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Chapter 16Equilibrium in the Aqueous Phase

HCl(aq) reacting with NH3(aq)

The color of hydrangea flowers depends on the acid content of the soil

Acid rain forms when volatile, nonmetal oxides react with water vapor.

SO3 + H2O ↔ H2SO4

This tutorial explores the effects of fossil fuel burning on the pH of rainwater, as well as the resulting environmental and industrial consequences. Includes practice exercises.

»PC version

Acid Rain Tutorial

Strong Acid

HNO3

Weak Acid

HNO2

Acids…A molecular view.

Acids (proton donors) react with bases (proton acceptors) forming a conjugate acid (H3O+) and conjugate base (Cl-). Note that the conjugate base of a strong acid like HCl is a weak base.

Water molecules in acid solutions cluster around the Hydronium ion.

These “species” have the formula:

H(H2O)n+

Autoionization of WaterWater molecules have the ability to ionize each other.

Keq = [H3O+][OH-] = 1.0E-14 (25°C)

This important equilibrium constant is usually denoted Kw

Ammonia is a weak base in water, Kb = 1.8E-5

conj. acid conj. base

Trends in acid strength relative to the strength of their conjugate bases.

In water the strongest base is OH-; stronger bases will ionize water to produce hydroxide ions.

Problem

Benzoic acid is used as a preservative in foods. Calculate the concentration of H+ ions at equilibrium in a 0.100 M solution of Benzoic acid. Ka = 6.5E-5

In Solution, to what degree is benzoic acid ionized?

The pH of a solution is defined as the negative logarithm of the hydronium ion concentration:

pH = -log([H+])

Note that Ka and Kb values are frequently reported as pKa and pKb. This avoids writing the values as exponentials.

16.30. Calculate the pH of a 0.00500 M solution of HNO3.

Answer: 2.301 (4 S.F.)

Auto-ionization and pH

H2O ↔ H+ + OH-

Kw = 1.0E-14; pKw = 14.00 at 298K

In pure water,

[H+] = [OH-]

So

[H+]2 = 1.0E-14

Thus

[H+] = 1.0E-7

Or pH = 7.0

16.31. Calculate the pH and pOH of a 0.0450 M solution of NaOH.

Answer: pOH=1.347; pH=14.000 – 1.347 = 12.65

Problem: A solution of HF has a pH=2.30. Calculate the equilibrium concentration of all species present in this solution, and the original concentration of the HF (i.e. before dissociation). pKa(HF) = 3.14

For the reaction: HA ↔ H+ + A-

The concentration of H+ is a function of the strength of the H-A bond

Acid Strength and Molecular Structure

Sulfuric acid is a stronger acid than sulfurous acid due to the decrease in electron density on the O-H bond.

The oxyacids of chlorine increase in strength (Ka) with increasing numbers of oxygen atoms bound to the central chlorine atom.

HClO HBrO HIO

rH-O= 0.961Å rH-O= 0.957Å rH-O= 0.955Å

Increased H-O bond distance is due to decreased electron density

Blue color indicates increasing positive charge on the proton

Problem

The pH of a 0.10M solution of chloroacetic acid is found to be 1.95. Calculate Ka for this acid and compare it to Ka for acetic acid.

Table 16.1. Ionization of Diprotic Acids

Acid Formula Ka1 Ka2

Carbonic H2CO3 4.3E-7 4.7E-11

Sulfurous H2SO3 4.3E-3 6.2E-8

Sulfuric H2SO4 >>1 1.2E-2

Polyprotic Acid Ionization

The H+ concentration due to the second dissociation is generally insignificant, i.e. compared with the first dissociation.

16.59. What is the pH of a 0.300 M solution of H2SO4 (Ka2 = 1.2 10–2)?

Problem

Methylamine is a weak base (Kb=4.4E-4). Calculate the OH- concentration in a 0.200M aqueous solution of CH3NH2.

What is the pH of this solution?

This tutorial explores the differences among Brønsted-Lowry acids, Brønsted-Lowry bases, Lewis acids and Lewis Bases. Includes practice exercises.

»PC version

Acid and Base Ionization Tutorial

Learn to determine relative acid strength based on the molecular and electronic structure of the acid. Includes practice exercises.

»PC version

Acid Strength and Molecular Structure Tutorial

This tutorial introduces the pH scale and uses interactive graphs to explain the relationship between pH, pOH [H3O

+], and [OH-]. Includes practice

exercises.

»PC version

pH Scale Tutorial

This tutorial illustrates the process by which water molecules act as both a proton acceptor (base) and a proton donor (acid), and explores the equilibrium constant (Kw) for

the self-ionization of water. Includes practice exercises.

»PC version

The Self-Ionization of Water Tutorial

Many naturally occurring compounds used as drugs act as weak bases (due to amine groups).

