Chapter 16 Aqueous Ionic Equilibrium

Copyright 2011 Pearson Education, Inc.

Chapter 16Aqueous Ionic

Equilibrium

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro


ethylene glycol(aka 1,2–ethandiol)

The Danger of Antifreeze• Each year, thousands of pets and wildlife

species die from consuming antifreeze• Most brands of antifreeze contain ethylene

glycolsweet tasteinitial effect drunkenness

• Metabolized in the liver to glycolic acidHOCH2COOH

2

glycolic acid(aka a-hydroxyethanoic acid)

Tro: Chemistry: A Molecular Approach, 2/e


Why Is Glycolic Acid Toxic?• If present in high enough concentration in the

bloodstream, glycolic acid overwhelms the buffering ability of the HCO3

− in the blood, causing the blood pH to drop

• When the blood pH is low, its ability to carry O2 is compromisedacidosis

HbH+(aq) + O2(g) HbO2(aq) + H+(aq)• One treatment is to give the patient ethyl

alcohol, which has a higher affinity for the enzyme that catalyzes the metabolism of ethylene glycol

3Tro: Chemistry: A Molecular Approach, 2/e


Buffers• Buffers are solutions that resist changes in pH

when an acid or base is added• They act by neutralizing acid or base that is

added to the buffered solution• But just like everything else, there is a limit to

what they can do, and eventually the pH changes• Many buffers are made by mixing a solution of a

weak acid with a solution of soluble salt containing its conjugate base anionblood has a mixture of H2CO3 and HCO3

−



Making an Acid Buffer



How Acid Buffers Work:Addition of Base

HA(aq) + H2O(l) A−(aq) + H3O+

(aq)

• Buffers work by applying Le Châtelier’s Principle to weak acid equilibrium

• Buffer solutions contain significant amounts of the weak acid molecules, HA

• These molecules react with added base to neutralize it

HA(aq) + OH−(aq) → A−(aq) + H2O(l)you can also think of the H3O+ combining with the OH−

to make H2O; the H3O+ is then replaced by the shifting equilibrium



H2O

HA

How Buffers Work

HA + H3O+

A−

AddedHO−

newA−

A−



How Acid Buffers Work:Addition of Acid

HA(aq) + H2O(l) A−(aq) + H3O+

(aq)

• The buffer solution also contains significant amounts of the conjugate base anion, A−

• These ions combine with added acid to make more HA

H+(aq) + A−(aq) → HA(aq)• After the equilibrium shifts, the concentration

of H3O+ is kept constant



H2O

How Buffers Work

HA + H3O+A−A−

AddedH3O+

newHA

HA



Common Ion Effect HA(aq) + H2O(l) A−

(aq) + H3O+(aq)

• Adding a salt containing the anion NaA, which is the conjugate base of the acid (the common ion), shifts the position of equilibrium to the left

• This causes the pH to be higher than the pH of the acid solutionlowering the H3O+ ion concentration



Common Ion Effect



Example 16.1: What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2?

12

write the reaction for the acid with water

construct an ICE table for the reaction

enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HC2H3O2 + H2O C2H3O2 + H3O+

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change

equilibrium



[HA] [A−] [H3O+]

initial 0.100 0.100 0

change

equilibrium


13

represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+xx0.100 x 0.100 + x x





14

determine the value of Ka

because Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init , then solve for x

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 x0.100 x 0.100 +x

Ka for HC2H3O2 = 1.8 x 10−5




15

check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init

Ka for HC2H3O2 = 1.8 x 10−5

the approximation is valid

x = 1.8 x 10−5

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 x




16

substitute x into the equilibrium concentration definitions and solve

x = 1.8 x 10−5

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 1.8E-50.100 + x x 0.100 x




17

substitute [H3O+] into the formula for pH and solve

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x

equilibrium 0.100 0.100 1.8E−5




18

check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka

the values match

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0

change −x +x +x


Ka for HC2H3O2 = 1.8 x 10−5



Practice − What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?



Practice − What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?

20



enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HF + H2O F + H3O+

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change

equilibrium



[HA] [A−] [H3O+]

initial 0.14 0.071 0

change

equilibrium

Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?

21




+x+xx0.14 x 0.071 + x x

HF + H2O F + H3O+



pKa for HF = 3.15Ka for HF = 7.0 x 10−4

Practice – What is the pH of a buffer that is 0.14 M and 0.071 M KF?

22


because Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x

equilibrium 0.012 0.100 x0.14 x 0.071 +x




23


Ka for HF = 7.0 x 10−4


x = 1.4 x 10−3

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x





24


x = 1.4 x 10−3

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x





25


[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x





26


the values are close enough

[HA] [A−] [H3O+]

initial 0.14 0.071 ≈ 0

change −x +x +x


Ka for HF = 7.0 x 10−4



Henderson-Hasselbalch Equation• Calculating the pH of a buffer solution can be

simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation

• The equation calculates the pH of a buffer from the pKa and initial concentrations of the weak acid and salt of the conjugate baseas long as the “x is small” approximation is valid



Deriving the Henderson-Hasselbalch Equation




29

assume the [HA] and [A−] equilibrium concentrations are the same as the initial

substitute into the Henderson-Hasselbalch Equation

check the “x is small” approximation


Ka for HC7H5O2 = 6.5 x 10−5







31

find the pKa from the given Ka


substitute into the Henderson-Hasselbalch equation


HF + H2O F + H3O+



Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?• The Henderson-Hasselbalch equation is

generally good enough when the “x is small” approximation is applicable

• Generally, the “x is small” approximation will work when both of the following are true:

a) the initial concentrations of acid and salt are not very dilute

b) the Ka is fairly small• For most problems, this means that the initial

acid and salt concentrations should be over 100 to 1000x larger than the value of Ka



How Much Does the pH of a Buffer Change When an Acid or Base Is Added?• Though buffers do resist change in pH when acid

or base is added to them, their pH does change• Calculating the new pH after adding acid or base

requires breaking the problem into two parts1. a stoichiometry calculation for the reaction of the

added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A− to make more HA added base reacts with the HA to make more A−

2. an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]



Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00

L that has 0.010 mol NaOH added to it?

