Chapter 16: Aqueous Ionic Equilibrium

Buffers• Solutions that resist changes in pH when an acid or base is

added• Act by neutralizing acid or base that is added to the

buffered solution• There is a limit to buffering capacity eventually the pH

changes• Standard Buffer Solutions =

– solution of a weak acid + solution of soluble salt containing the conjugate base anion

Making an Acid Buffer

How Acid Buffers Work:Addition of Base

HA(aq) + H2O(l) A−(aq) + H3O+

(aq)

• Buffers = Application of Le Châtelier’s Principle to weak acid equilibrium

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

• HA react with added A- to neutralize itHA(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 NaA, a salt containing the A−

– A− = conjugate base of the acid (the common ion)– This shifts the equilibrium to the left

• The new pH is higher than the pH of the acid solution– Lower H3O+ ion concentration

Common Ion Effect

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

write the reaction for the acid with waterconstruct an ICE table for the reactionenter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HF + H2O F + H3O+

[HA] [A−] [H3O+]initial 0.14 0.071 ≈ 0changeequilibrium

[HA] [A−] [H3O+]initial 0.14 0.071 0

changeequilibrium

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

represent the change in the concentrations in terms of xsum the columns to find the equilibrium concentrations in terms of xsubstitute into the equilibrium constant expression

+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?

determine the value of Ka

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 +xequilibrium 0.012 0.100 x0.14 x 0.071 +x


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

Ka for HF = 7.0 x 10−4

the approximation is valid

x = 1.4 x 10−3

[HA] [A−] [H3O+]initial 0.14 0.071 ≈ 0

change −x +x +x

equilibrium 0.14 0.071 x


substitute x into the equilibrium concentration definitions and solve

x = 1.4 x 10−3

[HA] [A−] [H3O+]initial 0.14 0.071 ≈ 0

change −x +x +x

equilibrium 0.14 0.072 1.4E-30.071 + x x 0.14 x


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

[HA] [A−] [H3O+]initial 0.14 0.071 ≈ 0

change −x +x +x

equilibrium 0.14 0.072 1.4E−3


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

the values are close enough

[HA] [A−] [H3O+]initial 0.14 0.071 ≈ 0

change −x +x +x

equilibrium 0.14 0.072 1.4E−3

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 base– as long as the “x is small” approximation is valid


find the pKa from the given Ka

assume the [HA] and [A−] equilibrium concentrations are the same as the initialsubstitute into the Henderson-Hasselbalch equationcheck the “x is small” approximation

HF + H2O F + H3O+

When should you 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?

• While buffers 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 parts

1. 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−]

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?

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

F− + H3O+ HF + H2O

F− H3O+ HFmols before 0.071 0 0.140

mols added – 0.020 –mols after


added?

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

F− + H3O+ HF + H2O

F− H3O+ HFmols before 0.071 0 0.140

mols added – 0.020 –mols after 0.051 0 0.160


added?

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 reactionenter the initial number of moles for each

F− + H3O+ HF + H2O

F− H3O+ HFmols 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

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?

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


added?

write the reaction for the acid with waterconstruct an ICE table.

assume the [HA] and [A−] equilibrium concentrations are the same as the initialsubstitute into the Henderson-Hasselbalch equation

HF + H2O F + H3O+

[HF] [F−] [H3O+]

initial 0.160 0.051 ≈ 0

change −x +x +xequilibrium 0.160 0.051 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

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 0mols added ─ ─ 0.010mols 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 before 0.100 0.100 0mols added ─ ─ 0.010

mols after 0.090 0.110 ≈ 0


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

HA A− OH−

mols before 0.50 0.500 0mols added ─ ─ 0.010

mols after 0.49 0.51 ≈ 0

HA A− OH−

mols before 0.050 0.050 0mols added ─ ─ 0.010mols 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

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 = 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: pH vs Amount of Titrant Added

• Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH

• The inflection point of the curve is the equivalence point of the titration– 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

HCl NaCl NaOHmols before 2.50E-3 0 5.0E-4mols change −5.0E-4 +5.0E-4 −5.0E-4mols end 2.00E-3 5.0E-4 0molarity, new 0.0667 0.017 0

HCl NaCl NaOHmols before 2.50E-3 0 5.0E-4mols change

mols end

molarity, new

HCl NaCl NaOHmols before 2.50E-3 0 5.0E-4mols change −5.0E-4 +5.0E-4 −5.0E-4mols end 2.00E-3 5.0E-4 0molarity, 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

42


• 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


mols end

molarity, new

HCl NaCl NaOHmols before 2.50E-3 0 2.5E-3mols change −2.5E-3 +2.5E-3 −2.5E-3mols end 0 2.5E-3 0molarity, new

HCl NaCl NaOHmols before 2.50E-3 0 2.5E-3mols change −2.5E-3 +2.5E-3 −2.5E-3mols end 0 2.5E-3 0molarity, new 0 0.050 0