For this reason they are often referred to as alkaloids…they produce alkaline solution.

Salts of weak acids and bases.

Problem 63. Which of the following salts produce an acidic solution in water?

Ammonium acetate

NH4Cl

Sodium formate

Problem 63. Which of the following salts produce an basic solution in water?

NaF

KCl

Sodium bicarbonate

Problem 66

Codeine is a widely-prescribed pain killer because it is much less addictive than morphine (which is much less addictive than heroin). Codeine contains a basic nitrogen atom that can be protonated to form the conjugate acid .

Calculate the pH of a 3.97E-4 M solution of codeine if the pKa of the conjugate acid is 8.21.

Problem. For a 6.75E-3 M solution of sodium benzoate, determine the following:

Identify the equilibrium reaction that determines the pH.

Calculate the pH. pKa(benzoic acid) = 4.20

Lewis Acids and Bases

• A Lewis Acid is a substance that accepts a pair of electrons.

• A Lewis Base is a substance that donates a pair of electrons.

A Lewis Acid/base adduct

Buffer Solutions are solutions that contain significant amounts of both an acid and it’s conjugate base.

For example the following solutions prepared by adding equivalent amounts of the acid/conj.base pairs:

CH3COOH/CH3COO-

H2PO4-/HPO4

2-

HCO3-/CO3

2-

The presence of both the acid and base means that the pH will resist change went additional acid or base are added to the solution.

Common Buffer SystemsCommon Buffer Systems

AcidAcid Conj.baseConj.base pKapKa pH rangepH range

HH33POPO44 H H22POPO44-- 2.162.16 1-3 1-3

CHCH33COCO22HH CH CH33COCO2 2 -- 4.754.75 4-6 4-6

HH22POPO44-- HPOHPO4 4

2-2- 7.21 7.21 6-8 6-8

HCOHCO3 3 -- COCO3 3

2-2- 10.3310.33 9-11 9-11

Acididosis can be caused by extreme changes in diet as well as chronic respiratory diseases

Problem 80 (pH buffer problem)

Determine the pH and pOH of 0.250 L of a buffer containing 0.0200M boric acid and 0.0250M sodium borate.

The pKa for B(OH)3 = 9.00 at 25°C.

Use the Henderson-Haselbach equation to predict the pH of a buffer. The tutorial concludes with practice exercises and an interactive titration experiment.

»PC version

Buffers Tutorial

Alkalinity Titrations to determine total CO32-

Acid/Base TitrationsStrong acid with a strong baseWeak acid with a strong base

Problem 93Problem 93. A 25.0 mL sample of 0.100 M . A 25.0 mL sample of 0.100 M acetic acid is titrated with 0.125 M NaOH. acetic acid is titrated with 0.125 M NaOH. Calculate the pH of the of the reaction Calculate the pH of the of the reaction solution after 10.0, 20.0, and 30.0 mL of base solution after 10.0, 20.0, and 30.0 mL of base have been added.have been added.

Acid/Base Titrations

Strong base with a strong acid

Weak base with a strong acid

Problem 100

In an alkalinity titration of 100.0 mL sample of water from a hot spring, 2.56mL of 0.0355 M HCl is needed to reach the first equivalence point (pH=8.3) and another 10.42mL is needed to reach the second equivalence point (pH=4). If the alkalinity of the spring is due only to the presence of carbonate and bicarbonate, what are the concentrations of each of them?

Problem. Which of the following solutions Problem. Which of the following solutions show buffer properties. Compute the pH of show buffer properties. Compute the pH of each solution that is bufferedeach solution that is buffered

(a)(a)0.100 L of 0.25 M NaCH0.100 L of 0.25 M NaCH33COCO22 + 0.150 L of + 0.150 L of 0.25 M HCl0.25 M HCl

(b)(b)0.100 L of 0.25 M NaCH0.100 L of 0.25 M NaCH33COCO22 + 0.050 L of + 0.050 L of 0.25 M HCl0.25 M HCl

(c)(c)0.100 L of 0.25 M NaCH0.100 L of 0.25 M NaCH33COCO22 + 0.050 L of + 0.050 L of 0.25 M NaOH0.25 M NaOH

Acid/base Indicators are weak organic acids that change color when ionized.

The pKas of the Indicators determine the pH range that they can be used in titrations

This interactive virtual titration lab introduces the titration apparatus and challenges you to determine the concentration of an unknown acid from the volume of basic solution added. Includes practice exercises.

»PC version

Strong Acid and Strong Base Titrations Tutorial

Learn to read and understand the different stages of a titration curve for a weak acid or polyprotic acid, and understand what is happening at a molecular level. Includes practice exercises.

