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Chemistry: The Molecular Science Moore, Stanitski and Jurs. Chapter 17: Additional Aqueous Equilibria. Buffer Solutions. # #. * *. # 5 pH units~100,000X change in [H 3 O + ] *added 10 mL of 1.0 M HCl (to 1.00L buffer (0.50mol HOAc/NaOAc). Buffer Solutions.

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buffer solutions
Buffer Solutions

#

#

*

*

# 5 pH units~100,000X change in [H3O+]

*added 10 mL of 1.0 M HCl

(to 1.00L buffer (0.50mol HOAc/NaOAc)

buffer solutions1
Buffer Solutions

A buffer is a solution characterized by the ability to resist changes in pH when limited amounts of acid or base are added to it.

Buffers contain either a weak acid and its conjugate base (or a weak base and its conjugate acid) in approximately equal proportion and in equilibrium (they do not react with each other).

buffer solutions2
Buffer Solutions
  • Consider a buffer with equal molar amounts of HA and its conjugate base A-.
  • When H3O+ is added to the buffer it reacts with the base A-.
  • When OH- is added to the buffer it reacts with the acid HA.
slide5

Human blood has a

normal pH of 7.40  0.05.

Below 7.35 - acidosis,

Above 7.45 - alkalosis

– both life threatening

conditions.

The most important blood buffer is provided by CO2

CO2(aq) + H2O(l) H2CO3(aq)

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)

the ph of buffer solutions
The pH of Buffer Solutions

Calculate the pH of a solution that contains 0.0025 mol carbonic acid

and 0.025 mol of hydrogen carbonate ion in 1.00L of water.

The Ka of carbonic acid is 4.2 × 10-7

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)

To calculate pH we need the concentration of hydronium ion

in the solution

slide7

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)

0.025

0.0025

0

-x +x +x

Initial

Change

Equilibrium

0.0025 - x x 0.025 + x

pH = - log [H3O+] = - log (4.2 x 10-8) = 7.38

the ph of buffer solutions the henderson hasselbalch equation

Consider a buffer of a weak acid HA and its conjugate base A-. The acid ionization equilibrium is:

The pH of Buffer Solutions The Henderson-Hasselbalch Equation
  • The acid ionization constant is:
the henderson hasselbalch equation
The Henderson-Hasselbalch Equation

OR

pH = pKa when [HA] = [A-], since log(1)=0

Useful buffer range: pH = pKa ±1

(10:1 or 1:10 ratio of [A-]/[HA]).

example
Example

Calculate the pH of a solution that contains 0.0025 mol carbonic acid and 0.025 mol of hydrogen carbonate ion in 1.00L of water. The Ka of carbonic acid is 4.2 × 10-7

H2CO3(aq) + H2O(l) H3O+(aq) + HCO3-(aq)

buffer solutions3
Buffer Solutions

A buffer that is needed for maintaining a desired pH can be chosen by examining pKa values

example prepare a ph 4 buffer
Example – Prepare a pH=4 buffer

From table 17.1 lactic acid, CH3CHOHCOOH has the closest value pKa = 3.85.

What proportion of CH3CHOHCOO-/CH3CHOHCOOH should you use?

CH3CHOHCOO-/CH3CHOHCOOH = 1, pH=pKa

CH3CHOHCOO-/CH3CHOHCOOH <1, more acidic – pH<3.85

CH3CHOHCOO-/CH3CHOHCOOH >1, more basic – pH>3.85

example prepare a ph 4 buffer1
Example – Prepare a pH=4 buffer

Using Henderson-Hasselbach equation…

example1
Example

The concentration of lactate ion has to be 1.4 times that of lactic acid.

For example if we use 0.10 molar lactic acid we will need 0.14 molar

lactate ion from a soluble salt such as sodium lactate,

CH3CHOHCOONa.

preparing buffers
Preparing Buffers

There are two ways of preparing a buffer:

  • Combine the conjugate acid/base pairs directly.

0.25 M CH3CHOHCOOH combined with about 0.25 M NaCH3CHOHCOO

2. Start with one of the conjugate pairs and add strong acid

or strong base to create the other.

