Unit 8 Chapter 17 notes B and L.ppt

Acid-Base Equilibria and

Solubility Equilibria

Strong Acids and Bases

• Reminder: (Memorize!)

HCl

NaOH

HBr

KOH

HI

*Ca(OH)

How do salts affect pH?

• Reminder: a salt will dissociate, but is not

considered an acid nor a base.

• Some do have the ability to interact with

water to produce OH

and H

3

O

in solution

and will create an acidic or basic solution.

(This is called hydrolysis)

Acid-Base Properties of Salts

•

Neutral Solutions:

Salts containing an alkali metal or alkaline earth metal ion (except Be2+) and the conjugate base of a strong acid

(e.g. Cl-, Br-, and NO 3-).

•

Basic Solutions:

Salts derived from a strong base and a weak acid. NaCl (s) + H₂O (l) Na+ (aq) + Cl- (aq)

NaCH₃COO (s) + H₂O (l) Na+ (aq) + CH

3COO- (aq)

CH₃COO- (aq) + H

A/B Properties Cont

•

Acid Solutions:

Salts derived from a strong acid and a weak base.

Example: Predict if the following salt solutions will be

acidic, basic or neutral:

1. NaI

2. BaClO

3. NH

NO

4. K

CO

NH₄Cl (s) + H₂O (l) NH₄+ (aq) + Cl- (aq)

NH₄+ (aq)+ H

Acid-Base Properties of Salts

Solutions in which both the cation and the anion hydrolyze:

The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance.

The presence of a common ion suppresses

the ionization of a weak acid or a weak base due to LE CHATELIER!

Consider mixture of CH₃COONa (strong electrolyte) and CH₃COOH (weak acid) - this is the idea behind BUFFERS!

CH₃COONa (s) Na+ _(aq) + CH

3COO- (aq)

CH₃COOH (aq) H+ _(aq) + CH

3COO- (aq)

A buffer solution is a solution of: 1. A weak acid or a weak base and

2. The salt of the weak acid or weak base 3. The weak acid neutralizes added base 4. The base neutralizes added acid

Both must be present!

A buffer solution has the ability to resist changes in pH upon the addition of SMALL amounts of either acid or base, it consists of a weak acid or base and its conjugate.

Add strong acid H+ _(aq) + CH

3COO- (aq) CH3COOH (aq)

Add strong base

OH- _(aq) + CH

3COOH (aq) CH3COO- (aq) + H2O (l)

Weak Acid/Salt Buffering Pairs

Weak Acid/Salt Buffering Pairs

Weak Acid of the acidFormula Example of a salt of the weak acid

 Hydrofluoric  HF   KF – Potassium fluoride 

 Formic   HCOOH   KHCOO – Potassium formate 

 Benzoic   C6H5COOH   NaC6H5COO – Sodium benzoate

 Acetic   CH3COOH   NaH3COO – Sodium acetate 

 Carbonic   H2CO3   NaHCO3 - Sodium bicarbonate

 Propanoic   HC3H5O2    NaC3H5O2  - Sodium propanoate

 Hydrocyanic   HCN   KCN - potassium cyanide 

The salt will contain the anion of the weak acid,

The salt will contain the anion of the weak acid,

and the cation of a strong base (NaOH, KOH)

Weak Base/Salt Buffering Pairs

Weak Base/Salt Buffering Pairs

The salt will contain the cation of the weak base,

The salt will contain the cation of the weak base,

and the anion of a strong acid (HCl, HNO

and the anion of a strong acid (HCl, HNO

)

)

Base Formula of the base Example of a salt of the weak acid

Ammonia   NH₃  NH₄Cl - ammonium chloride

 Methylamine  CH₃NH₂  CH₃NH₃Cl – methylammonium chloride  Ethylamine  C₂H₅NH₂  C₂H₅NH₃NO₃ -  ethylammonium nitrate

