chapter 17 additional aspects of aqueous...

Chapter 17

Additional Aspects of Aqueous Equilibria

Yonsei University

Lecture Presentation

© 2012 Pearson Education, Inc.

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17.1 The Common-Ion Effect

• Consider a solution of acetic acid:

• If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.

• “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO(aq)


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The Common-Ion EffectCalculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

[H3O+] [F][HF]

Ka = = 6.8 104


HF(aq) + H2O(l) H3O+(aq) + F(aq)

Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.

[HF], M [H3O+], M [F], M

Initially 0.20 0.10 0

Change

At equilibrium

x +x +x

0.20 x 0.20 0.10 + x 0.10 x

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The Common-Ion Effect

= x = 1.4 103

(0.10) (x)(0.20)6.8 104 =

(0.20) (6.8 104)(0.10)


• Therefore, [F] = x = 1.4 103

[H3O+] = 0.10 + x = 0.10 + 1.4 103 = 0.10 M

• So, pH = log (0.10)

pH = 1.00

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

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. Answer: [HCOO–] = 9.0 105, pH = 1.00

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17. 2 Buffered Solutions

• 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.


Buffers are composed of a weak acid and the salt of the acid (conjugate base)or weak base and the salt of the base (conjugate acid).

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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.

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Buffers

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Buffer Calculations

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

[H3O+] [A][HA]

Ka =

HA + H2O H3O+ + A


[A][HA]

Ka = [H3O+]

[A][HA]

log Ka = log [H3O+] + log

pKaacid

base

© 2012 Pearson Education Inc

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Buffer Calculations

• SopKa = pH log

[base][acid]

• Rearranging, this becomes

pH = pKa + log[base][acid]

• This is the Henderson–Hasselbalch eq.


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Ex) What is the pH of a buffer that is 0.12 M in lactic acid,CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka

for lactic acid is 1.4 104.


4 4

[ ][ ]

[ ]

[ ] (0.12 )[ ] 1.4 10 1.68 10

[ ] (0.10 )

a

a

RCOOH H RCOO

H RCOOK

RCOOH

RCOOH MH K

RCOO M

pH=3.77

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pH = 3.85 + (0.08)

pH = 3.77

Ex) What is the pH of a buffer that is 0.12 M in lactic acid,CH3CH(OH)COOH, and 0.10 M in sodium lactate? Ka

for lactic acid is 1.4 104. (Henderson–Hasselbalchequation)

pH = pKa + log[base][acid]

pH = log (1.4 104) + log(0.10)(0.12)


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

Solution

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?

x = 1.8 105 M = [H+]

pH = –log(1.8 105) = 4.74

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Sample Exercise 17.4 Preparing a Buffer

Solution

How 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.)

pOH = 14.00 – pH = 14.00 – 9.00 = 5.00[OH–] = 1.0 105 M

[NH3] = 0.10 M

(2.0 L)(0.18 mol NH4CI/L) = 0.36 mol NH4CI

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Buffer Capacity • Buffer capacity is the amount of acid or base

that can be neutralized by the buffer before there is a significant change in pH.

• 0.200M HCOOH+0.400M HCOONa:100mL + 1.00M HCl 50.0mL

3 ( ) ( ) ( ) 5.6 10

Initial 0.0400 0.0200

Added 0.0500

change -0.0400 -0.0400 +0.0400

Equil. 0

aHCOO aq H aq HCOOH aq K

0.0100 0.0600

0.0100[ ] 0.0667 1.18

0.1500

molH M pH

L

Ka=1.8x10-4

** Buffer capacity : 0.0200mol strong base, 0.0400mol strong acid

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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.Useful pH range

: pKa±1 pH unit


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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.


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Buffers

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Calculating pH Changes in Buffers

A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.


HC2H3O2(aq) + OH(aq) C2H3O2(aq) + H2O(l)

Before the reaction, since

mol HC2H3O2 = mol C2H3O2

pH = pKa = log (1.8 105) = 4.74

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The 0.020 mol NaOH will react with 0.020 mol of the acetic acid:

HC2H3O2(aq) + OH(aq) C2H3O2(aq) + H2O(l)

HC2H3O2 C2H3O2 OH

Before reaction 0.300 mol 0.300 mol 0.020 mol

After reaction


0.280 mol 0.320 mol 0.000 mol

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Now use the Henderson–Hasselbalch equation to calculate the new pH:

pH = 4.74 + log(0.320)(0.200)

pH = 4.74 + 0.06pH

pH = 4.80


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Importance of Buffers in Medicine?

Aspirin+ MgCO3 (antacid: increase stomach pH)

TetracyclineAn antibiotic drug

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17.3 Acid-Base Titrations

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

• Standard solution


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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.