  • Begin with 0.50 M CH3CHOHCOOHsolution.
  • Add 0.25 M NaOH.
  • This produces a solution containing:

0.25 M CH3CHOHCOOH + 0.25 M CH3CHOHCOO -

CH3CHOHCOOH + OH - CH3CHOHCOO - + H2 O

practice
Practice
  • 1. (17-21) To buffer a solution at a pH of 4.57, what mass of sodium acetate (NaCH3COO). would you add to 500. mL of a 0.150 M solution of acetic acid (CH3COOH).
  • A. Choose an acid/base combination to prepare a pH=7 buffer.
  • B. Describe how you would prepare the buffer by starting with 1.0L of a 0.10 M solution of the weak acid and adding 1.0M NaOH.
buffer capacity
Buffer Capacity

Buffer Capacity- the ability of the buffer to prevent changes in pH when small amounts of acid or base are added.

The amount of an acid or a base that can be added to a buffer without causing a significant pH change ( more than 1 unit).

The ratio of [A-]/[HA] determines the pH of the buffer whereas the magnitude of [A-] and [HA] determine the buffer capacity.

buffer capacity1
Buffer Capacity

1 mol of acetate ion + 1 mol of acetic acid

buffer capacity2
Buffer Capacity
  • A 1 liter solution contains:
  • 0.390 M hydrocyanic acid + 0.350 M sodium cyanide.
  • What is the pH of the buffer? Ka HCN = 4.0x10-10
  • How will the pH change if we add 0.200 moles of NaOH?
  • (neglect the change in volume)

HCN(aq) + H2O H3O+ + CN-(aq)

BEFORE: 0.390 M 0.350 M

buffer capacity3
Buffer Capacity

2. How will the pH change if we add 0.200 moles of NaOH?

(neglect the change in volume)

HCN(aq) + H2O H3O+ + CN-(aq)

BEFORE: 0.390 M 0.350 M

AFTER: -0.200 M +0.200 M

I

C

E

0.190

0

0.550

-x +x + x

0.190-x + x 0.550 + x

slide22

Buffer Capacity

HCN(aq) + H2O H3O+ + CN-(aq)

E 0.190-x + x 0.550 + x

[H3O+] = 1.4 x 10-10

The pH value changes slightly.

The buffer capacity has not been exceeded

pH = -log (1.4 x 10-10) = 9.86

buffer capacity4
Buffer Capacity

A 1 liter solution contains 0.390 M hydrocyanic acid and 0.350 M

sodium cyanide. Ka = 4.0 x 10-10 3. What will be pH if we add 0.450 moles of hydrochloric acid?

(neglect the change in volume)

4. Will the buffer capacity be exceeded?

HCN(aq) + H2O H3O+ + CN-((aq)

BEFORE: 0.390 M 0.350 M

AFTER: +0.450 -0.450

0

0.740

I

C

E

0.100

-x +x + x

0.740-x 0.100 + x + x

slide24

Buffer Capacity

HCN(aq) + H2O H3O+ + CN-(aq)

E 0.740-x 0.100 + x + x

[H3O+] = 0.100+3.0x10-9

The pH value changes significantly.

The buffer capacity has been exceeded.

pH = -log 0.100 = 1.000

example buffer capacity
Example - Buffer Capacity

A buffer is prepared using 0.25 mol H2PO4- and 0.15 mol HPO42- in 1000.0 mL solution. pKa of H2PO4- = 7.21

a.) What is the pH of the buffer?

b.) What will be the pH of a solution if 6.2 g KOH are added to this buffer. Will the buffer capacity be exceeded?

c.) What will be the pH of a solution if 20.0 mL of 6.00 M HCl are added to this buffer. Will the buffer capacity be exceeded?

H2PO4-(aq) + H2O(l) H3O+(aq)+ HPO42-(aq)

We can calculate the pH of the buffer using Henderson-Hasselbach equation:

slide26

Moles of KOH:

a.) What will be the pH of a solution if 6.2 g KOH are added to this buffer. Will the buffer capacity be exceeded?

H2PO4-(aq) + H2O(l) H3O+ + HPO42-(aq)

Before adding base:

Reaction with base:

After adding base:

0.25 0.15

-0.11 +0.11

0.14 0.26

The buffer capacity was not exceeded

slide27

What will be the pH of a solution if 20.0 mL of 6.00 M HCl

are added to this buffer. Will the buffer capacity be exceeded?