 Aniline  C6H5NH2  C6H5NH3Cl – aniline hydrochloride

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) NH₃/NH₄Cl

(a) HF is a weak acid and F- is its conjugate base

buffer solution (b) HBr is a strong acid

not a buffer solution

(c) NH₃ is a weak base and NH₄+ is its conjugate acid

[HF] / [F

-

] buffered pair

Summary:

Explained by Le Chat:

HF H

+ F

-•add base, reacts with H

to produce water, [H

]

decreases, rxn shift FORWARD [F

] increases

HF + OH

F

+ H

2

O

•Add acid, rxn shifts REVERSE [HF] increases

F

+ H

What is the pH of a solution containing 0.200 M

CH₃COOH and 0.100 M CH₃COONa? K_a = 1.8 x 10-5

CH₃COOH (aq) H+ _(aq) + CH

3COO- (aq)

Initial (M) Change (M)

Equilibrium (M)

0.200 0.00

-x +x

0.200 - x

0.100 +x

x 0.100 + x

Use Ka formula: Ka = [HCOO

-_{] [H}+]

[HCOOH]

1.8 x 10-5 = [0.100] [x]

[0.200]

Mixture of weak acid and conjugate base: BUFFER!

Consider mixture of salt NaA and weak acid HA.

HA (aq) H+ _(aq) + A- _(aq)

NaA (s) Na+ _(aq) + A- _(aq)

K_a = [H+][A-] [HA]

[H+] = Ka [HA]

[A-_]

-log [H+] = -log K

a - log

[HA] [A-_]

-log [H+] = -log K

a + log [A

-_]

[HA]

pH = pK_a + log [A-]

[HA] pKa = -log Ka

Henderson-Hasselbalch equation

pH = pK_a + log [conjugate base] [acid]

Calculate the pH of a solution of 0.250 M

HCN and 0.170 M KCN. The K

of HCN is

4.9 x 10

. Use both equilibrium approach

and Henderson-Hasselbalch approach.

Solve using

Henderson-Hasselbalch Approach

Calculate the pH of a solution that is

0.050 M in benzoic acid (HC

H

O

) and

0.150 M sodium benzoate (NaC

H

O

). K

=

6.5 x 10

Solve using

Henderson-Hasselbalch Approach

Calculate the pH of a solution that is 0.18

M in C

H

NH

and 0.18 M C

H

NH

Cl.

K

= 3.8 x 10

Answer: pH = 4.58

There is significance to [A

] = [HA],

Calculate the pH of the 0.30

M

NH

/0.36

M

NH

Cl buffer system. K

= 1.8 x 10

NH₃ (aq) OH- (aq) + NH₄+ (aq)

pH = pK_a + log [NH3]

[NH₄+] pKa = 9.25 pH = 9.25 + log

[0.30]

Buffer Capacity

•

Buffer Capacity

is the amount of acid or

base the buffer can neutralize before pH

begins to change to an appreciable degree

(higher conc. [M] more buffering capacity)

•

pH range

is the pH range over which a

buffer effectively resists change in pH in

either direction.

AP Check for Understanding

A solution containing HCl and the weak acid HClO

has

a pH of 2.4. Enough KOH(

aq

) is added to the solution

to increase the pH to 10.5. The amount of which of the

following species increases as the KOH(

aq

) is added?

A. Cl

(

aq

)

B. H

(

aq

)

C. ClO

(

aq

)

AP Check for Understanding

A solution is prepared by adding 100 mL of

1.0

M

HC

H

O

(

aq

) to 100 mL of 1.0

M

NaC

H

O

(

aq

).

The solution is stirred, and its pH is measured to be

4.73. After 3 drops of 1.0

M

HCl are added to the

solution, the pH of the solution is measured and is still

4.73. What chemical equation represents the chemical

reaction that accounts for the fact that acid was added

but there was no detectable change in pH?

H

O

(

aq

) + C

AP Check for Understanding

Mixtures that would be considered buffers include which

of the following?

•I. 0.10

M

HCl + 0.10

M

NaCl

•II. 0.10

M

HF + 0.10

M

NaF

•III. 0.10

M

HBr + 0.10

M

NaBr

A. I only

AP Check for Understanding

HC

H

O

(aq)+H

O(l) H

⇄

O

(aq)+C

4

H

O

(aq)

The chemical equation above represents the acid

ionization equilibrium for HC

H

O

for which pK

=4.8.

