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y 0.20 Kb 1.1x10 10 1081 y 0.20 Kb 1.1x10 10 y = 1.0 x 10-5 M [OH-] = 1.0 x 10-5 M 1082 y 0.20 Kb 1.1x10 10 Hence y = 1.0 x 10-5 M [OH-] = 1.0 x 10-5 M pOH = 5.0 1083 y 0.20 Kb 1.1x10 10 Hence y = 1.0 x 10-5 M [OH-] = 1.0 x 10-5 M pOH = 5.0 Since pOH + pH = 14.0, therefore pH = 9.0 1084 Acid-base titrations: The impact of hydrolysis 1085 Acid-base titrations: The impact of hydrolysis Salt hydrolysis has an important effect on the pH profile of acid-base titrations. 1086 Acid-base titrations: The impact of hydrolysis Salt hydrolysis has an important effect on the pH profile of acid-base titrations. The equivalence point may be above or below neutral conditions (i.e. pH = 7). 1087 Acid-base titrations: The impact of hydrolysis Salt hydrolysis has an important effect on the pH profile of acid-base titrations. The equivalence point may be above or below neutral conditions (i.e. pH = 7). For the titration of a strong acid and a strong base, the equivalence point should be at pH = 7. 1088 Acid-base titrations: The impact of hydrolysis Salt hydrolysis has an important effect on the pH profile of acid-base titrations. The equivalence point may be above or below neutral conditions (i.e. pH = 7). For the titration of a strong acid and a strong base, the equivalence point should be at pH = 7. Example: HCl(aq) + NaOH(aq) NaCl(aq) + H2O 1089 1090 1091 1092 1093 Indicators 1094 Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. 1095 Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. During a titration such as (where we assume NaOH is being added) 1096 Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. During a titration such as (where we assume NaOH is being added) HCl(aq) + NaOH(aq) NaCl(aq) + H2O 1097 Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. During a titration such as (where we assume NaOH is being added) HCl(aq) + NaOH(aq) NaCl(aq) + H2O The first drop of excess NaOH then reacts with the indicator that is present: 1098 Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. During a titration such as (where we assume NaOH is being added) HCl(aq) + NaOH(aq) NaCl(aq) + H2O The first drop of excess NaOH then reacts with the indicator that is present: HIn(aq) + OH-(aq) H2O + In-(aq) 1099 Indicators Indicators are often very weak organic acids. We will represent an indicator as HIn. During a titration such as (where we assume NaOH is being added) HCl(aq) + NaOH(aq) NaCl(aq) + H2O The first drop of excess NaOH then reacts with the indicator that is present: HIn(aq) + OH-(aq) H2O + In-(aq) Now HIn and In- have different colors, so we can detect that the acid-base reaction is complete. 1100 + For the equilibrium: HIn(aq) H + In (aq) (aq) 1101 + For the equilibrium: HIn(aq) H + In (aq) (aq) [H ][In ] KIn [HIn ] 1102 + For the equilibrium: HIn(aq) H + In (aq) (aq) [H ][In ] KIn [HIn ] Midway in the transition of the indicator color change: [HIn] = [In-], and hence KIn = [H+] (midway point). 1103 + For the equilibrium: HIn(aq) H + In (aq) (aq) [H ][In ] KIn [HIn ] Midway in the transition of the indicator color change: [HIn] = [In-], and hence KIn = [H+] (midway point). Take the log of both sides of this relationship, leads to pKIn = pH (midway point). 1104 1105 IONIC EQUILIBRIUM 1106 IONIC EQUILIBRIUM Buffers 1107 Buffers Buffer: A solution whose pH remains approximately constant despite the addition of small amounts of either acid or base. 1108 Buffers Buffer: A solution whose pH remains approximately constant despite the addition of small amounts of either acid or base. A buffer is a combination of species in solution that maintains an approximately constant pH by virtue of a pair of chemical reactions. 1109 Buffers Buffer: A solution whose pH remains approximately constant despite the addition of small amounts of either acid or base. A buffer is a combination of species in solution that maintains an approximately constant pH by virtue of a pair of chemical reactions. One reaction describes a reaction of a buffer component with added acid, the other reaction describes the reaction of a buffer component with added base. 1110 Example: acetic acid/sodium acetate buffer 1111 Example: acetic acid/sodium acetate buffer A solution containing these two substances has the ability to neutralize both added acid and added base. 1112 Example: acetic acid/sodium acetate buffer A solution containing these two substances has the ability to neutralize both added acid and added base. If base is added to the buffer, it will react with the acid component: CH3CO2H(aq) + OH-(aq) CH3CO2-(aq) + H2O 1113 Example: acetic acid/sodium acetate buffer A solution containing these two substances has the ability to neutralize both added acid and added base. If base is added to the buffer, it will react with the acid component: CH3CO2H(aq) + OH-(aq) CH3CO2-(aq) + H2O If acid is added to the buffer, it will react with the base component: CH3CO2-(aq) + H+(aq) CH3CO2H(aq) 1114 Example: acetic acid/sodium acetate buffer A solution containing these two substances has the ability to neutralize both added acid and added base. If base is added to the buffer, it will react with the acid component: CH3CO2H(aq) + OH-(aq) CH3CO2-(aq) + H2O If acid is added to the buffer, it will react with the base component: CH3CO2-(aq) + H+(aq) CH3CO2H(aq) Note that the Na+ is not directly involved in the buffer chemistry. 