19-1 copyright ©the mcgraw-hill companies, inc. permission required for reproduction or display....

19-1

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 19

Ionic Equilibria in Aqueous Systems

19-2


Ionic Equilibria in Aqueous SystemsIonic Equilibria in Aqueous Systems

19.1 Equilibria of Acid-Base Buffer Systems

19.2 Acid-Base Titration Curves

19.3 Equilibria of Slightly Soluble Ionic Compounds

19.4 Equilibria Involving Complex Ions

19-3


Figure 19.1

The effect of addition of acid or base to …

an unbuffered solution

or a buffered solution

acid added base added

acid added base added

Figure 19.2

19-4


Table 19.1

The Effect of Added Acetate Ion on the Dissociation of Acetic Acid

[CH3COOH]initial [CH3COO-]added % Dissociation* pH

* % Dissociation =[CH3COOH]dissoc

[CH3COOH]initial

x 100

0.10 0.00

0.10 0.050

0.10

0.10 0.10

0.15

1.3

0.036

0.018

0.012

2.89

4.44

4.74

4.92

19-5


Figure 19.3 How a buffer works.

Buffer with equal concentrations of

conjugate base and acid

OH-H3O+

Buffer after addition of H3O+

H2O + CH3COOH H3O+ + CH3COO-

Buffer after addition of OH-

CH3COOH + OH- H2O + CH3COO-

19-6


Sample Problem 19.1 Calculating the Effect of Added H3O+ or OH-

on Buffer pHPROBLEM: Calculate the pH:

(a) of a buffer solution consisting of 0.50M CH3COOH and 0.50M CH3COONa

(b) after adding 0.020mol of solid NaOH to 1.0L of the buffer solution in part (a)

(c) after adding 0.020mol of HCl to 1.0L of the buffer solution in part (a)

Ka of CH3COOH = 1.8x10-5. (Assume the additions cause negligible volume changes.

PLAN: We know Ka and can find initial concentrations of conjugate acid and base. Make assumptions about the amount of acid dissociating relative to its initial concentration. Proceed step-wise through changes in the system.

Initial

Change

Equilibrium

0.50+ x

0.50-x

--

-

0.50 0+ x

0.50 +x x

- x

SOLUTION:

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)Concentration (M)

(a)

19-7


Sample Problem 19.1 Calculating the Effect of Added H3O+ and OH-

on Buffer pHcontinued (2 of 4)

[CH3COOH]equil ≈ 0.50M [CH3COO-]initial ≈ 0.50M[H3O+] = x

Ka =[H3O+][CH3COO-]

[CH3COOH][H3O+] = x = Ka

[CH3COO-]

[CH3COOH]= 1.8x10-5M

Check the assumption: 1.8x10-5/0.50 X 100 = 3.6x10-3 %

CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O (l)Concentration (M)

Before addition

Addition

After addition

(b)[OH-]added =

0.020 mol

1.0L soln= 0.020M NaOH

0.50 - 0.50 -

-

-

- 0.020 -

0.48 0 0.52

19-8




Set up a reaction table with the new values.


InitialChange

Equilibrium

0.48 -- x

0.48 -x

-

-

0.52 0

x

+ x + x

0.52 +x

[H3O+] = 1.8x10-50.48

0.52= 1.7x10-5 pH = 4.77

CH3COO-(aq) + H3O+(aq) CH3COOH(aq) + H2O (l)Concentration (M)

Before addition

Addition

After addition

(c) [H3O+]added = 0.020 mol

1.0L soln= 0.020M H3O+

0.50 - 0.50 -

-

-

- 0.020 -

0.48 0 0.52

19-9




Set up a reaction table with the new values.


InitialChange

Equilibrium

0.52 -- x

0.52 -x

-

-

0.48 0

x

+ x + x

0.48 +x

[H3O+] = 1.8x10-5

0.48

0.52= 2.0x10-5 pH = 4.70

19-10


The Henderson-Hasselbalch Equation

HA + H2O H3O+ + A-

Ka = [H3O+] [A-]

[HA]

[H3O+] = Ka [HA]

[A-]

- log[H3O+] = - log Ka + log [A-]

[HA]

pH = pKa + log [base]

[acid]

19-11


Buffer Capacity and Buffer Range

Buffer capacity is the ability to resist pH change.

Buffer range is the pH range over which the buffer acts effectively.

The more concentrated the components of a buffer, the greaterthe buffer capacity.