Which of the following is the best estimate for the pH of

a buffer prepared by mixing 100.mL of 0.20M HC

H

O

with 100.mL of 0.10M NaC

H

O

?

Titrations

In a titration a solution of accurately known concentration (titrant) is added gradually to another solution of unknown

concentration (analyte) until the chemical reaction between the two solutions is complete.

Equivalence point – the point at which the reaction is complete

moles [H+] = moles [OH-]

End point – the point at which a color change is observed in soln.

Indicator – substance that changes color at (or near) the

equivalence point

Slowly add base to unknown acid

UNTIL

The indicator changes color

Match the Titration Curves

Match the Titration Curves

Match the Titration Curves

Match the Titration Curves

pH Refresher

Acids: 0-6

Bases: 8-14

Strong Acid with Strong Base

Features of curve:

1.Low pH starting point

means strong acid.

2.Equivalence pt at pH

= 7 means strong

acid/base pair

3.End of pH curve is

near pH = 14 means

strong base.

Strong Acid-Strong Base Titrations

NaOH (aq) + HCl (aq) H₂O (l) + NaCl (aq)

OH- _{(aq) + H}+ _(aq) H

2O (l)

Weak acid w/ strong base

Features of curve:

1.High starting pH value

means weak acid

2.Very small slope (flat) in

acid region (buffer)

3.Equivalence pt. higher

than 7 means strong base

4.End of pH curve is near

pH = 14 means strong

base.

Weak Acid-Strong Base Titrations

CH₃COOH (aq) + NaOH (aq) CH₃COONa (aq) + H₂O (l)

CH₃COOH (aq) + OH- (aq) CH₃COO- (aq) + H₂O (l)

CH₃COO- _{(aq) + H}

2O (l) OH- (aq) + CH3COOH (aq)

Strong base with strong acid

Features of curve:

1.Starts with high pH,

means strong base.

2.Equivalence at pH = 7,

means strong acid mixed

with strong base

3.Ends with low pH, means

strong acid.

Weak Base – Strong Acid Titrations

HCl (aq) + NH₃ (aq) NH₄Cl (aq)

NH₄+ _(aq) + H

2O (l) NH3 (aq) + H3O+ (aq)

At equivalence point (pH < 7):

16.4

H+ _(aq) + NH

Comparing Results

Comparing Results

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00

m illiliters NaOH (0.10 M)

p H 1 2 3 4 5 6 7 8 9 10 11 12 13

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00

m illiliters NaOH (0.10 M)

p

H

Unbuffered

Buffered

Polyprotic Acids

• Two Equivalence

points shown on

graph.

• You are not expected

to need to know or do

any calculations

regarding polyprotic

acids and titrations,

just be able to

Example:

Write out dissociation reactions to demonstrate

the following acid, base or salt reactions:

1. Hydrofluoric acid (HF) is mixed with NaF.

2. Formic acid (HCOOH) is mixed with NaOH.

3. Ammonia (NH

) is mixed with HCl.

4. Acetic acid (CH

COOH) is mixed with

KCH

COO.

pH Calcs using titration data

• Usually 2 types of calcs with one weak

and one strong (buffered titration):

– Find pH after small amount of titrant added

– Find pH at equivalence point.

1. Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH₃COOH (K_a = 1.8 x 10-5).

CH₃COOH (aq) + OH- (aq) CH₃COO- (aq) + H₂O (l)

start (moles)

end (moles)

0.0050 0.0045

0.0005 0.0 0.0045

Final volume = 95 mL _[CH3COO-_{] =} 0.0045

0.095 = 0.047 M [CH₃COOH] = 0.0005

0.095 = 0.0053 M

K_a =[H+][CH3COO-]

[CH₃COOH] =

[H+_][0.0045]

[0.0053] = 1.8 x 10-5

2. Exactly 100 mL of 0.10 M HNO₂ are titrated with 0.10 M

NaOH solution. The K_a = 4.0 x 10-4. What is the pH at the

equivalence point ?