»PC version

Titrations of Weak Acids Tutorial

16.9. Solubility of Minerals and other Compounds

Minerals in contact with ground water will dissolve to some extent.

CaCO3(s) ↔ Ca2+(aq) + CO32-(aq)

Write the equilibrium expression for this dissolution.

Solubility Equilibria

CaCO3(s) ↔ Ca2+(aq) + CO32-(aq)

K = [Ca2+(aq)][CO32-(aq)] = Ksp

The equilibrium expression is called the solubility product (sp), because it only involves products of the concentrations of the dissolved species and not the solid.

If Ksp is known, the solubility (at equilibrium) of the solid can be calculated

KKspsps of some common saltss of some common salts

HgS(s)HgS(s) KKspsp = [Hg = [Hg2+2+][S][S2-2-] = 4.0 x 10] = 4.0 x 10-53-53

Fe(OH)Fe(OH)33(s)(s) KKspsp = [Fe = [Fe3+3+][OH][OH--]]33 = 2.8 x 10 = 2.8 x 10-39-39

AgI(s)AgI(s) KKspsp = [Ag = [Ag1+1+][I][I1-1-] = 8.5 x 10] = 8.5 x 10-17-17

CaCOCaCO33(s)(s) KKspsp = [Ca = [Ca2+2+][CO][CO332-2-] = 9.8 x 10] = 9.8 x 10-9-9

CaSOCaSO44(s)(s) KKspsp = [Ca = [Ca2+2+][SO][SO442-2-] = 4.9 x 10] = 4.9 x 10-5-5

AgAg22SOSO33(s)(s)KKspsp = [Ag = [Ag1+1+]]22[SO[SO332-2-] = 1.2 x 10] = 1.2 x 10-5-5

NaCl(s)NaCl(s) KKspsp = [Na = [Na1+1+][Cl][Cl1-1-] = 6.2] = 6.2

Solubility Problem

2.75 grams of BaF2 is placed in enough water to make 1.00 L at 25°C. After equilibrium has been established…the F- concentration equal 0.0150 M, what is the Ksp for BaF2.

Solubility Problem

50 mg of PbSO4 is placed in 250 mL of pure water; What percentage of the solid dissolves?

Solubility Problem 116

Calculate the pH of a saturated solution of zinc hydroxide, Ksp = 4.0E-17

Solubility Problem 120

Calculate the solubility of silver chloride in seawater with a chloride concentration of 0.547 M. Ksp(AgCl) = 1.8E-10

16.10. Complex Ions16.10. Complex Ions

Dissolved metal ions are Lewis Acids, and form complexes with Lewis Bases

Metal ions as Lewis Acid also promote hydrolysis of water, and the formation of H3O+

Metal cations (Fe3+, Cr3+ and Al3+)with large positive charges are more likely to cause hydrolysis.

Formation Reactions

The equilibrium constants associated with complexation are called formation constants

Kf = [Cu(NH3)42+]/[Cu2+][NH3]4

= 5.0E+13

Chlorophylls such as Chl a, and Chl b absorb visible light in a process that creates an electrical potential that drives phosphorylation

Heme Group of Hemoglobin

 

CQ16-10.3a-Adding a Drop of HCl to a AgNO3 Solution

If a single drop of water containing 0.02 mmol of HCl is added to 1.0 L of a 10‑5 M solution of AgNO3, will a precipitate form?

AgCl(s) Ag+(aq) + Cl‑(aq) Ksp = 1.8×10-10

A) Yes B) No C) Can’t tell

CQ16-10.3b-Adding a Drop of HCl to a AgNO3 Solution

Consider the following arguments for each answer and

vote again:

A. After only 1 drop of HCl is added, the Cl-(aq) concentration will be 2×10-5 M, which is high enough to induce the precipitation of AgCl(s).

B. Far more than 1 drop of HCl is required to raise the Cl- concentration to the point where Cl‑(aq) and Ag+(aq) are in equilibrium with AgCl(s).

C. It is not clear whether the ΔG° of the drop increases or decreases when it enters the solution.

 

CQ16-10.4a-Concentration of Ag+ In Ionic Solutions

Given that Ksp(AgCl) > Ksp(AgBr), which of the following salts, when added in excess to an aqueous 0.1 M AgNO3 solution, will result in the lowest concentration of Ag+(aq)?

A) AgNO3 B) NaCl C) AgBr

CQ16-10.4b-Concentration of Ag+ In Ionic Solutions

Consider the following arguments for each answer

and vote again:

A. Because of the common-ion effect, the addition of AgNO3(s) will cause a net decrease in the concentration of Ag+(aq).

B. Adding NaCl will induce the precipitation of AgCl(s) from the solution, thus lowering the Ag+(aq) concentration.

C. AgBr(s) is less soluble than AgCl(s), and so its addition will cause the greatest decrease in the Ag+

(aq) concentration.

