Chemistry

The Science in Context

Chapter 16 Equilibrium in the Aqueous Phase HCl(aq) reacting with NH 3 (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.

SO 3 + H 2 O ↔ H 2 SO 4

Acid Rain Tutorial

» PC version

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.

Strong Acid

HNO 3

Weak Acid

HNO 2

Acids…A molecular view.

Acids (proton donors) react with bases (proton acceptors) forming a conjugate acid (H 3 O + ) and conjugate base (Cl a weak base.

). Note that the conjugate base of a strong acid like HCl is

Water molecules in acid solutions cluster around the Hydronium ion.

These “species” have the formula: H(H 2 O) n +

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

K eq = [H 3 O + ][OH ] = 1.0E-14 (25°C)

This important equilibrium constant is usually denoted K

w

Ammonia is a weak base in water, K b = 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 ions.

; stronger bases will ionize water to produce hydroxide

Problem

Benzoic acid is used as a preservative in foods. Calculate the concentration of H Benzoic acid. K a = 6.5E-5 + ions at equilibrium in a 0.100 M solution of 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 K a and K b values are frequently reported as pK a and pK b . This avoids writing the values as exponentials.

16.30. Calculate the pH of a 0.00500 M solution of HNO 3 .

Answer: 2.301 (4 S.F.)

Auto-ionization and pH

H 2 O ↔ H + + OH K w = 1.0E-14; pK w at 298K = 14.00 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). pK a (HF) = 3.14

For the reaction: HA ↔ H + + A The concentration of H + strength of the H-A bond is a function of the

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 (K a ) with increasing numbers of oxygen atoms bound to the central chlorine atom.

Blue color indicates increasing positive charge on the proton

HClO HBrO HIO r H-O = 0.961Å r H-O = 0.957Å r H-O = 0.955Å Increased H-O bond distance is due to decreased electron density

Problem

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

Polyprotic Acid Ionization Table 16.1. Ionization of Diprotic Acids Acid

Carbonic Sulfurous Sulfuric H

Formula

2 CO 3 H 2 SO 3 H 2 SO 4

K a1

4.3E-7 4.3E-3 >>1

K a2

4.7E-11 6.2E-8 1.2E-2 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 H 2 SO 4 (K a2 = 1.2  10 –2 )?

Problem

Methylamine is a weak base (K b =4.4E-4). Calculate the OH concentration in a 0.200M aqueous solution of CH 3 NH 2 .

What is the pH of this solution?

Acid and Base Ionization Tutorial

» PC version

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

Acid Strength and Molecular Structure Tutorial » PC version

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

pH Scale Tutorial

» PC version

This tutorial introduces the pH scale and uses interactive graphs to explain the relationship between pH, pOH [H 3 O + ], and [OH ]. Includes practice exercises.

The Self-Ionization of Water Tutorial

» PC version

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 (

K

w ) for the self-ionization of water. Includes practice exercises.

Salts of weak acids and bases.

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.

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

Ammonium acetate NH 4 Cl 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 pK a 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. pK a (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:

CH 3 COOH/CH 3 COO H 2 PO 4 /HPO 4 2 HCO 3 /CO 3 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.

Acid

H 3 PO 4

Common Buffer Systems Conj.base

H 2 PO 4 -

pKa

2.16

pH range

1-3 CH 3 CO 2 H CH 3 CO 2 H 2 PO 4 HPO 4 2 HCO 3 CO 3 2 4.75

7.21

10.33

4-6 6-8 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 pK a for B(OH) 3 = 9.00 at 25°C.

Buffers Tutorial

» PC version

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

Alkalinity Titrations to determine total CO 3 2-

Acid/Base Titrations Strong acid with a strong base Weak acid with a strong base

Problem 93. A 25.0 mL sample of 0.100 M acetic acid is titrated with 0.125 M NaOH. Calculate the pH of the of the reaction solution after 10.0, 20.0, and 30.0 mL of base 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 show buffer properties. Compute the pH of each solution that is buffered (a) 0.100 L of 0.25 M NaCH 3 CO 2 0.25 M HCl + 0.150 L of (b) 0.100 L of 0.25 M NaCH 3 CO 2 0.25 M HCl + 0.050 L of (c) 0.100 L of 0.25 M NaCH 3 CO 2 0.25 M NaOH + 0.050 L of

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

The pK a s of the Indicators determine the pH range that they can be used in titrations

Strong Acid and Strong Base Titrations Tutorial » PC version

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.

Titrations of Weak Acids Tutorial

» PC version

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.

16.9. Solubility of Minerals and other Compounds

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

CaCO 3 (s) ↔ Ca 2+ (aq) + CO 3 2 (aq) Write the equilibrium expression for this dissolution.

