Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved
Common Ion Effect Shift in equilibrium position that occurs because of the addition of an ion already involved in the equilibrium reaction. An application of Le Châtelier’s principle. Copyright © Cengage Learning. All rights reserved

The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. Consider a solution of acetic acid (HC2H3O2), in which you have the following equilibrium. 2

The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. If we were to add NaC2H3O2 to this solution, it would provide C2H3O2- ions which are present on the right side of the equilibrium. 2

The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. The equilibrium composition would shift to the left and the degree of ionization of the acetic acid is decreased. 2

The Common Ion Effect The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium. This repression of the ionization of acetic acid by sodium acetate is an example of the common-ion effect. 2

CH3COONa (s) Na+ (aq) + CH3COO- (aq)
The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. Consider mixture of CH3COONa (strong electrolyte) and CH3COOH (weak acid). CH3COONa (s) Na+ (aq) + CH3COO- (aq) common ion CH3COOH (aq) H+ (aq) + CH3COO- (aq)

A Problem To Consider An aqueous solution is M in formic acid, HCH2O and M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4. Consider the equilibrium below: Starting 0.025 0.018 Change -x +x Equilibrium 0.025-x x 0.018+x 2

A Problem To Consider An aqueous solution is M in formic acid, HCH2O and M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4. The equilibrium constant expression is: 2

A Problem To Consider An aqueous solution is M in formic acid, HCH2O and M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4. Substituting into this equation gives: 2

A Problem To Consider An aqueous solution is M in formic acid, HCH2O and M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4. Assume that x is small compared with and Then 2

A Problem To Consider An aqueous solution is M in formic acid, HCH2O and M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4. The equilibrium equation becomes 2

A Problem To Consider An aqueous solution is M in formic acid, HCH2O and M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4. Hence, 2

A Problem To Consider An aqueous solution is M in formic acid, HCH2O and M in sodium formate, NaCH2O. What is the pH of the solution. The Ka for formic acid is 1.7 x 10-4. Note that x was much smaller than or For comparison, the pH of M formic acid is 2.69. 2

Key Points about Buffered Solutions
Buffered Solution – resists a change in pH. They are weak acids or bases containing a common ion. After addition of strong acid or base, deal with stoichiometry first, then the equilibrium. Copyright © Cengage Learning. All rights reserved

Adding an Acid to a Buffer
To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Buffers To play movie you must be in Slide Show Mode
PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Solving Problems with Buffered Solutions
Copyright © Cengage Learning. All rights reserved

Buffering: How Does It Work?

Henderson–Hasselbalch Equation
For a particular buffering system (conjugate acid–base pair), all solutions that have the same ratio [A–] / [HA] will have the same pH. Copyright © Cengage Learning. All rights reserved

Henderson-Hasselbalch equation
Consider mixture of salt NaA and weak acid HA. NaA (s) Na+ (aq) + A- (aq) Ka = [H+][A-] [HA] HA (aq) H+ (aq) + A- (aq) [H+] = Ka [HA] [A-] Henderson-Hasselbalch equation -log [H+] = -log Ka - log [HA] [A-] pH = pKa + log [conjugate base] [acid] -log [H+] = -log Ka + log [A-] [HA] pH = pKa + log [A-] [HA] pKa = -log Ka

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK?
Mixture of weak acid and conjugate base! HCOOH (aq) H+ (aq) + HCOO- (aq) Initial (M) Change (M) Equilibrium (M) 0.30 0.00 0.52 -x +x +x x x x pH = pKa + log [HCOO-] [HCOOH] Common ion effect 0.30 – x  0.30 pH = log [0.52] [0.30] = 4.01 x  0.52 HCOOH pKa = 3.77

A buffer solution is a solution of:
A weak acid or a weak base and The salt of the weak acid or weak base Both must be present! A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. Consider an equal molar mixture of CH3COOH and CH3COONa Add strong acid H+ (aq) + CH3COO- (aq) CH3COOH (aq) Add strong base OH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)

HCl + CH3COO- CH3COOH + Cl-
HCl H+ + Cl- HCl + CH3COO CH3COOH + Cl-

Which of the following are buffer systems? (a) KF/HF
(b) KBr/HBr, (c) Na2CO3/NaHCO3 (a) KF is a weak acid and F- is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO32- is a weak base and HCO3- is its conjugate acid buffer solution

Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. Buffers contain either a weak acid and its conjugate base or a weak base and its conjugate acid. Thus, a buffer contains both an acid species and a base species in equilibrium. 2

Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. Consider a buffer with equal molar amounts of HA and its conjugate base A-. When H3O+ is added to the buffer it reacts with the base A-. 2

Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. Consider a buffer with equal molar amounts of HA and its conjugate base A-. When OH- is added to the buffer it reacts with the acid HA. 2

Buffers A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it. Two important characteristics of a buffer are its buffer capacity and its pH. Buffer capacity depends on the amount of acid and conjugate base present in the solution. The next example illustrates how to calculate the pH of a buffer. 2

The Henderson-Hasselbalch Equation
How do you prepare a buffer of given pH? A buffer must be prepared from a conjugate acid-base pair in which the Ka of the acid is approximately equal to the desired H3O+ concentration. To illustrate, consider a buffer of a weak acid HA and its conjugate base A-. The acid ionization equilibrium is: 2

How do you prepare a buffer of given pH? The acid ionization constant is: By rearranging, you get an equation for the H3O+ concentration. 2

How do you prepare a buffer of given pH? Taking the negative logarithm of both sides of the equation we obtain: The previous equation can be rewritten 2

How do you prepare a buffer of given pH? More generally, you can write This equation relates the pH of a buffer to the concentrations of the conjugate acid and base. It is known as the Henderson-Hasselbalch equation. 2

How do you prepare a buffer of given pH? So to prepare a buffer of a given pH (for example, pH 4.90) we need a conjugate acid-base pair with a pKa close to the desired pH. The Ka for acetic acid is 1.7 x 10-5, and its pKa is 4.77. You could get a buffer of pH 4.90 by increasing the ratio of [base]/[acid]. 2

Note: Should be Molarity
Calculate the pH of the 0.20M NH3 /0.20M NH4Cl buffer. What is the pH of the buffer after the addition of 10.0mL of 0.10M HCl to 65.0 mL of the buffer? Note: Should be Molarity

Calculate the pH of the 0. 30 M NH3/0. 36 M NH4Cl buffer system
Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of M NaOH to 80.0 mL of the buffer solution? NH4+ (aq) H+ (aq) + NH3 (aq) pH = pKa + log [NH3] [NH4+] pH = log [0.30] [0.36] pKa = 9.25 = 9.17 start (moles) 0.029 0.001 0.024 NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq) end (moles) 0.028 0.0 0.025 final volume = 80.0 mL mL = 100 mL [NH4+] = 0.028 0.10 [NH3] = 0.025 0.10 pH = log [0.25] [0.28] = 9.22

EXERCISE! What is the pH of a buffer solution that is 0.45 M acetic acid (HC2H3O2) and 0.85 M sodium acetate (NaC2H3O2)? The Ka for acetic acid is 1.8 × 10–5. pH = 5.02 pH = –logKa + log([C2H3O2–] / [HC2H3O2]) = –log(1.8 × 10–5) + log(0.85 M / 0.45 M) = 5.02 Copyright © Cengage Learning. All rights reserved

Buffered Solution Characteristics
Buffers contain relatively large concentrations of a weak acid and corresponding conjugate base. Added H+ reacts to completion with the weak base. Added OH- reacts to completion with the weak acid. Copyright © Cengage Learning. All rights reserved

Buffered Solution Characteristics
The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentrations of the buffering materials (HA and A– or B and BH+) are large compared with amounts of H+ or OH– added. Copyright © Cengage Learning. All rights reserved

The amount of protons or hydroxide ions the buffer can absorb without a significant change in pH. Determined by the magnitudes of [HA] and [A–]. A buffer with large capacity contains large concentrations of the buffering components. Copyright © Cengage Learning. All rights reserved

Optimal buffering occurs when [HA] is equal to [A–]. It is for this condition that the ratio [A–] / [HA] is most resistant to change when H+ or OH– is added to the buffered solution. Copyright © Cengage Learning. All rights reserved

Titration Curve Plotting the pH of the solution being analyzed as a function of the amount of titrant added. Equivalence (Stoichiometric) Point – point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated. Copyright © Cengage Learning. All rights reserved

Neutralization of a Strong Acid with a Strong Base
To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved

Equivalence point – the point at which the reaction is complete
Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink)

Acid-Ionization Titration Curves
An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid). Such curves are used to gain insight into the titration process. You can use titration curves to choose an appropriate indicator that will show when the titration is complete. 2

Strong Acid-Strong Base Titrations
NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq) OH- (aq) + H+ (aq) H2O (l) Note that the pH changes slowly until the titration approaches the equivalence point. At the equivalence point, the pH of the solution is 7.0 because it contains a salt, NaCl, that does not hydrolyze.

