19.1 The common Ion Effect19.2 Controlling pH : Buffer Solution19.3 Acid and Base Titration

19.4 Solubility of Salts19.5 Precipitation Reactions19.6 Solubility Constant

Ch 19 Reactivity of Aqueous Solution

The common-ion effect to describe the effect on a solution of two dissolved solutes that contain the same ion.The presence of a common ion suppresses the ionization of a weak acid or a weak base.

Formation of acidic precipitation.

A view inside Carlsbad Caverns, New Mexico

The effect of acid rain on marble statuary.

1944 1994

Location: New York City

A forest damaged by acid rain

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-

Buffer Solution

Buffer solution is defined as a mixture of a conjugate acid and base pair.

Buffer solution corresponds to approximately 10-90 % titration of a relative flat region so the titration curve.

Buffer solution tends to resist changes in pH when an acid or base is added into the system.

Buffer solution is commonly used when pH must be maintained at a relative constant or in many biological systems.

Sample ProblemCalculating the Effect of Added H3O+ or OH- on Buffer pH

PROBLEM: 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)

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]= 1.810-5M

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

1.0 L soln= 0.020M NaOH

0.50 - 0.50 -

-

-

- 0.020 -

0.48 0 0.52

Calculating the Effect of Added H3O+ and OH-

on Buffer pH

continued (3 of 4)

Set up a reaction table with the new values.

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

InitialChange

Equilibrium

0.48 -- x

0.48 -x

-

-

0.52 0

x

+ x + x

0.52 + x

[H3O+] = 1.810-50.48

0.52= 1.710-5 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

Calculating the Effect of Added H3O+ and OH- on Buffer pHcontinued (4 of 4)

Set up a reaction table with the new values.

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

InitialChange

Equilibrium

0.52 -- x

0.52 -x

-

-

0.48 0

x

+ x + x

0.48 +x

[H3O+] = 1.810-5

0.48

0.52= 2.010-5 pH = 4.70

The Henderson-Hasselbalch Equation

pH = pKa + log [base]

[acid]

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 greater the 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 of its acid component.

pH = pKa + log [base]

[acid]

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.

Sample Problem 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 4.710-11.

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

Fig. 1 Curve for a strong acid-strong base titration

Fig. 2 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

Fig. 3 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

pKa = 7.19

pKa = 1.85

Curve for the titration of a weak polyprotic acid.

Titration of 40.00mL of 0.1000M H2SO3 with 0.1000M NaOH

Homework

Calculate pH for (b), (c), and (d)

Sample Problem 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 = 1.310-5) after adding the following volumes of 0.1000M NaOH:

(b) 30.00mL (c) 40.00mL (d) 50.00mL

PLAN: 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 (1.3x105)(0.10) x = 1.110-3 ; pH = 2.96

(a) 0.00mL

(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)

Continued

[H3O+] = 1.3x10-50.001000 mol

0.003000 mol= 4.3x10-6M pH = 5.37

PROBLEM: Calculate the pH during the titration of 40.00 mL of 0.1000M propanoic acid (HPr; Ka = 1.310-5) after adding the following volumes of 0.1000M NaOH:

Change 0.03000-0.03000 --0.03000

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

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

pH =

-log [H+]

Calculating the pH During a Weak Acid-Strong Base Titration

(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 = 1.0x10-14/1.310-5 = 7.7x10-10

[H3O+] = Kw / = 1.610-9M

Kbx[Pr ] pH = 8.80

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

mol V of base = (0.1000M)(0.05000L - 0.04000L) = 0.00100mol M = (0.00100)(0.0900L)

M = 0.01111[H3O+] = 1.010-14/0.01111 = 9.010-11M

pH = 12.05

Solubility Rules For Ionic Compounds in Water

1. All common compounds of Group 1A(1) ions (Li+, Na+, K+, etc.) and ammonium ion (NH4

+) are soluble.

2. All common nitrates (NO3-), acetates (CH3COO- or C2H3O2

-) and most perchlorates (ClO4

-) are soluble.

3. All common chlorides (Cl-), bromides (Br-) and iodides (I-) are soluble, except those of Ag+, Pb2+, Cu+, and Hg2

2+.

1. All common metal hydroxides are insoluble, except those of Group 1A(1) and the larger members of Group 2A(2)(beginning with Ca2+).

2. All common carbonates (CO32-) and phosphates (PO4

3-) SO32-, SO4

2-, S2- are insoluble, except those of Group 1A(1) and NH4

+.

3. All common sulfides are insoluble except those of Group 1A(1), Group 2A(2) and NH4

+.

Soluble Ionic Compounds

Insoluble Ionic Compounds

Sample Problem 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

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

3-(aq)

Sample Problem Determining Ksp from Solubility

PROBLEM: 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 (net ionic equation); find moles of dissociated ions; convert solubility to M and substitute values into solubility product constant expression.

SOLUTION:

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

= 2.610 -3 M

Ksp = (2.610-3)(5.210-3)2 =

0.64g

L soln 245.2g PbF2

mol PbF2

7.010 -8

Sample Problem 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 6.510-6.

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 = 1.210-2M

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

The effect of a common ion on solubility

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

2-(aq)

CrO42- added

Sample Problem Calculating the Effect of a Common Ion on Solubility

PROBLEM: In Sample Problem above, 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 6

4

Sample Problem 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)

Sample Problem 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.

Sample Problem Separating Ions by Selective Precipitation

continued

Use the Ksp for Cu(OH)2 to find the amount of Cu remaining.

[Cu2+] = Ksp/[OH-]2 = 2.2x10-20/(5.6x10-5)2 = 7.0x10-12M

Since the solution was 0.10M CuCl2, virtually none of the Cu2+ remains in solution.