Solubility Equilibria

CaCO 3 (s) ↔ Ca 2+ (aq) + CO 3 2 (aq) K = [Ca 2+ (aq) ][CO 3 2 (aq) ] = K sp 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 K sp is known, the solubility (at equilibrium) of the solid can be calculated

K

sp

s of some common salts

HgS(s) Fe(OH) 3 (s) AgI(s) CaCO 3 (s) CaSO 4 (s) Ag 2 SO NaCl(s) 3 (s) K sp = [Hg 2+ ][S 2 ] = 4.0 x 10 -53 K sp = [Fe 3+ ][OH ] 3 = 2.8 x 10 -39 K sp = [Ag 1+ ][I 1 ] = 8.5 x 10 -17 K sp = [Ca 2+ ][CO 3 2 ] = 9.8 x 10 -9 K sp = [Ca 2+ ][SO 4 2 ] = 4.9 x 10 -5 K sp = [Ag 1+ ] 2 [SO 3 2 ] = 1.2 x 10 -5 K sp = [Na 1+ ][Cl 1 ] = 6.2

Solubility Problem

2.75 grams of BaF 2 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 K sp for BaF 2 .

Solubility Problem

50 mg of PbSO water; What percentage of the solid dissolves?

4 is placed in 250 mL of pure

Solubility Problem 116

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

Solubility Problem 120

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

16.10. 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 H 3 O + Metal cations (Fe hydrolysis.

3+ , Cr 3+ and Al 3+ )with large positive charges are more likely to cause

Formation Reactions The equilibrium constants associated with complexation are called formation constants

K f

= [Cu(NH 3 ) 4 2+ ]/[Cu 2+ ][NH 3 ] 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

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 AgNO 3 , will a precipitate form?

AgCl(s) Ag + (aq) + Cl (aq) K sp = 1.8×10 -10 A) Yes B) No C) Can’t tell CQ16-10.3a-Adding a Drop of HCl to a AgNO 3 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.

CQ16-10.3b-Adding a Drop of HCl to a AgNO 3 Solution

Given that Ksp(AgCl) > K sp (AgBr), which of the following salts, when added in excess to an aqueous 0.1 M AgNO 3 solution, will result in the lowest concentration of Ag + (aq)? A) AgNO 3 B) NaCl C) AgBr CQ16-10.4a-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 AgNO 3 (s) will cause concentration of Ag + (aq).

a net decrease in the 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.

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

Suppose water is slowly added to a vessel containing a speck of the sparingly soluble salt BaSO 4 (s). Which of the following plots shows the equilibrium concentration of Ba 2+ (aq) in the resulting solution versus the amount of water added?

B) C) A) CQ16-10.5a-Dissolution of a Speck of BaSO 4 in H 2 O

Consider the following arguments for each answer and vote again:

A. As water is added and more BaSO 4 (s) is dissolved, the concentration of Ba 2+ (aq) will increase until the solution becomes saturated. B. The concentration of Ba 2+ (aq) will increase until all the BaSO 4 (s) has dissolved, after which additional water will decrease the Ba 2+ (aq) concentration.

C. Until the BaSO 4 (s) has completely dissolved, the concentration of Ba 2+ (aq) will remain constant.

CQ16-10.5b-Dissolution of a Speck of BaSO 4 in H 2 O

A) The conductivity of an aqueous solution is directly concentration proportional of the ions to the present.

Given this fact, which of the following plots shows the conductivity of a NaCl solution as a function of the amount of AgNO 3 (s) added?

B) C) CQ16-10.6a-Conductivity of a NaCl + AgNO 3 Solution

Consider the following arguments for each answer and vote again:

A. Adding AgNO 3 (s) will increase the total ion concentration, so the conductivity will increase until the solution is saturated.

B. As AgNO 3 (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 AgNO 3 (s) is added, the total concentration of ions will remain constant until the Cl (aq) is depleted.

CQ16-10.6b-Conductivity of a NaCl + AgNO 3 Solution

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

A) < 7 B) 7 C) > 7 CQ16-11.5a-Dependence of pH on Temperature

Consider the following arguments for each answer and vote again:

A. At higher temperatures, the concentrations of H 3 O + and OH increase. Therefore, the pH of hot water is less than 7.

B. Regardless of temperature, the concentrations of H 3 O + 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. CQ16-11.5b-Dependence of pH on Temperature

Which of the following, when added to an NH 3 (aq) solution, will form a basic buffer?

A) NaOH B) HCl C) NaCl CQ16-11.6a-NH 3 Buffer Solution

Consider the following arguments for each answer and vote again:

A. NH 3 , a weak base, is normally an acidic buffer, so to create a basic buffer, one must add NaOH.

B. By adding HCl to the NH 3 solution to form some NH 4 + , the solution will become a basic buffer. C. NH 3 is already a weak base, so to create a basic buffer solution, one need only add a neutral buffering salt like NaCl.

CQ16-11.6b-NH 3 Buffer Solution

B) 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?

C) A) CQ16-11.7a-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.

CQ16-11.7b-Titration of a Diluted Weak Base

A) 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 the strong acid H 2 SO 4 ?

with B) C) CQ16-11.8a-Conductivity of a H 2 SO 4 /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 BaSO 4 (s) will precipitate as H 2 SO 4 is added, the total concentration of ions will remain constant until the Ba 2 + (aq) is depleted.

C. The conductivity will decrease as BaSO 4 (s) and H 2 O(λ) are formed, after which excess H 2 SO 4 will increase the conductivity.

CQ16-11.8b-Conductivity of a H 2 SO 4 /Ba(OH) 2 Solution

W. W. Norton & Company Independent and Employee-Owned This concludes the Norton Media Library slide set for chapter 16

Chemistry

The Science in Context by Thomas Gilbert, Rein V. Kirss, & Geoffrey Davies