34

If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for H3O+ with A−.

construct a stoichiometry table for the reaction

HC2H3O2 + OH− C2H3O2 + H2O

HA A− OH−

mols before 0.100 0.100 0

mols added ─ ─ 0.010

mols after



Example 16.3: What is the pH of a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in

1.00 L that has 0.010 mol NaOH added to it?

35

fill in the table – tracking the changes in the number of moles for each component


HA A− OH−

mols before 0.100 0.100 ≈ 0


mols after 0.090 0.110 ≈ 0





36

If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for it with A−.


enter the initial number of moles for each


HA A– OH−

mols before 0.100 0.100 0.010

mols change

mols end

new molarity





37

using the added chemical as the limiting reactant, determine how the moles of the other chemicals change

add the change to the initial number of moles to find the moles after reaction

divide by the liters of solution to find the new molarities


HA A− OH−

mols before 0.100 0.100 0.010

mols change

mols end

new molarity

−0.010−0.010 +0.010

00.1100.090

0.090 0.110 0





38



enter the initial concentrations – assuming the [H3O+] from water is ≈ 0, and using the new molarities of the [HA] and [A−]


[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change

equilibrium



[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change

equilibrium



39




+x+xx0.090 x 0.110 + x x






40


because Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x

[HA] [A−] [H3O+]

initial 0.100 0.100 ≈ 0change −x +x +x

equilibrium 0.090 0.110 x0.090 x 0.110 +x

Ka for HC2H3O2 = 1.8 x 10−5





41


Ka for HC2H3O2 = 1.8 x 10−5


x = 1.47 x 10−5

[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change −x +x +x






42


x = 1.47 x 10−5

[HA] [A-] [H3O+]

initial 0.090 0.110 ≈ 0

change −x +x +x






43


[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0change −x +x +x






44


the values match

[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change −x +x +x


Ka for HC2H3O2 = 1.8 x 10−5



or, by using the

Henderson-Hasselbalch

Equation




46



enter the initial concentrations – assuming the [H3O+] from water is ≈ 0, and using the new molarities of the [HA] and [A−]

fll in the ICE table


[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change

equilibrium

+x+xx0.090 x 0.110 + x x





47

find the pKa from the given Ka



Ka for HC2H3O2 = 1.8 x 10−5

[HA] [A−] [H3O+]

initial 0.090 0.110 ≈ 0

change −x +x +x






48




pKa for HC2H3O2 = 4.745



Example 16.3: Compare the effect on pH of adding 0.010 mol NaOH to a buffer that has 0.100 mol HC2H3O2 and 0.100 mol NaC2H3O2 in 1.00 L to

adding 0.010 mol NaOH to 1.00 L of pure water

49


pKa for HC2H3O2 = 4.745



Practice – What is the pH of a buffer that has 0.140 moles HF (pKa = 3.15) and 0.071 moles KF in 1.00 L of solution when 0.020 moles of

HCl is added? (The “x is small” approximation is valid)



Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of

solution when 0.020 moles of HCl is added?

51



F− + H3O+ HF + H2O

F− H3O+ HF

mols before 0.071 0 0.140

mols added – 0.020 –

mols after





52

fill in the table – tracking the changes in the number of moles for each component

F− + H3O+ HF + H2O

F− H3O+ HF

mols before 0.071 0 0.140

mols added – 0.020 –

mols after 0.051 0 0.160





53



enter the initial number of moles for each

F− + H3O+ HF + H2O

F− H3O+ HF

mols before 0.071 0.020 0.140

mols change

mols after

new molarity



F− H3O+ HFmols before 0.071 0.020 0.140

mols change

mols after

new molarity



54

using the added chemical as the limiting reactant, determine how the moles of the other chemicals change

add the change to the initial number of moles to find the moles after reaction

divide by the liters of solution to find the new molarities

0.16000.051

F− + H3O+ HF + H2O

+0.020−0.020 −0.020

0.16000.051





55


construct an ICE table.



HF + H2O F + H3O+

[HF] [F−] [H3O+]

initial 0.160 0.051 ≈ 0

change −x +x +x




Basic BuffersB:(aq) + H2O(l) H:B+

(aq) + OH−(aq)

• Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−

H2O(l) + NH3 (aq) NH4+

(aq) + OH−(aq)



• The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product

• The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product

B: + H2O H:B+ + OH−

• To apply the Henderson-Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reactionH:B+ + H2O B: + H3O+

this does not affect the concentrations, just the way we are looking at the reaction

Henderson-Hasselbalch Equation for Basic Buffers



Relationship between pKa and pKb

• Just as there is a relationship between the Ka of a weak acid and Kb of its conjugate base, there is also a relationship between the pKa of a weak acid and the pKb of its conjugate base

Ka Kb = Kw = 1.0 x 10−14

−log(Ka Kb) = −log(Kw) = 14

−log(Ka) + −log(Kb) = 14

pKa + pKb = 14



Example 16.4: What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl?

