chapter 17 additional aspects of aqueous equilibriaholcombslab.yolasite.com/resources/chapter...

7/3/2012

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Aqueous

Equilibria

© 2009, Prentice-Hall, Inc.

Chapter 17

Additional Aspects of

Aqueous Equilibria

Chemistry, The Central Science, 11th edition

Theodore L. Brown; H. Eugene LeMay, Jr.;

and Bruce E. Bursten

John D. Bookstaver

St. Charles Community College

Cottleville, MO

Aqueous

Equilibria


Chapter 17 Problems

• Problems 11, 15, 17, 19, 21, 27, 33, 35,

47, 49, 51, 55, 61

Aqueous

Equilibria


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.

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

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Aqueous

Equilibria



“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.”

Aqueous

Equilibria



Calculate the fluoride ion concentration and pH

of a solution that is 0.20 M in HF and 0.10 M in

HCl.

Ka for HF is 6.8 10−4.

[H3O+] [F−]

[HF] Ka = = 6.8 10-4

Aqueous

Equilibria



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 −x +x +x

At Equilibrium 0.20 − x 0.20 0.10 + x 0.10 x

HF(aq) + H2O(l) H3O+(aq) + F−(aq)

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Aqueous

Equilibria



= x

1.4 10−3 = x

(0.10) (x)

(0.20) 6.8 10−4 =

(0.20) (6.8 10−4)

(0.10)

Aqueous

Equilibria



• Therefore, [F−] = x = 1.4 10−3

[H3O+] = 0.10 + x = 0.10 + 1.4 10−3 = 0.10 M

• So, pH = −log (0.10)

pH = 1.00

Aqueous

Equilibria


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.

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Aqueous

Equilibria


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.

Aqueous

Equilibria


Buffers

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

Aqueous

Equilibria


Buffer Calculations

Consider the equilibrium constant

expression for the dissociation of a

generic acid, HA:

[H3O+] [A−]

[HA] Ka =

HA + H2O H3O+ + A−

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Aqueous

Equilibria


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

pH acid

base

Aqueous

Equilibria


Buffer Calculations

• So

pKa = pH − log [base]

[acid]

• Rearranging, this becomes

pH = pKa + log [base]

[acid]

• This is the Henderson–Hasselbalch equation.

Aqueous

Equilibria


Henderson–Hasselbalch Equation

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 10−4.

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Aqueous

Equilibria


Henderson–Hasselbalch Equation

pH = pKa + log [base]

[acid]

pH = −log (1.4 10−4) + log (0.10)

(0.12)

pH

pH = 3.77

pH = 3.85 + (−0.08)

pH

Aqueous

Equilibria


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.

Aqueous

Equilibria


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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Aqueous

Equilibria


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 the Henderson–Hasselbalch

equation to determine the new

pH of the solution.

Aqueous

Equilibria


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.

Aqueous

Equilibria



Before the reaction, since

mol HC2H3O2 = mol C2H3O2−

pH = pKa = −log (1.8 10−5) = 4.74

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Aqueous

Equilibria



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

Aqueous

Equilibria



Now use the Henderson–Hasselbalch equation to

calculate the new pH:

pH = 4.74 + log (0.320)

(0.200)

pH = 4.74 + 0.06 pH

pH = 4.80

Aqueous

Equilibria


Titration

In this technique a

known concentration of

base (or acid) is slowly

added to a solution of

acid (or base).

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Aqueous

Equilibria


Titration

A pH meter or

indicators are used to

determine when the

solution has reached

the equivalence point,

at which the

stoichiometric amount

of acid equals that of

base.

Aqueous

Equilibria


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.

Aqueous

Equilibria



Strong Base Just before (and

after) the equivalence

point, the pH

increases rapidly.

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Aqueous

Equilibria



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.

Aqueous

Equilibria



Strong Base As more base is

added, the increase

in pH again levels

off.

Aqueous

Equilibria


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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Aqueous

Equilibria



Strong Base

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.

Aqueous

Equilibria



Strong Base With weaker acids,

the initial pH is

higher and pH

changes near the

equivalence point

are more subtle.

Aqueous

Equilibria


Titration of a Weak Base with a

Strong Acid

• The pH at the

equivalence point in

these titrations is < 7.

• Methyl red is the

indicator of choice.

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Aqueous

Equilibria


Titrations of Polyprotic Acids

When one

titrates a

polyprotic acid

with a base

there is an

equivalence

point for each

dissociation.

Aqueous

Equilibria


The Solubility Product

Constant, Ksp

Many ionic compounds are only slightly soluble

in water: ex. Ag salts, sulfides

Equations are written to represent the equilibrium

between the compound and the ions present in a

saturated aqueous solution

AgCl(s) Ag+(aq) + Cl–(aq)

c c

[Ag ][Cl ]... [AgCl] [Ag ][Cl ]

[AgCl]K but K

EOS

Ksp = [Ag+][Cl–]

Aqueous

Equilibria

SAMPLE EXERCISE Writing Solubility-Product (Ksp) Expressions

Write the expression for the solubility-product constant for CaF2, and look up the corresponding Ksp

value in Appendix D.

In Appendix D we see that this Ksp has a value of 3.9 10–11.