The pH of a buffer is distinct from its buffer capacity.

A buffer has the highest capacity when the component concentrations are equal.

Buffers have a usable range within ± 1 pH unit of the pKa ofits acid component.

19-12


Figure 19.4 The relation between buffer capacity and pH change.

19-13


Preparing a Buffer

1. Choose the conjugate acid-base pair.

2. Calculate the ratio of buffer component concentrations.

3. Determine the buffer concentration.

4. Mix the solution and adjust the pH.

19-14


Sample Problem 19.2 Preparing a Buffer

SOLUTION:

PROBLEM: An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of the acid rain on limsetone-rich soils. How many grams of Na2CO3 must she add to 1.5L of freshly prepared 0.20M NaHCO3 to make the buffer? Ka of HCO3

- is 4.7x10-11.

PLAN: We know the Ka and the conjugate acid-base pair. Convert pH to [H3O+], find the number of moles of carbonate and convert to mass.

HCO3-(aq) + H2O(l) CO3

2-(aq) + H3O+(aq) Ka =

[CO32-][H3O+]

[HCO3-]

pH = 10.00; [H3O+] = 1.0x10-10 4.7x10-11 =[CO3

2-](0.20)

1.0x10-10[CO3

2-] = 0.094M

moles of Na2CO3 = (1.5L)(0.094mols/L) = 0.14

= 15 g Na2CO30.14 moles 105.99g

mol

19-15


pH

Figure 19.5Colors and approximate pH range of some

common acid-base indicators.

19-16


Figure 19.6 The color change of the indicator bromthymol blue.

acidic

basic

change occurs over ~2pH units

19-17


Figure 19.7 Curve for a strong acid-strong base titration.

19-18


Figure 19.8

Curve for a weak acid-

strong base titration.

Titration of 40.00mL of 0.1000M HPr with 0.1000M NaOH

[HPr] = [Pr-]

pH = 8.80 at equivalence point

pKa of HPr = 4.89

methyl red

19-19


Sample Problem 19.3 Calculating the pH During a Weak Acid-Strong Base Titration

PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; Ka = 1.3x10-5) after adding the following volumes of 0.1000M NaOH:

(a) 0.00mL (b) 30.00mL (c) 40.00mL (d) 50.00mLPLAN: The amounts of HPr and Pr- will be changing during the titration.

Remember to adjust the total volume of solution after each addition.

SOLUTION: (a) Find the starting pH using the methods of Chapter 18.

Ka = [Pr-][H3O+]/[HPr] [Pr-] = x = [H3O+] [Pr-] = x = [H3O+]

€

x= (1.3x10−5)(0.10) x = 1.1x10-3 ; pH = 2.96

(b)

Before addition

Addition

After addition

0.04000

0.03000

0.030000.01000

0 -

-

-0

-

- -

HPr(aq) + OH-(aq) Pr-(aq) + H2O (l)Amount (mol)

19-20


Sample Problem 19.3 Calculating the pH During a Weak Acid-Strong Base Titration

continued

[H3O+] = 1.3x10-50.001000 mol

0.003000 mol= 4.3x10-6M pH = 5.37

(c) When 40.00mL of NaOH are added, all of the HPr will be reacted and the [Pr -] will be (0.004000 mol)

(0.004000L) + (0.004000L)= 0.05000M

Ka x Kb = Kw Kb = Kw/Ka = 1.0x10-14/1.3x10-5 = 7.7x10-10

[H3O+] = Kw / = 1.6x10-9M

€

Kbx[Pr−] pH = 8.80

(d) 50.00mL of NaOH will produce an excess of OH-.

mol XS base = (0.1000M)(0.05000L - 0.04000L) = 0.00100molM = (0.00100)

(0.0900L)

M = 0.01111[H3O+] = 1.0x10-14/0.01111 = 9.0x10-11M

pH = 12.05

19-21


Figure 19.9

Curve for a weak base-strong acid

titration.

Titration of 40.00mL of 0.1000M NH3 with 0.1000M HCl

pH = 5.27 at equivalence point

pKa of NH4+ =

9.25

19-22


Ion-Product Expression (Qsp)and Solubility Product Constant (Ksp)

At equilibrium Qsp = [Mn+]p [Xz-]q = Ksp

For the hypothetical compound, MpXq

19-23


Sample Problem 19.4 Writing Ion-Product Expressions for SlightlySoluble Ionic Compounds

SOLUTION:

PROBLEM: Write the ion-product expression for each of the following:

(a) Magnesium carbonate (b) Iron (II) hydroxide(c) Calcium phosphate (d) Silver sulfide

PLAN: Write an equation which describes a saturated solution. Take note of the sulfide ion produced in part (d).