59

find the pKa of the conjugate acid (NH4

+) from the given Kb

assume the [B] and [HB+] equilibrium concentrations are the same as the initial



NH3 + H2O NH4+ + OH−



• The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product

• The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product

B: + H2O H:B+ + OH−

• We can rewrite the Henderson-Hasselbalch equation for the chemical equation of the basic buffer in terms of pOH

Henderson-Hasselbalch Equation for Basic Buffers



Example 16.4: What is the pH of a buffer that is 0.50 M NH3 (pKb = 4.75) and 0.20 M NH4Cl?

61

find the pKb if given Kb

assume the [B] and [HB+] equilibrium concentrations are the same as the initial

substitute into the Henderson-Hasselbalch equation base form, find pOH


calculate pH from pOH

NH3 + H2O NH4+ + OH−



Buffering Effectiveness• A good buffer should be able to neutralize

moderate amounts of added acid or base• However, there is a limit to how much can be

added before the pH changes significantly• The buffering capacity is the amount of acid or

base a buffer can neutralize• The buffering range is the pH range the buffer

can be effective• The effectiveness of a buffer depends on two

factors (1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base



HA A− OH−



mols after 0.17 0.030 ≈ 0

Effect of Relative Amounts of Acid and Conjugate Base

Buffer 10.100 mol HA & 0.100 mol A−

Initial pH = 5.00


Initial pH = 4.05pKa (HA) = 5.00

after adding 0.010 mol NaOHpH = 5.09

HA + OH− A + H2O

HA A− OH−



mols after 0.090 0.110 ≈ 0


A buffer is most effective with equal concentrations of acid and base


HA A− OH−



mols after 0.49 0.51 ≈ 0

HA A− OH−



mols after 0.040 0.060 ≈ 0

Effect of Absolute Concentrations of Acid and Conjugate Base


Initial pH = 5.00


Initial pH = 5.00pKa (HA) = 5.00


HA + OH− A + H2O


A buffer is most effective when the concentrations of acid and base are largest


Buffering Capacity

a concentrated buffer can neutralize more added acid or base than a dilute buffer



Effectiveness of Buffers

• A buffer will be most effective when the [base]:[acid] = 1equal concentrations of acid and base

• A buffer will be effective when 0.1 < [base]:[acid] < 10

• A buffer will be most effective when the [acid] and the [base] are large



Buffering Range• We have said that a buffer will be effective when

0.1 < [base]:[acid] < 10• Substituting into the Henderson-Hasselbalch

equation we can calculate the maximum and minimum pH at which the buffer will be effective

Lowest pH Highest pH

Therefore, the effective pH range of a buffer is pKa ± 1When choosing an acid to make a buffer, choose one whose is pKa closest to the pH of the buffer



Formic acid, HCHO2 pKa = 3.74Formic acid, HCHO2 pKa = 3.74

Example 16.5a: Which of the following acids would be the best choice to combine with its sodium salt to make a buffer with pH 4.25?

Chlorous acid, HClO2 pKa = 1.95

Nitrous acid, HNO2 pKa = 3.34

Hypochlorous acid, HClOpKa = 7.54

The pKa of HCHO2 is closest to the desired pH of the buffer, so it would give the most effective buffering range



Example 16.5b: What ratio of NaCHO2 : HCHO2 would be required to make a buffer with pH 4.25?

69

Formic acid, HCHO2, pKa = 3.74

To make a buffer with pH 4.25, you would use 3.24 times as much NaCHO2 as HCHO2



Practice – What ratio of NaC7H5O2 : HC7H5O2 would be required to make a buffer with pH 3.75?

Benzoic acid, HC7H5O2, pKa = 4.19



Practice – What ratio of NaC7H5O2 : HC7H5O2 would be required to make a buffer with pH 3.75?

Benzoic acid, HC7H5O2, pKa = 4.19

to make a buffer with pH 3.75, you would use 0.363 times as much NaC7H5O2 as HC7H5O2



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

that can be added to a buffer without causing a large change in pH

• The buffering capacity increases with increasing absolute concentration of the buffer components

• As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added acid and base improves

• Buffers that need to work mainly with added acid generally have [base] > [acid]

• Buffers that need to work mainly with added base generally have [acid] > [base]



Titration• In an acid-base titration, a solution of unknown

concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is completewhen the reaction is complete we have reached the

endpoint of the titration• An indicator may be added to determine the

endpointan indicator is a chemical that changes color when the

pH changes

• When the moles of H3O+ = moles of OH−, the titration has reached its equivalence point



Titration



Titration Curve• A plot of pH vs. amount of added titrant• The inflection point of the curve is the equivalence

point of the titration• Prior to the equivalence point, the known solution in

the flask is in excess, so the pH is closest to its pH• The pH of the equivalence point depends on the pH of

the salt solutionequivalence point of neutral salt, pH = 7 equivalence point of acidic salt, pH < 7 equivalence point of basic salt, pH > 7

• Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH



Titration Curve:Unknown Strong Base Added to

Strong Acid



Before Equivalence(excess acid)

