Acid Base EquilibriaDr. HarrisCh 20Suggested HW: Ch 20: 5, 9, 11*, 19*, 21, 29**, 35, 56** * Use rule of logs on slide 10** Use Ka and Kb tables in ch 20 text.

Introduction

• In chapter 10, we defined an acid as a substance capable of producing H+

(aq) ions in solution.

• For strong acids, this equilibrium lies so far to the right that you can make the mathematical assumption that [HX] ≈ 0.

• These “simple” acids are known as Arrhenius acids. The Arrhenius definition of an acid is restricted to aqueous acid solutions.

HXaq H+aq + X-aq

• An Arrhenius base is a substance that produces OH-(aq) ions in aqueous solution.

MOHaq M+aq + OH-aq

Bronsted-Lowry Acids and Bases

• Until now, H+(aq) has been used to represent the species formed when an acid is dissolved in water.

• In the Bronsted-Lowry definition of acids and bases, an acid is a H+ (proton) donor. A base is a proton acceptor.

• Here, we define the hydronium ion, H3O+(aq). This species is formed when a water molecule accepts a proton. In a chemical equation, H+ and H3O+ are interchangeable.

• For example:

• In the example above, HA is a Bronsted acid, and water acts as a Bronsted base.

Autoionization of Water

• Pure water contains very low concentrations of hydronium and hydroxide ions that arise from the equilibrium

• In the reaction above, one water molecule transfers a proton to another. Thus, water can act as BOTH a Bronsted acid and Bronsted base

• The equilibrium constant expression for the reaction above is:

• The subscript ‘w’ means water. At 25oC, the value of Kw is 1 x 10-14 M2.

• In pure water, the concentrations of hydronium and hydroxide are equal.

H2O(l) + H2O(l) H3O+(aq) + OH−(aq)

𝑲𝒘=¿

Strong Acids and Bases

• Acids and bases that dissociate completely are considered strong.

• Because of the high degree of dissociation, strong acids easily donate their protons, and strong bases easily accept them.

• By now, you should know the strong acids and bases. If not, make sure that you learn them. They are listed in the table to the right.

Conjugate Acid-Base Pairs

• In any acid-base equilibrium, the forward and reverse reactions involve proton transfers.

• When an acid, HA, loses its H+(aq), the A-(aq) left behind can behave as a base. Likewise, when a base accepts the H+(aq), it can now act as an acid.

• An acid and a base such as HA and A- that differ only in the presence of a proton are called a conjugate acid-base pair. Every acid has a conjugate base, and visa-versa.

Relative Strengths of Acids and Bases

• The more easily an acid donates a proton, the less easily its conjugate base accepts. In other words, the stronger an acid, the weaker the conjugate base. Same for bases and conjugate acids.

1. A strong acid completely transfers its protons into water. Its conjugate base has a negligible tendency to be protonated.

2. A weak acid partially dissociates, and so it exists as a mixture of molecules and ions. The conjugate of a weak acid/base is a weak base/acid

HCl(aq) + H2O(l) H3O+(aq) + Cl−(aq)x

The position of equilibrium is always favored in the direction that forms the weaker acid/base.

pH• The rates of many chemical reactions depend on the concentration of

[H3O+]. However, based on the value of Kw , we see that [H3O+] can range vastly.

• It is very difficult to plot such a large range of values, so we use the pH scale. In terms of [H3O+]:

• Recall that for pure water, we have the equilibrium:

• And in pure water:

• So in pure water, = 10-7 M, and pH = 7.00

𝒑𝑯=−𝒍𝒐𝒈 [ H 3O+ ]

𝑲𝒘=¿

¿

pOH and the PH scale

• In terms of basicity, we can introduce a quantity pOH.

• For a basic solution, you can not directly calculate pH. You must calculate pOH first. Referring back to the following relationship:

• We see that:

• Therefore:

𝒑𝑶𝑯=−𝒍𝒐𝒈 [ OH−]

𝑲𝒘=¿

− log 𝐾𝑤=−log ¿¿

𝒑𝑯 +𝒑𝑶𝑯=𝟏𝟒 * at 25oC

pH scale and Important Notes

• For this class, we will assume the temperature of all aqueous solutions to be exactly 25oC to that the previous expression holds true.