Ksp = [Mg2+][CO32-]

(a) MgCO3(s) Mg2+(aq) + CO3

2-(aq)

Ksp = [Fe2+][OH-] 2

(b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq) Ksp = [Ca2+]3[PO4

3-]2

(c) Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq)

(d) Ag2S(s) 2Ag+(aq) + S2-(aq)

S2-(aq) + H2O(l) HS-(aq) + OH-(aq)

Ag2S(s) + H2O(l) 2Ag+(aq) + HS-(aq) + OH-(aq)

Ksp = [Ag+]2[HS-][OH-]

19-24


Table 19.2 Solubility-Product Constants (Ksp) of Selected Ionic

Compounds at 250C

Name, Formula Ksp

Aluminum hydroxide, Al(OH)3

Cobalt (II) carbonate, CoCO3

Iron (II) hydroxide, Fe(OH)2

Lead (II) fluoride, PbF2

Lead (II) sulfate, PbSO4

Silver sulfide, Ag2S

Zinc iodate, Zn(IO3)2

3 x 10-34

1.0 x 10-10

4.1 x 10-15

3.6 x 10-8

1.6 x 10-8

4.7 x 10-29

8 x 10-48

Mercury (I) iodide, Hg2I2

3.9 x 10-6

19-25


Sample Problem 19.5 Determining Ksp from Solubility

PROBLEM: (a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is the Ksp of PbSO4?

(b) When lead (II) fluoride (PbF2) is shaken with pure water at 250C, the solubility is found to be 0.64g/L. Calculate the Ksp of PbF2.

PLAN: Write the dissolution equation; find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression.

Ksp = [Pb2+][SO42-]

= 1.40x10-4M PbSO4

Ksp = [Pb2+][SO42-] = (1.40x10-4)2 =

SOLUTION: PbSO4(s) Pb2+(aq) + SO42-(aq)(a)

1000mL

L

4.25x10-3g

100mL soln 303.3g PbSO4

mol PbSO4

1.96x10-8

19-26


Sample Problem 19.5 Determining Ksp from Solubility

continued

(b) PbF2(s) Pb2+(aq) + 2F-(aq) Ksp = [Pb2+][F-]2

= 2.6x10-3 M

Ksp = (2.6x10-3)(5.2x10-3)2 =

0.64g

L soln 245.2g PbF2

mol PbF2

7.0x10-8

19-27


Sample Problem 19.6 Determining Solubility from Ksp

PROBLEM: Calcium hydroxide (slaked lime) is a major component of mortar, plaster, and cement, and solutions of Ca(OH)2 are used in industry as a cheap, strong base. Calculate the solubility of Ca(OH)2 in water if the Ksp is 6.5x10-6.

PLAN: Write out a dissociation equation and Ksp expression; Find the molar solubility (S) using a table.

SOLUTION: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) Ksp = [Ca2+][OH-]2

-Initial

Change

Equilibrium

-

-

0 0

+S + 2S

S 2S

Ksp = (S)(2S)2 S =

€

6.5x10−6

43 = 1.2x10x-2M

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration (M)

19-28


Table 19.3 Relationship Between Ksp and Solubility at 250C

No. of Ions Formula Cation:Anion Ksp Solubility (M)

2 MgCO3 1:1 3.5 x 10-8 1.9 x 10-4

2 PbSO4 1:1 1.6 x 10-8 1.3 x 10-4

2 BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5

3 Ca(OH)2 1:2 5.5 x 10-6 1.2 x 10-2

3 BaF2 1:2 1.5 x 10-6 7.2 x 10-3

3 CaF2 1:2 3.2 x 10-11 2.0 x 10-4

3 Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5

19-29


Figure 19.10 The effect of a common ion on solubility.

PbCrO4(s) Pb2+(aq) + CrO42-(aq) PbCrO4(s) Pb2+(aq) + CrO4

2-(aq)

CrO42- added

19-30


Sample Problem 19.7 Calculating the Effect of a Common Ion on Solubility

PROBLEM: In Sample Problem 19.6, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.

PLAN: Set up a reaction equation and table for the dissolution of Ca(OH)2. The Ca(NO3)2 will supply extra [Ca2+] and will relate to the molar solubility of the ions involved.