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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.


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Titration of a Strong Acid with a Strong Base

-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.


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

Solution

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.

50.0 mL + 51.0 mL = 101.0 mL = 0.1010 LpOH = –log(1.0 103) = 3.00

pH = 14.00 – pOH = 14.00 – 3.00 = 11.00

–log(1.0 103) = 3.00

50.0 mL + 49.0 mL = 99.0 mL = 0.0990 L

(a) 49.0 mL

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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.


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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.


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

Solution

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).

45.0 mL + 50.0 mL = 95.0 mL = 0.0950 L

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Sample Exercise 17.8 Calculating the pH at the Equivalence Point

Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 MNaOH.

Kb = Kw/Ka = (1.0 1014)/(1.8 105) = 5.6 1010

x = [OH–] = 5.3 106 M, which gives pOH = 5.28 pH = 8.72.

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.Answers: (a) 8.21, (b) 5.28

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

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Indicators: conjugate acid pairs of organic compounds that change color in acid or base

• In a titration, an indicator is chosen with a pKa

close to the pH at the equivalence point. (EP≈pKa ±1.0).

• The visual color change indicates the endpoint of the titration.

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Titrating with an Acid-Base Indicator

HIn H+ + In-

HPpth H+ + Ppth-

HMRed H+ + MRed-

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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.


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Titrations of Polyprotic Acids

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


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17.4 Solubility Equilibria

Consider the equilibrium that exists in a saturated solution of BaSO4 in water:

Ksp = [Ba2+] [SO42]

BaSO4(s) Ba2+(aq) + SO42(aq)


Ksp : solubility product.

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Solubility Products

• Ksp is not the same as solubility.

• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL)of solution, or in mol/L (M).


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Solubility and Ksp

• Solubility is the amount of substance that dissolves to form a saturated solution.– This is often expressed as grams of solute that will

dissolve per liter (or 100g) of solution.

• Molar solubility is the number of moles of solutethat dissolve to form a liter of saturated solution.

• We can use the solubility to find Ksp and vice versa.

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Sample Exercise 17.10 Calculating Ksp from Solubility

SolutionFrom the chemical formula of silver chromate, we know that there must be 2 Ag+ ions in solution for each CrO4

2– ion in solution. Consequently, the concentration of CrO42– is half the

concentration of Ag+:

and Ksp is

Ksp = [Ag+]2[CrO42–] = (1.3 104)2(6.5 105) = 1.1 1012

Solid silver chromate is added to pure water at 25 C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 104 M. Assuming that Ag2CrO4 dissociates completely in water and that there are no other important equilibria involving Ag+ or CrO4

2– ions in the solution, calculate Ksp for this compound.

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Sample Exercise 17.11 Calculating Solubility from Ksp

Solution

The Ksp for CaF2 is 3.9 1011 at 25 C . Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.

Practice ExerciseThe Ksp for LaF3 is 2 1019. What is the solubility of LaF3 in water in moles per liter?Answer: 9 106 mol/L

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17.5 Factors Affecting Solubility

• The presence of a common ion,

• The pH of the solution, and

• The presence of complexing agents. – Amphoterism is related to the effects of

both pH and complexing agents.


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Factors Affecting Solubility

• The Common-Ion Effect– If one of the ions in a solution equilibrium

is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease:

– If NaF is added, ………….

CaF2(s) Ca2+(aq) + F(aq)


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Sample Exercise 17.12 Calculating the Effect of a Common Ion on Solubility

Calculate the molar solubility of CaF2 at 25 C in a solution that is (a) 0.010 M in Ca(NO3)2, (b) 0.010 Min NaF.

Solution

Ksp = 3.9 1011 = [Ca2+][F–]2 = (0.010 + x)(2x)2

[Ca2+] = x, [F–] = 0.010 + 2x

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• pH– If a substance has a basic anion, it will be more

soluble in an acidic solution.

– Substances with acidic cations are more soluble in basic solutions.


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Mg(OH)2(s) Mg2+(aq) + 2OH–(aq) Ksp = 8.9x10-12

OH–(aq) + H+(aq) H2O(aq) K = 1.0x1014

PbF2(s) Pb2+(aq) + 2F–(aq) Ksp = 4.0 x 10-8

F–(aq) + H+(aq) HF(aq) K = 1/7.2x10-4 = 1.4x103

• In general: – The solubility of slightly soluble salts containing

basic ions increases as pH decreases.

– The more basic the anion is, the greater the effect.

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Sample Exercise 17.13 Predicting the Effect of Acid on Solubility

Which of these substances are more soluble in acidic solution than in basic solution: (a)Ni(OH)2(s), (b) CaCO3(s), (c) BaF2(s), (d) AgCl(s)?