Solution

Analyze and Plan: We are asked to write an equilibrium-constant expression for the

process by which CaF2 dissolves in water. We apply the same rules for writing any

equilibrium-constant expression, making sure to exclude the solid reactant from the

expression. We assume that the compound dissociates completely into its component

ions.

Solve: Following the italicized rule stated previously, the expression for

Ksp is

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Aqueous

Equilibria

Ksp and Molar Solubility

The solubility product constant is related to

the solubility of an ionic solute

Ksp = [Ag+][Cl–]; solubility given by [Ag+]

From stoichiometry, the ion ratio is 1:1, so

[Ag+] = [Cl–], both of which are unknown

(x)

EOS

Ksp = x2 and [Ag+] = (Ksp)1/2

Ag+ Cl– Ag+ Cl– +

Aqueous

Equilibria

SAMPLE EXERCISE Calculating Ksp from Solubility

Solid silver chromate is added to pure water at 25°C. Some of the solid remains undissolved at the

bottom of the flask. 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 10–4 M. Assuming that Ag2CrO4 dissociates completely in water and

that there are no other important equilibria involving the Ag+ or CrO42– ions in the solution, calculate

Ksp for this compound. Solution

Analyze: We are given the equilibrium concentration of Ag+ in a saturated solution of Ag2CrO4. From

this, we are asked to determine the value of the solubility-product constant for the dissolution of

Ag2CrO4.

Plan: The equilibrium equation and the expression for Ksp are

To calculate Ksp, we need the equilibrium concentrations of Ag+ and CrO42–. We know that at

equilibrium [Ag+] = 1.3 10–4 M. All the Ag+ and CrO42– ions in the solution come from the Ag2CrO4

that dissolves. Thus, we can use [Ag+] to calculate [CrO42–].

Solve: From the chemical formula of silver chromate, we know that there must be 2 Ag+ ions in

solution for each CrO42– ion in solution. Consequently, the concentration of CrO4

2– is half the

concentration of Ag+.

We can now calculate the value of Ksp.

Aqueous

Equilibria

SAMPLE EXERCISE Calculating Solubility from Ksp

The Ksp for CaF2 is 3.9 10–11 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. Solution

Analyze: We are given Ksp for CaF2 and are asked to determine solubility. Recall that the solubility of

a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, Ksp, is

an equilibrium constant.

Plan: We can approach this problem by using our standard techniques for solving equilibrium

problems. We write the chemical equation for the dissolution process and set up a table of the initial

and equilibrium concentrations. We then use the equilibrium-constant expression. In this case we

know Ksp, and so we solve for the concentrations of the ions in solution.

Solve: Assume initially that none of the salt has dissolved, and then allow x moles/liter of CaF2 to

dissociate completely when equilibrium is achieved.

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Aqueous

Equilibria

SAMPLE EXERCISE continued

The stoichiometry of the equilibrium dictates that 2x moles/liter of F– are produced for each x

moles/liter of CaF2 that dissolve. We now use the expression for Ksp and substitute the equilibrium

concentrations to solve for the value of x:

(member that to calculate the cube root of a number, you can use the yx function on your

calculator, with ) Thus, the molar solubility of CaF2 is 2.1 10–4 mol/L. The mass of CaF2 that

dissolves in water to form a liter of solution is

Aqueous

Equilibria


Solubility Products

Consider the equilibrium that exists in a

saturated solution of BaSO4 in water:

BaSO4(s) Ba2+(aq) + SO42−(aq)

Aqueous

Equilibria


Solubility Products

The equilibrium constant expression for

this equilibrium is

Ksp = [Ba2+] [SO42−]

where the equilibrium constant, Ksp, is

called the solubility product.

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Aqueous

Equilibria


Ksp’s (25 oC)

EOS

Aqueous

Equilibria


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).

Aqueous

Equilibria


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.

BaSO4(s) Ba2+(aq) + SO42−(aq)

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Aqueous

Equilibria



• 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.

Aqueous

Equilibria



• Complex Ions

– Metal ions can act as Lewis acids and form

complex ions with Lewis bases in the solvent.

Aqueous

Equilibria



• Complex Ions

– The formation

of these

complex ions

increases the

solubility of

these salts.

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Aqueous

Equilibria



• 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+.

Aqueous

Equilibria


Will a Precipitate Form?

• In a solution,

– 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.

Aqueous

Equilibria


Applying the Criteria for Precipitation of a Slightly Soluble

Solute. Three drops of 0.20 M KI are added to 100.0 mL of 0.010

M Pb(NO3)2. Will a precipitate of lead iodide form?

(1 drop 0.05 mL)

PbI2(s) → Pb2+(aq) + 2 I-(aq) Ksp= 7.110-9

Determine the amount of I- in the solution:

= 310-5 mol I-

nI- = 3 drops 1 drop

0.05 mL

1000 mL

1 L

1 L

0.20 mol KI

1 mol KI

1 mol I-

EXAMPLE

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Aqueous

Equilibria


[I-] = 0.1000 L

310-5 mol I- = 310-4 M I-

Determine the concentration of I- in the solution:

Apply the Precipitation Criteria:

Q = [Pb2+][I-]2 = (0.010)(310-4)2

= 910-10 < Ksp = 7.110-9

EXAMPLE

Aqueous

Equilibria


Selective Precipitation of Ions

One can use

differences in

solubilities of

salts to separate

ions in a

mixture.

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