SOLUTION: Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration(M)

Initial

Change

Equilibrium

-

-

-

0.10 0

+S +2S

0.10 + S 2S

Ksp = 6.5x10-6 = (0.10 + S)(2S)2 = (0.10)(2S)2 S << 0.10

S = = 4.0x10-3 Check the assumption:

4.0%

0.10M

4.0x10-3 x 100 =

€

6.5x10−64

19-31


Figure 19.11 Test for the presence of a carbonate.

19-32


Sample Problem 19.8 Predicting the Effect on Solubility of Adding Strong Acid

PROBLEM: Write balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds:

(a) Lead (II) bromide (b) Copper (II) hydroxide (c) Iron (II) sulfide

PLAN: Write dissolution equations and consider how strong acid would affect the anion component.

Br- is the anion of a strong acid.

No effect.

SOLUTION: (a) PbBr2(s) Pb2+(aq) + 2Br-(aq)

(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)

OH- is the anion of water, which is a weak acid. Therefore it will shift the solubility equation to the right and increase solubility.

(c) FeS(s) Fe2+(aq) + S2-(aq) S2- is the anion of a weak acid and will react with water to produce OH-.

Both weak acids serve to increase the solubility of FeS.

FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq)

19-33


Sample Problem 19.9 Predicting Whether a Precipitate Will Form

PROBLEM: A common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100L of 0.30M Ca(NO3)2 is mixed with 0.200L of 0.060M NaF?

PLAN: Write out a reaction equation to see which salt would be formed. Look up the Ksp valus in a table. Treat this as a reaction quotient, Q, problem and calculate whether the concentrations of ions are > or < Ksp. Remember to consider the final diluted solution when calculating concentrations.

SOLUTION: CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2x10-11

mol Ca2+ = 0.100L(0.30mol/L) = 0.030mol [Ca2+] = 0.030mol/0.300L = 0.10M

mol F- = 0.200L(0.060mol/L) = 0.012mol [F-] = 0.012mol/0.300L = 0.040M

Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6x10-4

Q is >> Ksp and the CaF2 WILL precipitate.

19-34


Figure 19.12 Formation of acidic precipitation.

19-35


Figure 19.13 Cr(NH3)63+, a typical complex ion.

19-36


Figure 19.14

M(H2O)42+

M(H2O)3(NH3)2+

M(NH3)42+

NH3

3NH3

The stepwise exchange of NH3 for H2O in M(H2O)42+.

19-37


Sample Problem 19.10 Calculating the Effect of Complex-Ion Formation on Solubility

SOLUTION:

PROBLEM: In black-and-white film developing, excess AgBr is removed from the film negative by “hypo”, an aqueous solution of sodium thiosulfate (Na2S2O3), through formation of the complex ion Ag(S2O3)2

3-. Calculate the solubility of AgBr in (a) H2O; (b) 1.0M hypo. Kf of Ag(S2O3)2

3- is 4.7x1013 and Ksp AgBr is 5.0x10-13.

PLAN: Write equations for the reactions involved. Use Ksp to find S, the molar solubility. Consider the shifts in equilibria upon the addition of the complexing agent.

AgBr(s) Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 5.0x10-13

S = [AgBr]dissolved = [Ag+] = [Br-] Ksp = S2 = 5.0x10-13 ; S = 7.1x10-7M(a)

(b) AgBr(s) Ag+(aq) + Br-(aq)

Ag+(aq) + 2S2O32-(aq) Ag(S2O3)2

3-(aq)

AgBr(s) + 2S2O32-(aq) Br -(aq) + Ag(S2O3)2

3-(aq)

19-38


Sample Problem 19.10 Calculating the Effect of Complex-Ion Formation on Solubility

continued

Koverall = Ksp x Kf = [Br-][Ag(S2O3]2

3-

[AgBr][S2O32-]2

= (5.0x10-13)(4.7x1013) = 24

AgBr(s) + 2S2O32-(aq) Br-(aq) + Ag(S2O3)2

3-(aq)Concentration(M)

Initial

Change

Equilibrium

-

-

-

1.0

-2S

1.0-2S

0 0

+S +S

S S

Koverall =S2

(1.0-2S)2

= 24 S

1.0-2S= (24)1/2

S = [Ag(S2O3)23-] = 0.45M

19-1 copyright ©the mcgraw-hill companies, inc. permission required for reproduction or display....

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