After Equivalence(excess base)

Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH

Equivalence Point equal moles of HCl and NaOH

pH = 7.00

Because the solutions are equal concentration, and 1:1 stoichiometry, the equivalence point is at equal volumes



HCl NaCl NaOH

mols before 2.50E-3 0 5.0E-4

mols change −5.0E-4 +5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new 0.0667 0.017 0

HCl NaCl NaOH


mols change

mols end

molarity, new

HCl NaCl NaOH


mols change −5.0E-4 +5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new

5.0 x 10−4 mole NaOH added


• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

• Initial pH = −log(0.100) = 1.00• Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalence point added 5.0 mL NaOH




• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)

• To reach equivalence, the added moles NaOH = initial moles of HCl = 2.50 x 10−3 moles

• At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH left over

• Because the NaCl is a neutral salt, the pH at equivalence = 7.00



HCl NaCl NaOH


mols change

mols end

molarity, new

HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 0

molarity, new

HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 0

molarity, new 0 0.050 0

2.5 x 10−3 mole NaOH added




• At equivalence point added 25.0 mL NaOH



HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new 0 0.045 0.0091

HCl NaCl NaOH


mols change

mols end

molarity, new

HCl NaCl NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new




• After equivalence point added 30.0 mL NaOH3.0 x 10−3 mole NaOH added

L of added NaOH0.100 mol NaOH

1 Lmoles added NaOH


added 5.0 mL NaOH0.00200 mol HClpH = 1.18


added 25.0 mL NaOHequivalence pointpH = 7.00

added 30.0 mL NaOH0.00050 mol NaOHpH = 11.96






Adding 0.100 M NaOH to 0.100 M HCl

82

25.0 mL 0.100 M HCl0.00250 mol HClpH = 1.00



Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to

50.0 mL of 0.25 M HNO3



HNO3 NaNO3 NaOH


mols change −1.5E-3 +1.5E-3 −1.5E-3

mols end 1.1E-3 1.5E-3 0

molarity, new

HNO3 NaNO3 NaOH


mols change −1.5E-3 +1.5E-3 −1.5E-3

mols end 1.1E-3 1.5E-3 0

molarity, new 0.018 0.025 0


50.0 mL of 0.25 M HNO3 • HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)

• Initial pH = −log(0.250) = 0.60• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2

• Before equivalence point added 10.0 mL NaOH



Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M

HNO3 to reach equivalence



Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M

HNO3 to reach equivalence

• HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)

• Initial pH = −log(0.250) = 0.60

• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2

• At equivalence point: moles of NaOH = 1.25 x 10−2




50.0 mL of 0.25 M HNO3



HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new 0 0.0833 0.017

HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new



• Initial pH = −log(0.250) = 0.60• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2

• After equivalence point added 100.0 mL NaOH


HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new 0 0.0833 0.017

HNO3 NaNO3 NaOH


mols change −1.25E-2 +1.25E-2 −1.25E-2

mols end 0 1.25E-2 0.0025

molarity, new



• Initial pH = −log(0.250) = 0.60• initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2

• After equivalence point added 100.0 mL NaOH



Titration of a Strong Base with a Strong Acid

• If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown



Titration of a Weak Acid with a Strong Base• Titrating a weak acid with a strong base results in

differences in the titration curve at the equivalence point and excess acid region

• The initial pH is determined using the Ka of the weak acid

• The pH in the excess acid region is determined as you would determine the pH of a buffer

• The pH at the equivalence point is determined using the Kb of the conjugate base of the weak acid

• The pH after equivalence is dominated by the excess strong base the basicity from the conjugate base anion is negligible



Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH

• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)

• Initial pH

[HCHO2] [CHO2−] [H3O+]

initial 0.100 0.000 ≈ 0

change −x +x +x

equilibrium 0.100 − x x x

Ka = 1.8 x 10−4




• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)

• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalence added 5.0 mL NaOHHA A− OH−

mols before 2.50E-3 0 0

mols added – – 5.0E-4

mols after 2.00E-3 5.0E-4 ≈ 0



HA A− OH−

mols before 2.50E-3 0 0

mols added – – 2.50E-3

mols after 0 2.50E-3 ≈ 0

[HCHO2] [CHO2−] [OH−]

initial 0 0.0500 ≈ 0

change +x −x +x

equilibrium x 5.00E-2-x x


added 25.0 mL NaOHCHO2

−(aq) + H2O(l) HCHO2(aq) + OH−

(aq)

Kb = 5.6 x 10−11

[OH−] = 1.7 x 10−6 M

• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)


• At equivalence


HA A− NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new

HA A− NaOH


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new 0 0.045 0.0091

HA A− NaOH


mols change

mols end

molarity, new


95

added 30.0 mL NaOH3.0 x 10−3 mole NaOH added

• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)


• After equivalence


added 35.0 mL NaOH0.00100 mol NaOH xspH = 12.22

initial HCHO2 solution0.00250 mol HCHO2

pH = 2.37

added 5.0 mL NaOH0.00200 mol HCHO2

pH = 3.14


pH = 3.56

added 25.0 mL NaOHequivalence point0.00250 mol CHO2

−

[CHO2−]init = 0.0500 M

[OH−]eq = 1.7 x 10−6

pH = 8.23



pH = 4.34


pH = 3.92


pH = 3.74 = pKa

half-neutralization

Adding NaOH to HCHO2





Titrating Weak Acid with a Strong Base

• The initial pH is that of the weak acid solutioncalculate like a weak acid equilibrium problem

e.g., 15.5 and 15.6

• Before the equivalence point, the solution becomes a buffercalculate mol HAinit and mol A−

init using reaction stoichiometry

calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−

init

• Half-neutralization pH = pKa



Titrating Weak Acid with a Strong Base

• At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is establishedmol A− = original mole HA

calculate the volume of added base as you did in Example 4.8

[A−]init = mol A−/total literscalculate like a weak base equilibrium problem

e.g., 15.14

• Beyond equivalence point, the OH is in excess[OH−] = mol MOH xs/total liters[H3O+][OH−]=1 x 10−14



Example 16.7a: A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the volume

of KOH at the equivalence point.

