl red indicator is added, then the solution turns yellow. As the ph changes from 4 to 7, the color of the indicator changes from red, to reddishorange, to orange and to yellow. Thus, through changes in color, an indicator displays the acidity, or the basicity of a solution. Another very common acid-base indicator is phenolphthalein. Phenolphthalein is colorless in solutions with a ph below 8. Phenolphthalein turns bright pink when the ph of a solution is above 10. Acid-base indicators are generally weak organic acids. Hence, an indicator is often represented as HIn. HIn is a weak acid and its dissociation is represented as: HIn (aq) H (aq) In - (aq) Since HIn is a weak acid, it is characterized by its K a value. Assume that the K a value for the indicator HIn is 1.0 x Thus, the equilibrium constant is expressed as: K a = [ H ] [ In ] [HIn] Remember: HIn is a weak acid; thus, its conjugate base, In -, is a weak base. Rearrange this equation as:

31 K a [In- ] = [H ] [HIn] Assume that we add one or two drops of indicator to a solution of ph = 1.0. If ph = 1.0, then [H ] = 0.10 M. a ] -8 K 1.0 x 10 Thus, = = [H x , 000, 000 [ In ] = [ HIn] [HIn] = 10,000,000 x [In-] This means that the concentration of HIn is much greater than that of In -, and that the predominant species in solution is HIn. Hence, if HIn was methyl red, we know that the indicator would turn the solution red since the ph is below 4. Thus, a solution of ph = 1 with a few drops of methyl red appears red. As a base is added, the concentration of H decreases. This causes the equilibrium HIn (aq) H (aq) In - (aq) to shift to the right. At some point during the titration, [In - ] would increase to an extent that a change in color will be noticed. The question is what should be the concentration of In - present in solution for the human eye to detect a color change. In practice, the color of In - is observed if it is present in much higher concentration (ten fold) than HIn: [ In ] = 10 [ HIn] 1 Conversely, the color of HIn is observed if the concentration of HIn is ten times larger than that of In -. [ In ] [ HIn] = 1 10 Note: When choosing an indicator for an acid-base titration, make sure that the indicator end point (i.e. the point at which the color changes) is as close as possible to the equivalence point of the titration (i.e. point where the number of moles of acid and base are the same).

32 Example: Two drops of indicator HIn (K a = 1.0 x 10-9 ) are added to 50.0 ml of 0.10 M HNO 3. Solutions of HIn and In - are yellow and blue, respectively. a) What is the color of the solution? HIn (aq) H (aq) In - (aq) Yellow Blue x 10 [In ] -8 [In ] 8 - = or 1.0 x 10 = or [HIn] = 10 x [ In ] 0.10 [HIn] [HIn] Since HIn is the predominant species, the solution will be Yellow. - b) If the nitric acid solution is titrated with 0.10 M NaOH, at what ph will we observe the color change? Recall: To see a color change from that of HIn (yellow) to that of In - (blue). [ In ] = 10 [ HIn] 1 Use the Henderson-Hasselbalch equation to calculate the ph. base ph = pk a log [ ] [ acid] ph = pk a In log [ ] [ HIn] Thus, we will observe a color change for the indicator at ph of If the ph of the solution is above 10, the solution color is blue. Universal Indicators

33 These are mixtures of several acid-base indicators that display a continuous range of colors over a wide ph range. For example: A naturally occurring universal indicator is the juice of red (purple) cabbage. This indicator can be used in solution for ph ranging from 1 to 13. Previous Chapter Table of Contents Next Chapter