1115 Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation 1116 Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation Consider the equilibrium: CH3CO2H(aq) CH3CO2-(aq) + H+(aq) 1117 Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation Consider the equilibrium: CH3CO2H(aq) CH3CO2-(aq) + H+(aq) [H ][CH3CO2- ] Ka [CH3CO2H] 1118 Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation Consider the equilibrium: CH3CO2H(aq) CH3CO2-(aq) + H+(aq) [H ][CH3CO2- ] Ka [CH3CO2H] Now take the log of both sides of the preceding equation, to obtain 1119 Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation Consider the equilibrium: CH3CO2H(aq) CH3CO2-(aq) + H+(aq) [H ][CH3CO2- ] Ka [CH3CO2H] Now take the log of both sides of the preceding equation, to obtain -] [CH CO 3 2 logKa log [H ] [CH3CO2H] 1120 That is, [CH CO ] 3 2 logKa log[H ] log [CH3CO2H] 1121 That is, [CH CO ] 3 2 logKa log[H ] log [CH3CO2H] [CH CO ] 3 2 log[H ] logKa log [CH3CO2H] 1122 That is, [CH CO ] 3 2 logKa log[H ] log [CH3CO2H] [CH CO ] 3 2 log[H ] logKa log [CH3CO2H] [CH3CO2- ] pH pKa log [CH3CO2H] 1123 That is, [CH CO ] 3 2 logKa log[H ] log [CH3CO2H] [CH CO ] 3 2 log[H ] logKa log [CH3CO2H] [CH3CO2- ] pH pKa log [CH3CO2H] This is the Henderson-Hasselbalch equation for the acetic acid system. 1124 We could repeat the previous approach for the weak acid HA to obtain: 1125 We could repeat the previous approach for the weak acid HA to obtain: [A ] pH pKa log [HA] 1126 We could repeat the previous approach for the weak acid HA to obtain: [A ] pH pKa log [HA] In a more general form it would be: pH pKa log [conjugate base] [conjugate acid] 1127 We could repeat the previous approach for the weak acid HA to obtain: [A ] pH pKa log [HA] In a more general form it would be: pH pKa log [conjugate base] [conjugate acid] Either of the preceding two equations are called the Henderson-Hasselbalch equation. 1128 Example: Calculate the pH of a buffer system containing 1.0 M CH3CO2H and 1.0 M NaCH3CO2. What is the pH of the buffer after the addition of 0.10 moles of gaseous HCl to 1.00 liter of the buffer solution? The Ka for acetic acid is 1.8 x 10-5. 1129 Example: Calculate the pH of a buffer system containing 1.0 M CH3CO2H and 1.0 M NaCH3CO2. What is the pH of the buffer after the addition of 0.10 moles of gaseous HCl to 1.00 liter of the buffer solution? The Ka for acetic acid is 1.8 x 10-5. Because acetic acid is a weak acid, we can ignore the small amount of dissociation and assume at equilibrium that [CH3CO2H] = 1.0 M 1130 Example: Calculate the pH of a buffer system containing 1.0 M CH3CO2H and 1.0 M NaCH3CO2. What is the pH of the buffer after the addition of 0.10 moles of gaseous HCl to 1.00 liter of the buffer solution? The Ka for acetic acid is 1.8 x 10-5. Because acetic acid is a weak acid, we can ignore the small amount of dissociation and assume at equilibrium that [CH3CO2H] = 1.0 M It is also important to keep in mind that there is a lot of acetate ion present, and this will suppress the dissociation of the acetic acid (Le Châtelier’s Principle). 1131 A similar situation applies to the acetate ion, that is, we can ignore the hydrolysis of this ion. Also, the acetic acid present will suppress the hydrolysis of the acetate ion (Le Châtelier’s Principle), so that [CH3CO2-] = 1.0 M 1132 A similar situation applies to the acetate ion, that is, we can ignore the hydrolysis of this ion. Also, the acetic acid present will suppress the hydrolysis of the acetate ion (Le Châtelier’s Principle), so that [CH3CO2-] = 1.0 M Now Ka = 1.8 x 10-5 so that pKa = 4.7 1133 A similar situation applies to the acetate ion, that is, we can ignore the hydrolysis of this ion. Also, the acetic acid present will suppress the hydrolysis of the acetate ion (Le Châtelier’s Principle), so that [CH3CO2-] = 1.0 M Now Ka = 1.8 x 10-5 so that pKa = 4.7 Hence, from the Henderson-Hasselbalch equation: [CH3CO2 ] pH pKa log [CH3CO2H] 4.7 log 1.0 M 1.0 M = 4.7 1134 Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs: 1135 Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs: CH3CO2- + H+ CH3CO2H 1136 Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs: CH3CO2- + H+ CH3CO2H 0.10 mols 0.10 mols 0.10 mols (from the HCl) 1137 Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs: CH3CO2- + H+ CH3CO2H 0.10 mols 0.10 mols 0.10 mols (from the HCl) At equilibrium, the concentrations of the buffer components are (keep in mind the total volume is 1.0 liter): 1138 Upon addition of 0.10 moles of HCl to 1.0 liter of the buffer solution (we make the assumption that the total volume does not change), the following neutralization reaction occurs: CH3CO2- + H+ CH3CO2H 0.10 mols 0.10 mols 0.10 mols (from the HCl) At equilibrium, the concentrations of the buffer components are (keep in mind the total volume is 1.0 liter): [CH3CO2H] = 1.0 + 0.10 = 1.10 M [CH3CO2-] = 1.0 - 0.10 = 0.90 M 1139 To calculate the new pH, use the HendersonHasselbalch equation: 1140