99

write an equation for the reaction for B with HA

use stoichiometry to determine the volume of added B

HNO2 + KOH NO2 + H2O



Example 16.7b: A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after

adding 5.00 mL KOH.

100


determine the moles of HAbefore & moles of added B

make a stoichiometry table and determine the moles of HA in excess and moles A made


HNO2 NO2− OH−

mols before 0.00400 0 ≈ 0

mols added – – 0.00100

mols after ≈ 00.00300 0.00100



Example 16.7b: A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH after

adding 5.00 mL KOH.

101

write an equation for the reaction of HA with H2O

determine Ka and pKa for HA

use the Henderson-Hasselbalch equation to determine the pH

HNO2 + H2O NO2 + H3O+

HNO2 NO2− OH−



mols after 0.00300 0.00100 ≈ 0

Table 15.5 Ka = 4.6 x 10−4



Example 16.7b: A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at

the half-equivalence point.

102


determine the moles of HAbefore & moles of added B

make a stoichiometry table and determine the moles of HA in excess and moles A made


HNO2 NO2− OH−



mols after ≈ 00.00200 0.00200

at half-equivalence, moles KOH = ½ mole HNO2



Example 16.7b: A 40.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the pH at

the half-equivalence point.

103

write an equation for the reaction of HA with H2O

determine Ka and pKa for HA

use the Henderson-Hasselbalch equation to determine the pH

HNO2 + H2O NO2 + H3O+

HNO2 NO2− OH−



mols after 0.00200 0.00200 ≈ 0

Table 15.5 Ka = 4.6 x 10-4



Titration Curve of a Weak Base with a Strong Acid



Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the initial pH of the

NH3 solution.



Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M HCl. Calculate the initial pH of the NH3(aq)

106

• NH3(aq) + HCl(aq) NH4Cl(aq)

• Initial: NH3(aq) + H2O(l) NH4+

(aq) + OH−(aq)

[HCl] [NH4+] [NH3]

initial 0 0 0.10

change +x +x −x

equilibrium x x 0.10−x

pKb = 4.75Kb = 10−4.75 = 1.8 x 10−5



Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M HCl. Calculate the initial pH of the NH3(aq)

107


• Initial: NH3(aq) + H2O(l) NH4+

(aq) + OH−(aq)

[HCl] [NH4+] [NH3]

initial 0 0 0.10

change +x +x −x

equilibrium x x 0.10−x

pKb = 4.75Kb = 10−4.75 = 1.8 x 10−5



Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution

after adding 5.0 mL of HCl.



Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the

solution after adding 5.0 mL of HCl.

109


• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3

• Before equivalence: after adding 5.0 mL of HCl

NH3 NH4Cl HCl


mols change −5.0E-4 −5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new 0.0667 0.017 0

NH4+

(aq) + H2O(l) NH4+

(aq) + H2O(l) pKb = 4.75pKa = 14.00 − 4.75 = 9.25



Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution

after adding 5.0 mL of HCl.• NH3(aq) + HCl(aq) NH4Cl(aq)


• Before equivalence: after adding 5.0 mL of HCl

NH3 NH4Cl HCl


mols change −5.0E-4 −5.0E-4 −5.0E-4

mols end 2.00E-3 5.0E-4 0

molarity, new 0.0667 0.017 0




solution at equivalence.





112



• At equivalence mol NH3 = mol HCl = 2.50 x 10−3

added 25.0 mL HClNH3 NH4Cl HCl


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 0

molarity, new 0 0.050 0





113

NH3(aq) + HCl(aq) NH4Cl(aq)

at equivalence [NH4Cl] = 0.050 M

[NH3] [NH4+] [H3O+]

initial 0 0.050 ≈ 0

change +x −x +x

equilibrium x 0.050−x x

NH4+

(aq) + H2O(l) NH3(aq) + H3O+(aq)









115



• After equivalence: after adding 30.0 mL HClNH3 NH4Cl HCl


mols change −2.5E-3 +2.5E-3 −2.5E-3

mols end 0 2.5E-3 5.0E-4

molarity, new 0 0.045 0.0091

when you mix a strong acid, HCl, with a weak acid, NH4

+, you only need to consider the strong acid



Titration of a Polyprotic Acid

• If Ka1 >> Ka2, there will be two equivalence points in the titrationthe closer the Ka’s are to each other, the less

distinguishable the equivalence points are

titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH



Monitoring pH During a Titration

• The general method for monitoring the pH during the course of a titration is to measure the conductivity of the solution due to the [H3O+]using a probe that specifically measures just H3O+

