chapter 17 sections 1-3 common ions, buffers and titration © 2012 pearson education, inc

Chapter 17Sections 1-3

Common ions, Buffers and Titration

© 2012 Pearson Education, Inc.

AqueousEquilibria

Sample Exercise 17.1 Calculating the pH When a Common Ion is Involved

What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?

Practice ExerciseCalculate the pH of a solution containing 0.085 M nitrous acid (HNO2, Ka = 4.5 104) and 0.10 M potassium nitrite (KNO2).

AqueousEquilibria

Sample Exercise 17.2 Calculating Ion Concentrations When a Common Ion is Involved

Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

Practice ExerciseCalculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid (HCOOH, Ka = 1.8 104) and 0.10 M in HNO3.

AqueousEquilibria

(b)

A. NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

B. NH4+(aq) + Cl-(aq) NH3(aq) + HCl(l)

C. H3O+(aq) + OH–(aq) 2H2O(l)

D. NH4+(aq) + H2O(l) H3O

+(aq) + NH3(aq)

(a)A. There are no spectator ions.B. Cl–

C. Both NH4+ and Cl–

D. NH4+

AqueousEquilibria

Buffers

• Buffers are solutions of a weak conjugate acid–base pair.

• They are particularly resistant to pH changes, even when strong acid or base is added.


AqueousEquilibria

A. All listed pairs will not function as buffers.

B. HCHO2 and CHO2– will not work as a buffer

because HCHO2 is a weak acid and CHO2– is a

spectator ion.

C. HCO3– and CO3

2– will not work as a buffer because HCO3

– is a weak acid and HCO3– is a

spectator ion.

D. HNO3 and NO3– will not work as a buffer because

HNO3 is a strong acid and NO3– is a spectator

ion.

AqueousEquilibria

Buffers

If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH to make F and water.

Similarly, if acid is added, the F reacts with it to form HF and water.


AqueousEquilibria

A. There is no reaction because C2H3O2– and HC2H3O2

form a buffer.

B. The NaOH reacts with C2H3O2– converting some of it

into HC2H3O2.

C. The NaOH reacts with HC2H3O2 converting some of it into C2H3O2

–. D. The NaOH is neutralized and all concentrations

(HC2H3O2 and C2H3O2–) remain unchanged.

AqueousEquilibria

A. There is no reaction because C2H3O2– and HC2H3O2

form a buffer.

B. The HCl reacts with C2H3O2– converting some of it

into HC2H3O2.C. The HCl is neutralized and all concentrations

(HC2H3O2 and C2H3O2–) remain unchanged.

D. The HCl reacts with HC2H3O2 converting some of it into C2H3O2

–.

AqueousEquilibria

Sample Exercise 17.3 Calculating the pH of a Buffer

What is the pH of a buffer that is 0.12 M in lactic acid [CH3CH(OH)COOH, or HC3H5O3] and 0.10 M insodium lactate [CH3CH(OH)COONa, or NaC3H5O3]? For lactic acid, Ka = 1.4 104.

AqueousEquilibria

Buffer Calculations

Consider the equilibrium constant expression for the dissociation of a generic acid, HA:

[H3O+] [A][HA]

Ka =

HA + H2O H3O+ + A


AqueousEquilibria

Buffer Calculations

Rearranging slightly, this becomes

[A][HA]

Ka = [H3O+]

Taking the negative log of both side, we get

[A][HA]

log Ka = log [H3O+] + log

pKa

pHacid

base


AqueousEquilibria

Buffer Calculations

• SopKa = pH log

[base][acid]

• Rearranging, this becomes

pH = pKa + log[base][acid]

• This is the Henderson–Hasselbalch equation.


AqueousEquilibria

Continued

Practice ExerciseCalculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. (Refer to Appendix D.)

Sample Exercise 17.3 Calculating the pH of a Buffer

AqueousEquilibria

Sample Exercise 17.4 Preparing a BufferHow many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00?(Assume that the addition of NH4Cl does not change the volume of the solution.)

Practice ExerciseCalculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (C6H5COOH) to produce a pH of 4.00.

AqueousEquilibria

pH Range

• The pH range is the range of pH values over which a buffer system works effectively.

• It is best to choose an acid with a pKa close to the desired pH.


AqueousEquilibria

A. HClO because it is a stronger weak acid. A salt containing ClO– is also needed.

B. HNO2 because it is a stronger weak acid. A salt containing NO2

– is also needed.

C. HClO because its pKa is closer to pH = 7.0. A salt containing ClO– is also needed.

D. HNO2 because its pKa is closer to pH = 7.0. A salt containing NO2

– is also needed.