 

CQ16-10.5a-Dissolution of a Speck of BaSO4 in H2O

Suppose water is slowly added to a vessel containing a speck of the sparingly soluble salt BaSO4(s). Which of the following plots shows the equilibrium concentration of Ba2+

(aq) in the resulting solution versus the amount of water added?

A) B) C)

CQ16-10.5b-Dissolution of a Speck of BaSO4 in H2O

Consider the following arguments for each answer and vote again:

A. As water is added and more BaSO4(s) is dissolved, the concentration of Ba2+(aq) will increase until the solution becomes saturated.

B. The concentration of Ba2+(aq) will increase until all the BaSO4(s) has dissolved, after which additional water will decrease the Ba2+(aq) concentration.

C. Until the BaSO4(s) has completely dissolved, the concentration of Ba2+(aq) will remain constant.

 

CQ16-10.6a-Conductivity of a NaCl + AgNO3 Solution

The conductivity of an aqueous solution is directly proportional to the concentration of the ions present. Given this fact, which of the following plots shows the conductivity of a NaCl solution as a function of the amount of AgNO3(s) added?

A) B) C)

CQ16-10.6b-Conductivity of a NaCl + AgNO3 Solution

Consider the following arguments for each answer

and vote again:

A. Adding AgNO3(s) will increase the total ion concentration, so the conductivity will increase until the solution is saturated.

B. As AgNO3(s) is added, the conductivity of the solution will decrease because of the precipitation of AgCl(s) until all of the Cl-(aq) is consumed.

C. Although AgCl(s) will precipitate as AgNO3(s) is added, the total concentration of ions will remain constant until the Cl-(aq) is depleted.

 

CQ16-11.5a-Dependence of pH on Temperature

To the left is a plot of the autoionization constant, Kw, versus temperature. What is the pH of hot water?

A) < 7 B) 7 C) > 7

CQ16-11.5b-Dependence of pH on Temperature

Consider the following arguments for each answer and vote again:

A. At higher temperatures, the concentrations of H3O+ and OH- increase. Therefore, the pH of hot water is less than 7.

B. Regardless of temperature, the concentrations of H3O+ and OH- remain equal, so the pH remains 7, which is neutral.

C. At higher temperatures, H+ ions acquire enough kinetic energy to escape the solution, leaving a predominance of OH- ions.

 

CQ16-11.6a-NH3 Buffer Solution

Which of the following, when added to an NH3(aq) solution, will form a basic buffer?

A) NaOH B) HCl C) NaCl

CQ16-11.6b-NH3 Buffer Solution

Consider the following arguments for each answer

and vote again:

A. NH3, a weak base, is normally an acidic buffer, so to create a basic buffer, one must add NaOH.

B. By adding HCl to the NH3 solution to form some NH4

+, the solution will become a basic buffer.

C. NH3 is already a weak base, so to create a basic buffer solution, one need only add a neutral buffering salt like NaCl.

 

CQ16-11.7a-Titration of a Diluted Weak Base

To the left is a plot that shows the pH of a weak acid as it is titrated with 0.01 M NaOH. Which of the following plots would correspond to the same titration if the same weak acid were diluted with water and then titrated with 0.01 M NaOH?

A) B) C)

CQ16-11.7b-Titration of a Diluted Weak Base

Consider the following arguments for each answer

and vote again:

A. Diluting a weak acid with water will increase the initial pH of the solution and decrease the final pH of the solution.

B. The dilution would have little effect on the initial pH of the weak acid, especially in the buffer region. However, the pH after the equivalence point will be lower.

C. If the weak acid is diluted, the titration will reach the equivalence point sooner, since the concentration of the acid will be lower.

 

CQ16-11.8a-Conductivity of a H2SO4/Ba(OH)2 Solution

Given that the conductivity of an aqueous solution depends on the concentration of the ions present, which of the following graphs shows conductivity (y-axis) plotted against the acid added (x-axis) for the titration of the strong base Ba(OH)2 with the strong acid H2SO4?

A) B) C)

CQ16-11.8b-Conductivity of a H2SO4/Ba(OH)2 Solution

Consider the following arguments for each answer

and vote again:

A. This is a titration of a strong base with a strong acid, so the conductivity will track the pH of the solution.

B. Although BaSO4(s) will precipitate as H2SO4 is added, the total concentration of ions will remain constant until the Ba2

+(aq) is depleted.

C. The conductivity will decrease as BaSO4(s) and H2O(λ) are formed, after which excess H2SO4 will increase the conductivity.

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ChemistryChemistryThe Science in Context

byThomas Gilbert,Rein V. Kirss, &Geoffrey Davies