Solution(a) Ni(OH)2(s) is more soluble in acidic solution:

(b) Similarly, CaCO3(s) dissolves in acid solutions because CO32– is a basic anion:

The reaction between CO32– and H+ occurs in steps, with HCO3

– forming first and H2CO3 forming in appreciable amounts only when [H+] is sufficiently high.

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Continued

(c) The solubility of BaF2 is enhanced by lowering the pH because F– is a basic anion:

(d) The solubility of AgCl is unaffected by changes in pH because Cl– is the anion of a strong acid and therefore has negligible basicity.

Practice ExerciseWrite the net ionic equation for the reaction between an acid and (a) CuS, (b) Cu(N3)2.

Answers: (a)

(b)

Sample Exercise 17.13 Predicting the Effect of Acid on Solubility

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• Complex Ions– Metal ions can act as Lewis acids and form

complex ions with Lewis bases in the solvent.


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• Complex Ions– The formation

of these complex ions increases the solubility of these salts.


AgCl(s) Ag+(aq) + Cl–(aq) Ksp=1.8 10–10

Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) Kf=1.7 107

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Sample Exercise 17.14 Evaluating an Equilibrium Involving a Complex Ion

Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO3 to give an equilibrium concentration of [NH3] = 0.20 M. Neglect the small volume change that occurs when NH3 is added.

Solution

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• Amphoterism– Amphoteric metal oxides and hydroxides are soluble in

strong acid or base, because they can act either as acids or bases.

– Examples of such cations are Al3+, Zn2+, and Sn2+.


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17.6 Precipitation and Separation of Ions

Will a Precipitate Form?

• In a solutionWill a Precipitate Form?,– If Q = Ksp, the system is at equilibrium

and the solution is saturated.

– If Q < Ksp, more solid can dissolve until Q = Ksp.

– If Q > Ksp, the salt will precipitate until Q = Ksp.


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Sample Exercise 17.15 Predicting Whether a Precipitate Forms

Does a precipitate form when 0.10 L of 8.0 103 M Pb(NO3)2 is added to 0.40 L of 5.0 103 MNa2SO4?

Solution

Q = [Pb2+][SO42] = (1.6 103)(4.0 103) = 6.4 106

Because Q > Ksp, PbSO4 precipitates.

Practice ExerciseDoes a precipitate form when 0.050 L of 2.0 102 M NaF is mixed with 0.010 L of 1.0 102 M Ca(NO3)2?

Answer: Yes, CaF2 precipitates because Q = 4.6 108 is larger than Ksp = 3.9 10 11

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Selective Precipitation of Ions

One can use differences in solubilities of salts to separate ions in a mixture.


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17.7 Qualitative Analysis for Metallic Elements

• Separation :Five major groups – insoluble chlorides

– acid-insoluble sulfides

– base-insoluble sulfides and hydroxides

– insoluble phosphates

– alkali metals and ammonium ion

• Specific tests

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• Group I : form insoluble chloride

: Ag+, Pb2+, Hg22+ : AgCl, PbCl2 , Hg2Cl2

• Add hot water : PbCl2(s) Pb2+(aq) + 2Cl-(aq)

• Add NH3 : AgCl(s) + 2NH3(aq) Ag(NH3) 2+(aq) + Cl-(aq)

• Group II : acid-insoluble sulfide (at pH=0.5)

: Cu2+, Bi3+, Hg2+, Cd2+, Sn4+, Sb3+

• Cu2+(aq) + H2S(aq) CuS(s) + 2H+ (aq) at pH=0.5

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• Group III : forms base-insoluble sulfides (at pH=9.0) – Al3+, Cr3+, Co2+, Fe2+, Mn2+, Ni2+, Zn2+

– Al(OH)3, Cr(OH)3

– Ksp (CuS )= 1.0 10-16, Ksp (NiS )= 1.0 10–1

– Cu2+ + H2S ==CuS(s) + 2H+(aq) K=1x1016

– Ni2+ + H2S ==NiS(s) + 2H+(aq) K=2x10-7

• Group IV : insoluble phosphate (or carbonate): Ba2+ (Ca2+, Mg2+)+ CO3

2-(aq) Ba CO3(s)

• Group IV : alkali metals and ammonium ion– Na+, K+, NH4

+

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Seperation & Identification

Step1 : Cu2+, Ag+, Pb2++ Cl- (aq) Cu2+ (aq), AgCl (s), PbC12(s) Step 2 : Add hot water : AgCl (s), PbC12(s) AgCl (s), Pb2+(aq) + 2C1-(aq)Step 3 : Pb2+(aq) + CrO4

2-(aq) PbCrO4 (s)

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Problems

• 10,24,32, 50, 64,84,97

chapter 17 additional aspects of aqueous...

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