A Problem To Consider Calculate the pH of a solution in which 10.0 mL of M NaOH is added to 25.0 mL of M HCl. Because the reactants are a strong acid and a strong base, the reaction is essentially complete. 2

A Problem To Consider Calculate the pH of a solution in which 10.0 mL of M NaOH is added to 25.0 mL of M HCl. We get the amounts of reactants by multiplying the volume of each (in liters) by their respective molarities. 2

A Problem To Consider Calculate the pH of a solution in which 10.0 mL of M NaOH is added to 25.0 mL of M HCl. All of the OH- reacts, leaving an excess of H3O+ 2

A Problem To Consider Calculate the pH of a solution in which 10.0 mL of M NaOH is added to 25.0 mL of M HCl. You obtain the H3O+ concentration by dividing the mol H3O+ by the total volume of solution (= L L= L) 2

A Problem To Consider Calculate the pH of a solution in which 10.0 mL of M NaOH is added to 25.0 mL of M HCl. Hence, 2

Titration of a Weak Acid by a Strong Base
The titration of a weak acid by a strong base gives a somewhat different curve. The pH range of these titrations is shorter. The equivalence point will be on the basic side since the salt produced contains the anion of a weak acid. 2

Weak Acid-Strong Base Titrations
CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l) CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l) At equivalence point (pH > 7): CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq)

A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5. At the equivalence point, equal molar amounts of acetic acid and sodium hydroxide react to give sodium acetate. 2

A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5. First, calculate the concentration of the acetate ion. In this case, 25.0 mL of 0.10 M NaOH is needed to react with 25.0 mL of 0.10 M acetic acid. 2

A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5. The molar amount of acetate ion formed equals the initial molar amount of acetic acid. 2

A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5. The total volume of the solution is 50.0 mL. Hence, 2

A Problem To Consider Calculate the pH of the solution at the equivalence point when 25.0 mL of 0.10 M acetic acid is titrated with 0.10 M sodium hydroxide. The Ka for acetic acid is 1.7 x 10-5. The hydrolysis of the acetate ion follows the method given in an earlier section of this chapter. You find the Kb for the acetate ion to be 5.9 x and that the concentration of the hydroxide ion is 5.4 x The pH is 8.73 2

Strong Acid-Weak Base Titrations
HCl (aq) + NH3 (aq) NH4Cl (aq) H+ (aq) + NH3 (aq) NH4Cl (aq) At equivalence point (pH < 7): NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)

Titration of a Weak Base by a Strong Acid
The titration of a weak base with a strong acid is a reflection of our previous example. In this case, the pH declines slowly at first, then falls abruptly from about pH 7 to pH 3. Methyl red, which changes color from yellow at pH 6 to red at pH 4.8, is a possible indicator. 2

Curve for the titration of a weak base by a strong acid.
2

Exactly 100 mL of 0. 10 M HNO2 are titrated with a 0
Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ? start (moles) 0.01 0.01 HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l) end (moles) 0.0 0.0 0.01 [NO2-] = 0.01 0.200 = 0.05 M Final volume = 200 mL NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq) Initial (M) Change (M) Equilibrium (M) 0.05 0.00 0.00 -x +x +x x x x Kb = [OH-][HNO2] [NO2-] = x2 0.05-x pOH = 5.98 = 2.2 x 10-11 pH = 14 – pOH = 8.02 0.05 – x  0.05 x  1.05 x 10-6 = [OH-]

Acid-Base Indicators Marks the end point of a titration by changing color. The equivalence point is not necessarily the same as the end point (but they are ideally as close as possible). Copyright © Cengage Learning. All rights reserved

Acid-Base Indicators HIn (aq) H+ (aq) + In- (aq) [HIn]  10
Color of acid (HIn) predominates  10 [HIn] [In-] Color of conjugate base (In-) predominates

The titration curve of a strong acid with a strong base.

Which indicator(s) would you use for a titration of HNO2 with KOH ?
Weak acid titrated with strong base. At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein

The Acid and Base Forms of the Indicator Phenolphthalein

Useful pH Ranges for Several Common Indicators

Worked Example 16.1a

Worked Example 16.1b

Worked Example 16.4

Worked Example 16.5a

Worked Example 16.7

Copyright © Cengage Learning. All rights reserved

Similar presentations

Presentation on theme: "Copyright © Cengage Learning. All rights reserved"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Copyright © Cengage Learning. All rights reserved

Similar presentations

Presentation on theme: "Copyright © Cengage Learning. All rights reserved"— Presentation transcript:

Similar presentations

About project

Feedback