• It is important to know the following:

An acidic solution is defined as one in which [H3O+] > [OH-] (pH < 7)

A neutral solution is defined as one in which [H3O+] = [OH-] (pH = 7)

A basic solution is defined as one in which [OH-] > [H3O+] (pH > 7)

Also, please note that H+ and H3O+ plus are the same, and may be used interchangeably in the book.

𝒓𝒖𝒍𝒆𝒐𝒇 𝒍𝒐𝒈𝒔 :𝟏𝟎𝐥𝐨𝐠 (𝒙 )=𝒙 this will come in handy

Example

• What is the pH of a 0.040M solution of HClO4 at 25oC?– Perchloric acid is a strong acid, so we can assume complete

dissociation.

• What is the pH of a 0.028M solution of NaOH at 25oC?– NaOH is a strong base. We can therefore assume that it completely

dissociates, so [OH-] = 0.028 M. We must calculate pOH first.

𝑝𝑂𝐻=− log ( .028 )=1.55𝑝𝐻=12.45

𝑝𝐻=− log ( .040 )=1.40

.040M .040M .040M

.028M .028M .028M

Example

• The pH of a 0.40 M formic acid solution (HCOOH) is 2.08. What is the concentration of H3O+? What percentage of the formic acid dissociates?– Formic acid is a weak acid. Weak acids do not fully dissociate

HCOOH (aq)+ H2O(L) H3O+(aq) + HCOO-(aq)I .40 M 0 0C -x +x +xE .40-x x x

2.08=− log ¿¿

10− 2.08=¿¿

.0083𝑀=𝑥

𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛=[ .0083𝑀 ][0.40𝑀 ]

𝑥 100=2.07%

Acid Dissociation Constant, Ka

• Let’s consider acetic acid, CH3COOH, a weak acid. The equilibrium expression of this dissociation is defined by Ka:

CH3COOH (aq)+ H2O(L) H3O+(aq) + CH3COO-(aq)

𝐾 𝑎=¿ ¿• The size of Ka describes mathematically how readily an acid

dissociates. The stronger the acid, the higher the value of Ka.

Example

1. We set up the ICE table to determine the concentration of H3O+

Concentration CH3COOH H3O+ CH3COO-

Initial 0.050 M 0 0

Change - x + x + x

Equilibrium 0.050 – x x x

CH3COOH (aq) + H2O(L) H3O+(aq) + CH3COO-(aq)

2. Solve for x. We see that Ka is very small because CH3COOH is a weak acid.

𝑥20.050−𝑥=1.8𝑥 10−5𝑀 𝑥=9.4 𝑥 10−4

3. Calculate pH𝑝𝐻=− log ( .00094 )=3.03

Calculate the pH of a 0.050 M acetic acid solution given that Ka = 1.8 x 10-5 M

Method of Successive Approximations

• Weak acids tend to dissociate quite poorly, so the percentage of dissociation is very small. Hence, Ka is very small as well. Let’s revisit the previous example.

• Since x is so small:

• Therefore:

𝑥20.050−𝑥=1.8𝑥 10−5𝑀

• For the dissociation of any acid/base, the percent dissociation is given by:

𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑑𝑖𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑜𝑛=𝑥

[ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 ]𝑥100

0.050−𝑥 ≅ 0.050

𝑥20.050

=1.8 𝑥10−5𝑀 𝑥=9.4 𝑥 10−4

Use this method when K < 10-4

pKa

• Ka values range over many orders of magnitude. Therefore, you may find it more convenient to use the quantity pKa

𝑝 𝐾𝑎=−log 𝐾 𝑎

• As Ka increases, pKa decreases.

Amines

• Similar to acids, we can describe the protonation of bases. For example, lets look at the equilibrium involving ammonia in water.

• The lone electron pair of electrons in the ammonia molecule allows it to accept an H+ ion (proton).

• Ammonia falls under a class of bases known as amines, which are composed of a central nitrogen atom with 3 bonding domains.

• Amines react with acids to form ammonium salts (positively charged central N atom with 4 bonding domains). The right hand side of the equilibrium above shows the dissociated ammonium hydroxide salt (ammonium cation: NH4

+; hydroxide anion: OH-)

Kb and pKb

• We can use the base protonation constant Kb, or pKb to describe the degree of the dissociation of a base.