HNO₂ (aq) + OH- (aq) NO₂- (aq) + H₂O (l)

start (moles)

end (moles)

0.01 0.01

0.0 0.0 0.01

NO₂- _{(aq) + H}

2O (l) OH- (aq) + HNO2 (aq)

Initial (M) Change (M)

Equilibrium (M)

0.05 0.00

-x +x

0.05 - x

0.00 +x

x x

[NO₂-_{] =} 0.01

0.200 = 0.05 M Final volume = 200 mL

K_b = [OH-][HNO2]

[NO₂-] =

x2

0.05-x = 2.2 x 10-11

0.05 – x  0.05 x  1.05 x 10

-6 _{= [OH-]}

pOH = 5.98

Buffer/Titration Prob Check List:

• Always start with dissociation equilibrium

equation or acid added to base (depends on

what is asked)

• Be sure to note whether this substance is an

acid or base (K

or K

will tell you that!)

• Only three total types of weak/buffer

reactions:

– 1. Initial pH of weak acid or base

– 2. pH after addition of base to acid (or vice

versa)

Helpful Hints:

1. Initial pH: use K

or K

equilibrium eq. and

ICE box, then pH or pOH formula.

2. pH after adding OH-: begin w/ b, add, a

chart using moles, then convert to M and

use Henderson-Hasselbalch to find pH.

3. pH at equivalence: if don’t know vol. of

base added, use M

V

=M

V

to figure out

V

. Use b, add, a chart, then use conj.

Hydrolysis eq. in ICE, then use K

or K

expression to find OH

or H

then pOH or

3. A 25 mL solution of 0.175

M

CH

NH

is

titrated with 0.150

M

HBr solution. The

K

= 4.4 x

10

. What is the pH at the equivalence point ?

Solution:

step 1: b, add, a chart

4. Consider the titration of 25.0 mL of 0.100 M

HCOOH with 0.100 M NaOH.

K

= 1.8x10

Calculate the initial pH of formic acid.

5. Consider the titration of 25.0 mL of 0.100

M HCOOH with 0.100 M NaOH.

K

= 1.8x10

Calculate the pH at the

equivalence point.

What is the pH at the half Equivalence point?

Answer: pH = 8.23

6. A student creates a solution by mixing

0.200 M NH

with 0.250 M NH

Cl. What is

the resulting pH of the solution?

K

= 1.8 x 10

What type of solution has the student

created?

Answer: pH = 9.36

What about strong acid titrated

into weak base? YOU TRY!

Consider the titration of 25.0 mL of 0.100 M

NH

with 0.100 M HCl.

K

= 1.8 x 10

*This is a base, be careful when asked for pH

1.Calculate the initial pH of ammonia.

pH = 11.2

2. Calculate the pH after the addition of 10

mL of HCl.

pH = 9.43

3.Calculate the pH at the equivalence point.

Acid-Base Indicators

HIn (aq) H+ _(aq) + In- _(aq)  10

[In-]

[HIn] Color of base (In-) predominates

 0.1 [In-]

[HIn] Color of acid (HIn) predominates

16.5

Usually pick indicator

whose pK_a

is equal to pH at

equivalence pt.

Phenolphthalien

• Acid-base nature of indicators, each with

different K

values, different forms

Which indicator(s) would you use for a titration of HNO₂ with KOH ?

Weak acid titrated with strong base.

At equivalence point, will have conjugate base of weak acid. At equivalence point, pH > 7

Use cresol red or phenolphthalein

*Remember, pick an

indicator which will turn colors within pH

range of

A solution of a weak

monoprotic acid is titrated

with a solution of a strong

base, KOH. Consider the

points labeled (A) through

(E) on the titration curve

that results, as shown

below.

The point at which the

moles of the added strong

base are equal to the

moles of the weak acid

initially present______?

A solution of a weak

monoprotic acid is

titrated with a solution

of a strong base, KOH.

