1081
y = 1.0 x M [OH - ] = 1.0 x M 1082
y = 1.0 x M [OH - ] = 1.0 x M Hence pOH =
y = 1.0 x M [OH - ] = 1.0 x M Hence pOH = 5.0 Since pOH + pH = 14.0, therefore pH =
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 =
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) + H 2 O 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) + H 2 O 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) + H 2 O 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) + H 2 O The first drop of excess NaOH then reacts with the indicator that is present: HIn (aq) + OH - (aq) H 2 O + 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) + H 2 O The first drop of excess NaOH then reacts with the indicator that is present: HIn (aq) + OH - (aq) H 2 O + 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 + (aq) + In - (aq) 1101
For the equilibrium: HIn (aq) H + (aq) + In - (aq) 1102
For the equilibrium: HIn (aq) H + (aq) + In - (aq) Midway in the transition of the indicator color change: [HIn] = [In - ], and hence K In = [H + ] (midway point). 1103
For the equilibrium: HIn (aq) H + (aq) + In - (aq) Midway in the transition of the indicator color change: [HIn] = [In - ], and hence K In = [H + ] (midway point). Take the log of both sides of this relationship, leads to pK In = 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: CH 3 CO 2 H (aq) + OH - (aq) CH 3 CO 2 - (aq) + H 2 O 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: CH 3 CO 2 H (aq) + OH - (aq) CH 3 CO 2 - (aq) + H 2 O If acid is added to the buffer, it will react with the base component: CH 3 CO 2 - (aq) + H + (aq) CH 3 CO 2 H (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: CH 3 CO 2 H (aq) + OH - (aq) CH 3 CO 2 - (aq) + H 2 O If acid is added to the buffer, it will react with the base component: CH 3 CO 2 - (aq) + H + (aq) CH 3 CO 2 H (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: CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) 1117
Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation Consider the equilibrium: CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) 1118
Quantitative treatment of Buffers: The Henderson-Hasselbalch Equation Consider the equilibrium: CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) 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: CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) Now take the log of both sides of the preceding equation, to obtain 1120
That is, 1121
That is, 1122
That is, 1123
That is, 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: 1126
We could repeat the previous approach for the weak acid HA to obtain: In a more general form it would be: 1127
We could repeat the previous approach for the weak acid HA to obtain: In a more general form it would be: 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 CH 3 CO 2 H and 1.0 M NaCH 3 CO 2. 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 K a for acetic acid is 1.8 x
Example: Calculate the pH of a buffer system containing 1.0 M CH 3 CO 2 H and 1.0 M NaCH 3 CO 2. 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 K a for acetic acid is 1.8 x Because acetic acid is a weak acid, we can ignore the small amount of dissociation and assume at equilibrium that [CH 3 CO 2 H] = 1.0 M 1130
Example: Calculate the pH of a buffer system containing 1.0 M CH 3 CO 2 H and 1.0 M NaCH 3 CO 2. 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 K a for acetic acid is 1.8 x Because acetic acid is a weak acid, we can ignore the small amount of dissociation and assume at equilibrium that [CH 3 CO 2 H] = 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 [CH 3 CO 2 - ] = 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 [CH 3 CO 2 - ] = 1.0 M Now K a = 1.8 x so that pK a =
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 [CH 3 CO 2 - ] = 1.0 M Now K a = 1.8 x so that pK a = 4.7 Hence, from the Henderson-Hasselbalch equation: =
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: CH 3 CO H + CH 3 CO 2 H 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: CH 3 CO H + CH 3 CO 2 H 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: CH 3 CO H + CH 3 CO 2 H 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: CH 3 CO H + CH 3 CO 2 H 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): [CH 3 CO 2 H] = = 1.10 M [CH 3 CO 2 - ] = = 0.90 M 1139
To calculate the new pH, use the Henderson- Hasselbalch equation: 1140