AqueousEquilibria

When Strong Acids or Bases Are Added to a Buffer

When strong acids or bases are added to a buffer, it is safe to assume that all of the strong acid or base is consumed in the reaction.


AqueousEquilibria

Addition of Strong Acid or Base to a Buffer

1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.

2. Use equilibrium to determine the new pH of the solution.


AqueousEquilibria

Sample Exercise 17.5 Calculating pH Changes in BuffersA buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH3COONa to enough water to make 1.000 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this solution after 5.0 mL of 4.0 M NaOH (aq) solution is added. (b) For comparison, calculate the pH of a solution made by adding 5.0 mL of 4.0 M NaOH( aq) solution to 1.000 L of pure water.

Practice ExerciseDetermine (a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of

0.020 mol HCl

(b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.000 L of pure water.

AqueousEquilibria

Titration

In this technique, a known concentration of base (or acid) is slowly added to a solution of acid (or base).


AqueousEquilibria

A. No changeB. IncreaseC. DecreaseD. Need volumes and molarities to answer the question

AqueousEquilibria

Titration

A pH meter or indicators are used to determine when the solution has reachedthe equivalence point, at which the stoichiometric amount of acid equals that of base.


AqueousEquilibria

Titration of a Strong Acid with a Strong Base


From the start of the titration to near the equivalence point, the pH goes up slowly.

Just before (and after) the equivalence point, the pH increases rapidly.

At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

As more base is added, the increase in pH again levels off.

AqueousEquilibria

A. 100.00 mLB. 50.00 mLC. 25.00 mLD. 12.50 mL

AqueousEquilibria

Sample Exercise 17.6 Calculating pH for a Strong Acid–Strong Base Titration

Calculate the pH when (a) 49.0 mL and (b) 51.0 mL of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.

Practice ExerciseCalculate the pH when (a) 24.9 mL and (b) 25.1 mL of 0.100 M HNO3 havebeen added to 25.0 mL of 0.100 M KOH solution.

AqueousEquilibria

A. pH = 4.2

B. pH = 7.0

C. pH = 8.2

D. pH = 9.8

AqueousEquilibria

Titration of a Weak Acid with a Strong Base

• Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.

• At the equivalence point the pH is >7.

• Phenolphthalein is commonly used as an indicator in these titrations.


AqueousEquilibria

A. The volume of base does not change but the equivalence point pH increases.B. The volume of base increases but the equivalence point pH does not change.C. The volume of base decreases and the equivalence point pH does not change.D. The volume of base does not change but the equivalence point pH decreases.

AqueousEquilibria


At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.


AqueousEquilibria

Sample Exercise 17.7 Calculating pH for a Weak Acid–Strong Base Titration

Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH (Ka = 1.8 105).

Practice Exercise(a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (C6H5COOH, Ka = 6.3 105. (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3.

AqueousEquilibria

Sample Exercise 17.8 Calculating the pH at the Equivalence PointCalculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.

Practice ExerciseCalculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (C6H5COOH, Ka = 6.3 105) is titrated with 0.050 M NaOH; (b) 40.0 mL of 0.100 M NH3 is titrated with 0.100 M HCl.

AqueousEquilibria


With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.


AqueousEquilibria

A. The volume of base does not change but the equivalence point pH increases.

B. The volume of base increases but the equivalence point pH does not change.

C.The volume of base decrease sand the equivalence point does not change.

D.The volume of base does not change but the equivalence point pH increases.

AqueousEquilibria

Titrations of Polyprotic Acids

When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.


AqueousEquilibria

Titration of a Weak Base with a Strong Acid

• The pH at the equivalence point in these titrations is <7, so using phenolphthalein would not be a good idea.

• Methyl red is the indicator of choice.


AqueousEquilibria

A. No, methyl red does not change color at equivalence point.

B. No methyl red’s color change is not sharp and gives only a rough estimate of the equivalence point pH.

C. Yes, methyl red changes color in the pH region between 4 and 6 in a steeply sloping region.

D. Yes, methyl red changes color in the pH region between 8–10 in a steeply sloping region.

AqueousEquilibria

A. The nearly vertical equivalence point portion of the titration curve is large for a weak acid-strong base titration, and fewer indicators undergo their color change so quickly because the change is difficult to monitor.

B. The nearly vertical equivalence point portion of the titration curve is smaller for a weak acid-strong base titration, and fewer indicators undergo their color change within this narrow range.

C. Many indicators do not change colors at the equivalence points of weak acid-strong base titrations.

D. Equivalence points at pH’s other than 7.00 are difficult to determine.

chapter 17 sections 1-3 common ions, buffers and titration © 2012 pearson education, inc

Documents