Chapter 16: Aqueous Ionic Equilibria CHE 124: General Chemistry II Dr. Jerome Williams, Ph.D. Saint Leo University

Chapter 16: Aqueous Ionic Equilibria

CHE 124: General Chemistry IIDr. Jerome Williams, Ph.D.

Saint Leo University

Overview

• Basic Buffers & Hendersen-Hasselbalch • Buffering Effectiveness (Buffer Capacity)• Titration Fundamentals• Indicators• Solubility Equilibria

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)

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

• 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?

6

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

substitute into the Henderson-Hasselbalch equation

check the “x is small” approximation

NH3 + H2O NH4+ + OH−

Tro: Chemistry: A Molecular Approach, 2/e

• 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?

8

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

check the “x is small” approximation

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 before 0.18 0.020 0

mols added ─ ─ 0.010

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

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

after adding 0.010 mol NaOHpH = 5.09


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] = 1– equal 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?

16

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 complete– when the reaction is complete we have reached the

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

– an 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

solution– equivalence 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


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


Titrating Weak Acid with a Strong Base

• The initial pH is that of the weak acid solution– calculate like a weak acid equilibrium problem

• e.g., 15.5 and 15.6

• Before the equivalence point, the solution becomes a buffer– calculate 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 established– mol A− = original mole HA

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

– [A−]init = mol A−/total liters– calculate 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


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 solutionHInd(aq) + H2O(l) Ind

(aq) + H3O+(aq)

• The color of the solution depends on the relative concentrations of Ind:HInd– when 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 pH– pKa 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



Molar Solubility• Solubility is the amount of solute that will dissolve in

a given amount of solution– at a particular temperature

• The molar solubility is the number of moles of solute that will dissolve in a liter of solution– the 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

38

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

39

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


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

40

substitute into the Ksp expression,

assume S is small

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

[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


Documents

Chapter 16: Aqueous Ionic Equilibria CHE 124: General Chemistry II Dr. Jerome Williams, Ph.D. Saint Leo University