• At equivalence point added 25.0 mL NaOH

HCl NaCl NaOHmols before 2.50E-3 0 3.0E-3mols change −2.5E-3 +2.5E-3 −2.5E-3mols end 0 2.5E-3 5.0E-4molarity, new 0 0.045 0.0091


mols end

molarity, new

HCl NaCl NaOHmols before 2.50E-3 0 3.0E-3mols change −2.5E-3 +2.5E-3 −2.5E-3mols end 0 2.5E-3 5.0E-4molarity, new




• After equivalence point added 30.0 mL 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

25.0 mL 0.100 M HCl0.00250 mol HClpH = 1.00

HNO3 NaNO3 NaOHmols before 1.25E-2 0 1.5E-3mols change −1.5E-3 +1.5E-3 −1.5E-3mols end 1.1E-3 1.5E-3 0molarity, new

HNO3 NaNO3 NaOHmols before 1.25E-2 0 1.5E-3mols change −1.5E-3 +1.5E-3 −1.5E-3mols end 1.1E-3 1.5E-3 0molarity, new 0.018 0.025 0

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(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


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

HNO3 NaNO3 NaOH

mols before 1.25E-2 0 1.5E-2mols 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 end 0 1.25E-2 0.0025

molarity, new

Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25 M

HNO3



HNO3 NaNO3 NaOH


mols end 0 1.25E-2 0.0025

molarity, new 0 0.0833 0.017

HNO3 NaNO3 NaOH


mols end 0 1.25E-2 0.0025

molarity, new

Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 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


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 NaOH

HA A− OH−

mols before 2.50E-3 0 0mols 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− NaOHmols before 2.50E-3 0 3.0E-3mols change −2.5E-3 +2.5E-3 −2.5E-3mols end 0 2.5E-3 5.0E-4molarity, new

HA A− NaOHmols before 2.50E-3 0 3.0E-3mols change −2.5E-3 +2.5E-3 −2.5E-3mols end 0 2.5E-3 5.0E-4molarity, new 0 0.045 0.0091

HA A− NaOHmols before 2.50E-3 0 3.0E-3mols change

mols end

molarity, new


56

added 30.0 mL NaOH


• 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 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

Titration Curve of a Weak Base with a Strong Acid

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

• 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 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl.


• 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 before 2.50E-3 0 5.0E-4mols 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




• Before equivalence: after adding 5.0 mL of HCl

NH3 NH4Cl HCl

mols before 2.50E-3 0 5.0E-4mols 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

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 at equivalence.



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

added 25.0 mL HClNH3 NH4Cl HCl


mols end 0 2.5E-3 0

molarity, new 0 0.050 0

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 at equivalence.

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)




• After equivalence: after adding 30.0 mL HClNH3 NH4Cl HCl


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 titration– the 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 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

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)

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

write the dissociation reaction and Ksp expressioncreate an ICE table defining the change in terms of the solubility of the solid

[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

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

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

PbCl2(s) Pb2+(aq) + 2 Cl−(aq)addition of Cl− shifts the

equilibrium to the left

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

write the dissociation reaction and Ksp expressioncreate an ICE table defining the change in terms of the solubility of the solid

[Ag+] [Cl−]

initial 0 0.55

change +S +S

equilibrium S 0.55 + S

AgCl(s) Ag+(aq) + Cl−(aq)

Ksp = [Ag+][Cl−]

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

substitute into the Ksp expression,assume S is smallfind the value of Ksp from Table 16.2, plug into the equation, and solve for S

[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 hydroxide– and the lower the pH, the higher the solubility– higher 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 solubilityM2(CO3)n(s) 2 Mn+(aq) + nCO3

2−(aq)H3O+(aq) + CO3

2− (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 occur– Q = Ksp, the solution is saturated, no precipitation– Q < Ksp, the solution is unsaturated, no precipitation– Q > 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

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 reactiondetermine the ion concentrations of the original saltsdetermine the Ksp for any “insoluble” productwrite 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

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

precipitating may just occur when Q = Ksp

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

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

Ca(OH)2(s) Ca2+(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)

PbCl2(s) Pb2+(aq) + 2 Cl−(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 scheme– wet 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 water– as long as the concentration is large enough– PbCl2 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 ion– H2O is the Lewis base, donating electron pairs to form

coordinate covalent bondsAg+(aq) + 2 H2O(l) Ag(H2O)2

+(aq)• Ions that form by combining a cation with several

anions or neutral molecules are called complex ions– e.g., Ag(H2O)2

+

• The attached ions or molecules are called ligands– e.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 ligandAg(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

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)

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 xconfirm 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

solution– shifting the equilibrium to the right by removing OH−

• Some metal hydroxides also become more soluble in basic solution– acting 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 molecules

Al(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)

Chapter 16: Aqueous Ionic Equilibrium

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Transcript of Chapter 16: Aqueous Ionic Equilibrium