• As with acids, higher Kb (lower pKb) means stronger base

Example: Determine the pH of a 0.75 M hydroxylamine, HONH2(aq) solution given that Kb = 8.7 x 10-9 M

1. We can write out the equilibrium expression showing the base protonation.

HONH2 (aq) + H2O(L) HONH3+ (aq) + OH-(aq)

𝐾 𝑏=¿¿

Example, contd.

2. Set up ICE table

Concentration HONH2 HONH3+ OH-

Initial 0.75 M 0 0

Change - x + x + x

Equilibrium 0.75 – x x x

3. Solve x𝑥2

0.75−𝑥=8.7 𝑥10− 9𝑀 𝑥20.75

=8.7 𝑥10− 9𝑀approximate

𝑥=8.07 𝑥10−5

4. THIS IS A BASE. That means that you must calculate pOH FIRST. Then, determine pH.

𝑝𝑂𝐻=− log (8.07 𝑥10−5 )=4.09

𝑝𝐻=14−𝑝𝑂𝐻=9.91

Acidic and Basic Salts

• Suppose we dissolve NaF(s) in water. The result would be an aqueous solution of Na+(aq) and F-(aq). You would expect this solution to be neutral.

• However, F-(aq) is the conjugate base of a weak acid, HF

𝐻𝐹 (𝑎𝑞 )+𝐻2𝑂 (𝐿 )𝐻3𝑂+¿ (𝑎𝑞 )+𝐹 −(𝑎𝑞)¿

acid conjugate basebase conjugate acid

• Therefore, a solution of F-(aq) ions will be weakly basic.

𝐹 − (𝑎𝑞 )+𝐻2𝑂 (𝐿 ) 𝐻𝐹 (𝑎𝑞 )+𝑂𝐻 −(𝑎𝑞)weak base weak acid strong basenegligible

acid

• Na+(aq) is a cation of a strong base, NaOH. Thus, it has negligible acidity and does not affect the pH. So it is considered a neutral cation.

Acidic and Basic Salts

1. Neutral cations are the cations of strong bases.

2. Neutral anions are the anions of strong acids.

3. Basic anions are the conjugate bases of weak acids.

4. Acidic cations are the conjugate acids of weak bases, and hydrated metal ions, as shown on pg 760.

Polyprotic Acids

• A polyprotic acid is an acid than can donate more than one H+.

• For example, sulfuric acid, H2SO4, is a strong, diprotic acid. Thus, it can undergo two deprotonations:

• We can assume 100% dissociation of the H2SO4 in step 1). However, the remaining HSO4

- is a much weaker acid. EACH PROTON IS SUBSTANTIALLY HARDER TO REMOVE THAN THE PREVIOUS.

• Therefore, the degree of dissociation of the weaker acid is determined from its Ka

(1 ) 𝐻2𝑆𝑂4 (𝑎𝑞 )→𝐻+¿ (𝑎𝑞 )+𝐻𝑆𝑂4−¿

Polyprotic Acids

• Ka values of polyprotic acids are labeled Ka1, Ka2, and so forth to denote each deprotonation.

• The table to the left shows pKa values of common polyprotic acids.

• Remember, pKa = -log Ka

Example• Calculate the pH of a 0.10 M H2SO4(aq) solution.

1.) We first have the initial deprotonation. Because sulfuric acid is a strong acid, you can assume that all of it is dissociated at equilibrium.

𝐻2𝑆𝑂4 (𝑎𝑞 )+𝐻2𝑂(𝐿)→𝐻3𝑂+¿ (𝑎𝑞 )+𝐻𝑆𝑂4

− (𝑎𝑞 )¿

2.) Now, we have the second deprotonation. Use the concentrations of H3O+ and HSO4

- from above as your initial values. Use pKa values from the table to determine the final concentration of [H3O+]

E

Concentration HSO4- H3O+ SO4

2-

Initial (0.10 M) (0.10 M) 0

Change - x + x + x

Equilibrium 0.10 – x 0.10+ x x

𝐻𝑆𝑂4− (𝑎𝑞 )+𝐻 2𝑂 (𝐿) 𝐻3𝑂

+¿ (𝑎𝑞 )+𝑆𝑂 42−¿ 𝑝𝐾𝑎2=1.99

𝐾 𝑎2=10−1.99=.01

𝑥=.0085𝑝𝐻=− log [ .1085 ]= .96