Consider the points

labeled (A) through (E)

on the titration curve

that results, as shown

below.

The point at which the pH is

closest to that of the strong

base being added

______?

A solution of a weak

monoprotic acid is titrated

with a solution of a strong

base, KOH. Consider the

points labeled (A) through

(E) on the titration curve

that results, as shown

below.

The point at which the

concentrations of the weak

acid and its conjugate base

are approximately

equal

______?

AP Check for Understanding

A student mixes 40.mL of 0.10M

HBr

(aq)

with 60.mL of 0.10M KOH

(aq)

at 25°C. What

is the [OH

] of the resulting solution?

A

. [OH

]=0.060M

B.

[OH

]=0.033M

C.

[OH

]=0.020M

AP Check for Understanding

The pH of a solution made by combining 150.0

mL of 0.10

M

KOH(

aq

) with 50.0 mL of

0.20

M

HBr(

aq

) is closest to which of the

following?

A.

2

B.

4

C.

7

is the number of moles of solute dissolved in 1 L of a saturated solution.

is the number of grams of solute dissolved in 1 L of a saturated solution.

Solubility

(g/L)

Molar solubility

(mol/L)

Solubility Equilibria,

K

– solubility product -

degree of solubility

AgCl (s) Ag+ _(aq) + Cl- _(aq)

K_sp = [Ag+][Cl-] K

sp is the solubility product constant

MgF₂ (s) Mg2+ (aq) + 2F- (aq) K_sp = [Mg2+][F-]2

Ag₂CO₃ (s) 2Ag+ _(aq) + CO

32- (aq) Ksp = [Ag+]2[CO32-]

Ca₃(PO₄)₂ (s) 3Ca2+ _(aq) + 2PO

43- (aq) Ksp = [Ca2+]3[PO43-]2

*Just like any other equilibrium dissociation constant, K.

*

Remember based on equilibrium and K:

Which salt has a greater solubility: AgCl, K_sp = 1.6x10-10 or

AgBr, K_sp = 7.7x10-13 ?

Solving Solubility Problems

Solving Solubility Problems

For the salt PbCl

at 25



C, K

= 1.6 x 10

Find the molar solubility of PbCl

.

PbCl

(s)



Pb

(aq) + 2Cl

(aq)

I

C

E

O

O

+x

+2x

x

2x

1.6 x 10

= (

x

)(

2x

)

= 4

x

Try another:

For the salt AgCl at 25



C, K

= 1.77 x 10

. Find the molar solubility of AgCl.

Calculate K

sp

from measured ion

molarities:

Solid silver chromate (Ag

CrO

) is added to

water at 25°C and allowed to reach

equilibrium. The amount of silver [Ag

]

measured in the solution is 1.3x10

M

.

Calculate the

K

of silver chromate.

What is the solubility of silver chloride in g/L ?

AgCl (s) Ag+ _(aq) + Cl- _(aq)

K_sp = [Ag+][Cl-]

I (M) C (M) E (M)

0.00 +s

0.00 +s

s s

K_sp = s2

s =



s = 1.3 x 10-5

[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M Solubility of AgCl = 1.3 x 10-5 mol AgCl

1 L soln

143.35g AgCl 1 mol AgCl

x = 1.9 x 10-3 _g/L

K_sp = 1.6 x 10-10

16.6

Some books use “s” instead of “x” for solubility measured in grams – it is not a big deal if you do not, just remember you need to convert to g at the

end of the prob using stoich

What factors can affect

solubility?

• Other ions present

in solution

(

common ion

effect

)

Solving Solubility with a Common Ion

Solving Solubility with a Common Ion

For the salt AgI at 25



C, K

= 1.5 x 10

What is its solubility in pure water?

What is its solubility in 0.05 M NaI?

AgI(s)



Ag

(aq) + I

(aq)

I

C

E

0.05

O

+x

+x

x

0.05+x

1.5 x 10

= (

x

)(

0.05+x

)



(

x

)(

0.05

)

Why is solubility

higher in water?

Common ion effect:

*The solubility of a

slightly soluble salt is

decreased by the

presence of a second

solute with a

common ion.