• The endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve

• If you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator



Monitoring pH During a Titration



Indicators• Many dyes change color depending on the pH of the

solution• These dyes are weak acids, establishing an

equilibrium with the H2O and H3O+ in the solution

HInd(aq) + H2O(l) Ind(aq) + H3O+

(aq)

• The color of the solution depends on the relative concentrations of Ind:HIndwhen Ind:HInd ≈ 1, the color will be mix of the colors of

Ind and HInd when Ind:HInd > 10, the color will be mix of the colors

of Ind

when Ind:HInd < 0.1, the color will be mix of the colors of HInd



Phenolphthalein



Methyl Red

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N NH

NaOOC

C

C CH

CH

CH

CH

C

CH

CH

C

CH

CH

(CH3)2N N N N

NaOOC

H3O+ OH-



Monitoring a Titration with an Indicator

• For most titrations, the titration curve shows a very large change in pH for very small additions of titrant near the equivalence point

• An indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pHpKa of HInd ≈ pH at equivalence point



Acid-Base Indicators



Solubility Equilibria

• All ionic compounds dissolve in water to some degree however, many compounds have such low solubility

in water that we classify them as insoluble

• We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water



Solubility Product• The equilibrium constant for the dissociation of a

solid salt into its aqueous ions is called the solubility product, Ksp

• For an ionic solid MnXm, the dissociation reaction is:MnXm(s) nMm+(aq) + mXn−(aq)

• The solubility product would be Ksp = [Mm+]n[Xn−]m

• For example, the dissociation reaction for PbCl2 isPbCl2(s) Pb2+(aq) + 2 Cl−(aq)

• And its equilibrium constant is Ksp = [Pb2+][Cl−]2


Copyright 2011 Pearson Education, Inc.126Tro: Chemistry: A Molecular Approach, 2/e


Molar Solubility• Solubility is the amount of solute that will

dissolve in a given amount of solutionat a particular temperature

• The molar solubility is the number of moles of solute that will dissolve in a liter of solutionthe molarity of the dissolved solute in a saturated

solution

for the general reaction MnXm(s) nMm+(aq) + mXn−(aq)



Example 16.8: Calculate the molar solubility of PbCl2 in pure water at 25 C

128

write the dissociation reaction and Ksp expression

create an ICE table defining the change in terms of the solubility of the solid

[Pb2+] [Cl−]

Initial 0 0

Change +S +2S

Equilibrium S 2S

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)

Ksp = [Pb2+][Cl−]2



Example 16.8: Calculate the molar solubility of PbCl2 in pure water at 25 C

129

substitute into the Ksp expression

find the value of Ksp from Table 16.2, plug into the equation, and solve for S

[Pb2+] [Cl−]

Initial 0 0

Change +S +2S

Equilibrium S 2S

Ksp = [Pb2+][Cl−]2

Ksp = (S)(2S)2



Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25 C is 1.05 x 10−2 M




131



[Pb2+] [Br−]

initial 0 0

change +(1.05 x 10−2) +2(1.05 x 10−2)

equilibrium (1.05 x 10−2) (2.10 x 10−2)

PbBr2(s) Pb2+(aq) + 2 Br−(aq)

Ksp = [Pb2+][Br−]2




132

substitute into the Ksp expression

plug into the equation and solve

Ksp = [Pb2+][Br−]2

Ksp = (1.05 x 10−2)(2.10 x 10−2)2

[Pb2+] [Br−]

initial 0 0

change +(1.05 x 10−2) +2(1.05 x 10−2)

equilibrium (1.05 x 10−2) (2.10 x 10−2)



Ksp and Relative Solubility

• Molar solubility is related to Ksp

• But you cannot always compare solubilities of compounds by comparing their Ksps

• To compare Ksps, the compounds must have the same dissociation stoichiometry



The Effect of Common Ion on Solubility

• Addition of a soluble salt that contains one of the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt

• For example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2


addition of Cl− shifts the equilibrium to the left



Example 16.10: Calculate the molar solubility of CaF2 in 0.100 M NaF at 25 C

135



[Ca2+] [F−]

initial 0 0.100

change +S +2S

equilibrium S 0.100 + 2S

CaF2(s) Ca2+(aq) + 2 F−(aq)

Ksp = [Ca2+][F−]2



Example 16.10: Calculate the molar solubility of CaF2 in 0.100 M NaF at 25 C

136

substitute into the Ksp expression,assume S is small


[Ca2+] [F−]

initial 0 0.100

change +S +2S

equilibrium S 0.100 + 2S

Ksp = [Ca2+][F−]2

Ksp = (S)(0.100 + 2S)2

Ksp = (S)(0.100)2



Practice – Determine the concentration of Ag+ ions in seawater that has a [Cl−] of 0.55 M

Ksp of AgCl = 1.77 x 10−10






[Ag+] [Cl−]

initial 0 0.55

change +S +S

equilibrium S 0.55 + S

AgCl(s) Ag+(aq) + Cl−(aq)

Ksp = [Ag+][Cl−]




139

substitute into the Ksp expression,assume S is small


[Ag+] [Cl−]

Initial 0 0.55

Change +S +S

Equilibrium S 0.55 + S

Ksp = [Ag+][Cl−]

Ksp = (S)(0.55 + S)

Ksp = (S)(0.55)



The Effect of pH on Solubility

• For insoluble ionic hydroxides, the higher the pH, the lower the solubility of the ionic hydroxideand the lower the pH, the higher the solubilityhigher pH = increased [OH−]