The Common Ion Effect and Solubility

The presence of a common ion decreases

the solubility of the salt.

What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr? K_sp = 7.7 x 10-13

AgBr (s) Ag+ (aq) + Br- (aq)

K_sp = 7.7 x 10-13

s2 = K sp

s = 8.8 x 10-7

The Common Ion Effect and Solubility

The presence of a common ion decreases

the solubility of the salt.

What is the molar solubility of AgBr in (a) pure water and (b) 0.0010 M NaBr?

NaBr (s) Na+ _(aq) + Br- _(aq)

[Br-] = 0.0010 M

AgBr (s) Ag+ _(aq) + Br- _(aq)

+ s

0.0010 + s = 0.0010 K_sp = 0.0010 x s

s = 7.7x 10-10

0.0010 + s

0

s

pH and Solubility

Consider the substance Mg(OH)

, K

= 1.8 x 10

Is Mg(OH)

more soluble in acid or base?

Mg(OH)₂ (s) Mg2+ _(aq) + 2OH- _(aq)

K_sp = [Mg2+][OH-]2 = 1.2 x 10-11 Add acid (H

+)

Lower [OH-]

OH- _{(aq) + H}+ _(aq) H

2O (l) remove

Increase solubility of Mg(OH)₂ Add base (OH-)

Raise [OH-] add

Decrease solubility of Mg(OH)₂

Predicting if ppt will form from chemical reaction:

When asked to predict whether the mixing of chemical substances will create ppts, consider the following:

1. Write out chemical equation, reactants and products.

2. Predict which product will be the ppt based on solubility rules.

4. Compare to K_sp for that particular salt.

Dissolution of an ionic solid in aqueous solution:

Q = K_sp Saturated solution

Q < K_sp Unsaturated solution No precipitate

Q > K_sp Supersaturated solution Precipitate will form

*Remember based on equilibrium and K, that large K favors FORWARD (dissolves) and small K favors REVERSE (precipitate)

If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M

CaCl₂, will a precipitate form? Ca(OH)₂ K_sp = 8.0 x 10-6

The ions present in solution are Na+, OH-, Ca2+, Cl-.

Only possible precipitate is Ca(OH)₂ (solubility rules). Is Q > K_sp for Ca(OH)₂? K_sp = 8.0 x 10-6

[Ca2+]

0 = 0.100 M [OH-]0 = 4.0 x 10-4 M

K_sp = [Ca2+][OH-]2 = 8.0 x 10-6

Q = [Ca2+]

0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8

AP Check for Understanding

Equimolar samples of Pb(OH)

(

s

), PbI

(

s

), and

PbF

(

s

) are placed in three separate beakers,

each containing 250 mL of water at 25°C. After

the solutions are stirred, solid remains in the

bottom of each beaker. Based on the

K

values

for the compounds listed in the table above, a

solution of which of the compounds will have the

lowest [Pb

] ?

AP Check for Understanding

How many moles of NaF must be dissolved in 1.00

liter of a saturated solution of PbF

at 25˚C to

reduce the [Pb

] to 1 x 10

molar?

(K

PbF

at 25˚C = 4.0 x 10

)

A

. 0.020 mole

B.

0.040 mole

C.

0.10 mole

D.

0.20 mole

AP Check for Understanding

In a saturated solution of Zn(OH)

at 25°C, the value

of [OH

] is 2.0 x 10

M

. What is the value of the

solubility-product constant,

K

, for Zn(OH)

at 25°C ?

A

. 4.0 x 10

B.

8.0 x 10

C.

1.6 x 10

D.

4.0 x 10

AP Check for Understanding

A 1.0 L solution AgNO

(

ag

) of and Pb(NO

)

(

aq

)

has a Ag

concentration of 0.020

M

and a

Pb

concentration of 0.0010

M

.

A 0.0010 mol sample of K

SO

(

s

) is added to the

solution.

Based on the information in the table

will a precipitate form? Which substance?

(

Assume that the volume change of the solution

is negligible.)