M(OH)n(s) Mn+(aq) + nOH−(aq)

• For insoluble ionic compounds that contain anions of weak acids, the lower the pH, the higher the solubility

M2(CO3)n(s) 2 Mn+(aq) + nCO32−(aq)

H3O+(aq) + CO32− (aq) HCO3

− (aq) + H2O(l)



Precipitation• Precipitation will occur when the concentrations of

the ions exceed the solubility of the ionic compound

• If we compare the reaction quotient, Q, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occurQ = Ksp, the solution is saturated, no precipitationQ < Ksp, the solution is unsaturated, no precipitationQ > Ksp, the solution would be above saturation, the salt

above saturation will precipitate

• Some solutions with Q > Ksp will not precipitate unless disturbed – these are called supersaturated solutions



precipitation occurs if Q > Ksp

a supersaturated solution will precipitate if a seed crystal is

added



Selective Precipitation

• A solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others

• A successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different



Example 16.12: Will a precipitate form when we mix Pb(NO3)2(aq) with NaBr(aq) if the concentrations after mixing are 0.0150 M and 0.0350 M respectively?

write the equation for the reaction

determine the ion concentrations of the original salts

determine the Ksp for any “insoluble” product

write the dissociation reaction for the insoluble product

calculate Q, using the ion concentrations

compare Q to Ksp. If Q > Ksp, precipitation

Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq)

Ksp of PbBr2 = 4.67 x 10–6

PbBr2(s) Pb2+(aq) + 2 Br−

(aq)

Pb(NO3)2 = 0.0150 M Pb2+ = 0.0150 M,

NO3− = 2(0.0150 M)

NaBr = 0.0350 M Na+ = 0.0350 M,

Br− = 0.0350 M

Q < Ksp, so no precipitation144Tro: Chemistry: A Molecular Approach, 2/e


Practice – Will a precipitate form when we mix Ca(NO3)2(aq) with NaOH(aq) if the concentrations after

mixing are both 0.0175 M? Ksp of Ca(OH)2 = 4.68 x 10−6



Practice – Will a precipitate form when we mix Ca(NO3)2(aq) with NaOH(aq) if the concentrations after

mixing are both 0.0175 M? write the equation for the reaction

determine the ion concentrations of the original salts

determine the Ksp for any “insoluble” product

write the dissociation reaction for the insoluble product

calculate Q, using the ion concentrations

compare Q to Ksp. If Q > Ksp, precipitation

Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)

Ksp of Ca(OH)2 = 4.68 x 10–6

Ca(OH)2(s) Ca2+(aq) + 2 OH−

(aq)

Ca(NO3)2 = 0.0175 M Ca2+ = 0.0175 M, NO3

− = 2(0.0175 M)

NaOH = 0.0175 M Na+ = 0.0175 M, OH− = 0.0175 M

Q > Ksp, so precipitation


Copyright 2011 Pearson Education, Inc.147

Example 16.13: What is the minimum [OH−] necessary to just begin to precipitate Mg2+ (with

[0.059]) from seawater?

precipitating may just occur when Q = Ksp

Mg(OH)2(s) Mg2+(aq) + 2 OH−

(aq)



Practice – What is the minimum concentration of Ca(NO3)2(aq) that will precipitate Ca(OH)2 from 0.0175

M NaOH(aq)? Ksp of Ca(OH)2 = 4.68 x 10−6



Practice – What is the minimum concentration of Ca(NO3)2(aq) that will precipitate Ca(OH)2 from

0.0175 M NaOH(aq)?


[Ca(NO3)2] = [Ca2+] = 0.0153 M

Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)


(aq)



Example 16.14: What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from

seawater?



(aq)



Example 16.14: What is the [Mg2+] when Ca2+ (with [0.011]) just begins to precipitate from

seawater?precipitating Mg2+ begins when [OH−] = 1.9 x 10−6 M

precipitating Ca2+ begins when [OH−] = 2.06 x 10−2 M

when Ca2+ just begins to precipitate out, the [Mg2+] has dropped from 0.059 M to 4.8 x 10−10 M

Mg(OH)2(s) Mg2+(aq) + 2 OH−

(aq)



Practice – A solution is made by mixing Pb(NO3)2(aq) with AgNO3(aq) so both compounds have a concentration of

0.0010 M. NaCl(s) is added to precipitate out both AgCl(s) and PbCl2(aq). What is the [Ag+] concentration when the

Pb2+ just begins to precipitate?



Practice – What is the [Ag+] concentration when the Pb2+(0.0010 M) just begins to precipitate?


AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)




Practice – What is the [Ag+] concentration when the Pb2+(0.0010 M) just begins to precipitate?


Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)




Practice – What is the [Ag+] concentration when the Pb2+(0.0010 M) just begins to precipitate

precipitating Ag+ begins when [Cl−] = 1.77 x 10−7 M

precipitating Pb2+ begins when [Cl−] = 1.08 x 10−1 M

when Pb2+ just begins to precipitate out, the [Ag+] has dropped from 0.0010 M to 1.6 x 10−9 M




Qualitative Analysis

• An analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis schemewet chemistry

• A sample containing several ions is subjected to the addition of several precipitating agents

• Addition of each reagent causes one of the ions present to precipitate out



Qualitative Analysis




Group 1

• Group one cations are Ag+, Pb2+ and Hg22+

• All these cations form compounds with Cl− that are insoluble in wateras long as the concentration is large enoughPbCl2 may be borderline

molar solubility of PbCl2 = 1.43 x 10−2 M

• Precipitated by the addition of HCl



Group 2

• Group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+, Pb2+, Sb3+, and Hg2+

• All these cations form compounds with HS− and S2− that are insoluble in water at low pH

• Precipitated by the addition of H2S in HCl



Group 3

• Group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides

• All these cations form compounds with S2− that are insoluble in water at high pH

• Precipitated by the addition of H2S in NaOH



Group 4

• Group four cations are Mg2+, Ca2+, Ba2+ • All these cations form compounds with PO4

3− that are insoluble in water at high pH

• Precipitated by the addition of (NH4)2HPO4



Group 5

• Group five cations are Na+, K+, NH4+

• All these cations form compounds that are soluble in water – they do not precipitate

• Identified by the color of their flame



Complex Ion Formation

• Transition metals tend to be good Lewis acids• They often bond to one or more H2O molecules to

form a hydrated ionH2O is the Lewis base, donating electron pairs to form

coordinate covalent bonds

Ag+(aq) + 2 H2O(l) Ag(H2O)2+(aq)

• Ions that form by combining a cation with several anions or neutral molecules are called complex ionse.g., Ag(H2O)2

+

• The attached ions or molecules are called ligandse.g., H2O



Complex Ion Equilibria

• If a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand

Ag(H2O)2+

(aq) + 2 NH3(aq) Ag(NH3)2+

(aq) + 2 H2O(l) generally H2O is not included, because its complex

ion is always present in aqueous solution

Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq)



Formation Constant

• The reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction


+(aq)

• The equilibrium constant for the formation reaction is called the formation constant, Kf



Formation Constants



Example 16.15: 200.0 mL of 1.5 x 10−3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3.

What is the [Cu2+] at equilibrium? Write the formation reaction and Kf expression.Look up Kf value

determine the concentration of ions in the diluted solutions

Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)



Example 16.15: 200.0 mL of 1.5 x 10−3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M

NH3. What is the [Cu2+] at equilibrium?

Create an ICE table. Because Kf is large, assume all the Cu2+ is converted into complex ion, then the system returns to equilibrium.

[Cu2+] [NH3] [Cu(NH3)22+]

initial 6.7E-4 0.11 0

change ≈−6.7E-4 ≈−4(6.7E-4) ≈+6.7E-4

equilibrium x 0.11 6.7E-4

Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)



Example 16.15: 200.0 mL of 1.5 x 10-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M

NH3. What is the [Cu2+] at equilibrium?

Cu2+(aq) + 4 NH3(aq) Cu(NH3)22+(aq)substitute in

and solve for x

confirm the “x is small” approximation

[Cu2+] [NH3] [Cu(NH3)22+]

initial 6.7E-4 0.11 0

change ≈−6.7E-4 ≈−4(6.7E-4) ≈+6.7E-4

equilibrium x 0.11 6.7E-4

2.7 x 10−13 << 6.7 x 10−4, so the approximation is valid



Practice – What is [HgI42−] when 125 mL of 0.0010 M

KI is reacted with 75 mL of 0.0010 M HgCl2?4 KI(aq) + HgCl2(aq) 2 KCl(aq) + K2HgI4(aq)





Write the formation reaction and Kf expression.Look up Kf value

determine the concentration of ions in the diluted solutions

Hg2+(aq) + 4 I−(aq) HgI42−(aq)





Create an ICE table. Because Kf is large, assume all the lim. rgt. is converted into complex ion, then the system returns to equilibrium.

[Hg2+] [I−] [HgI42−]

initial 3.75E-4 6.25E-4 0

change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)

equilibrium 2.19E-4 x 1.56E-4


I− is the limiting reagent





substitute in and solve for x

confirm the “x is small” approximation

2 x 10−8 << 1.6 x 10−4, so the approximation is valid


[HgI42−] = 1.6 x 10−4

[Hg2+] [I−] [HgI42−]

initial 3.75E-4 6.25E-4 0

change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)

equilibrium 2.19E-4 x 1.56E-4



The Effect of Complex Ion Formation on Solubility

• The solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands

AgCl(s) Ag+(aq) + Cl−(aq) Ksp = 1.77 x 10−10


+(aq) Kf = 1.7 x 107

• Adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+




Solubility of Amphoteric Metal Hydroxides

• Many metal hydroxides are insoluble• All metal hydroxides become more soluble in acidic

solutionshifting the equilibrium to the right by removing OH−

• Some metal hydroxides also become more soluble in basic solutionacting as a Lewis base forming a complex ion

• Substances that behave as both an acid and base are said to be amphoteric

• Some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+



Al3+

• Al3+ is hydrated in water to form an acidic solutionAl(H2O)6

3+(aq) + H2O(l) Al(H2O)5(OH)2+

(aq) + H3O+(aq)

• Addition of OH− drives the equilibrium to the right and continues to remove H from the moleculesAl(H2O)5(OH)2+

(aq) + OH−(aq) Al(H2O)4(OH)2

+(aq) + H2O

(l)

Al(H2O)4(OH)2+

(aq) + OH−(aq) Al(H2O)3(OH)3(s) + H2O

(l)



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Chapter 16 Aqueous Ionic Equilibrium