Moles of HCl:

b.)

mol = M × V = 6.0M × 0.0200L = 0.12 mol HCl

H2PO4-(aq) + H2O(l) H3O+ + HPO42-(aq)

Before adding acid:

Reaction with acid:

After adding acid:

0.25mol 0.15mol

+ 0.12mol - 0.12mol

0.37mol 0.03mol

The buffer capacity was not exceeded

buffer capacity5
Buffer Capacity

A pH=4.0 buffer was prepared with 0.10 mol lactic acid and 0.14 mole sodium lactate in 1.00L solution. The Ka of lactic acid is 1.4x10-4.

  • Calculate the pH after addition of 4.2mL of 12M HCl.
  • Calculate the pH of the buffer after 6.25mL of 8M NaOH to the original buffer.

(In both cases, neglect volume changes.)

acid base titration
Acid-Base Titration

Equivalence point – is reached when

the stoichimetric amount of titrant

(solution in the buret) has been added.

End point – occurs when the indicator

changes color. We want to use

indicators that give an end point

as close as possible to the equivalence

point.

acid base titration curves
Acid-Base Titration Curves

An acid-base titration curve is a plot of the pH of a solution of acid (or base) against the volume of added base (or acid).

  • Such curves are used to gain insight into the titration process.
  • We can use titration curves to choose an appropriate indicator that will show when the titration is complete.
example strong acid and strong base
Example – strong acid and strong base

A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCl. Calculate the pH:

a.) before the titrant is added

b.) After 40.0mL of titrant has been added

c.) After 50.0mL of titrant has been added

d.) After 50.2mL of titrant has been added

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

H+(aq) + OH-(aq)  H2O(l)

slide33

Example – strong acid and strong base

A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCl. Calculate the pH:

a.) before the titrant is added

HCl is a strong acid, the H3O+ concentration = HCl concentration

pH = -log[H3O+] = -log(0.100) = 1.00

slide34

Calculate moles of acid:

nacid = 0.0500L × 0.100 M = 5.00 x 10-3 mol

Calculate moles of base:

nbase = 0.0400L × 0.100 M = 4.00 x 10-3 mol

b.) After 40.0mL of titrant (NaOH) has been added

0.00400 moles of base reacts with 0.00400 mol of acid

0.00500 molacid- 0.00400 molbase = 1.00 x 10-3mol acid left

pH = -log[H3O+] = -log(0.0111) = 1.95

slide35
A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCl. Calculate the pH:

c.) After 50.0mL of titrant has been added

Calculate moles of acid:

nacid = 0.0500L × 0.100 M = 5.00 x 10-3 mol

Calculate moles of base:

nbase = 0.0500L × 0.100 M = 5.00 x 10-3 mol

5.00 x 10-3 mol of base react with 5.00 x 10-3 mol mol of acid

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

This is the equivalence point.

The salt produced is a neutral salt

so the pH is 7.00

slide36
A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCl. Calculate the pH:

d.) After 50.2mL of titrant has been added.

Calculate moles of acid:

nacid = 0.0500L × 0.100 M = 5.00 x 10-3 mol

Calculate moles of base:

nbase = 0.0502L × 0.100 M = 5.02 x 10-3 mol

5.02 x 10-3 mol of base react with 5.00 x 10-3 mol of acid

5.02 x 10-3 molbase- 5.00 x 10-3 molacid = 0.02 x 10-3mol base

pOH = -log(2.00 x 10-4)=3.70

pH = 14.00 - 3.70 = 10.3

example weak acid and strong base
Example – weak acid and strong base

A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCH3COO. Calculate the pH:

a.) before the titrant is added

b.) After 40.0mL of titrant has been added

c.) After 50.0mL of titrant has been added

d.) After 50.2mL of titrant has been added

example weak acid and strong base1
Example – weak acid and strong base

A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M HCH3COO. Calculate the pH:

a.) before the titrant is added

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

b.) After 40.0mL of titrant has been added

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

Buffer!

slide41
c.) After 50.0mL of titrant has been added

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

d.) After 50.2mL of titrant has been added

solution has xs OH-

slide42
a.) CH3COOH is a weak acid:

-x +x +x

0.100-x x x

1.8 × 10 -5 =

x = [H3O+] = 1.34 × 10-3

pH = -log(1.34 × 10-3) = 2.87

slide43
b.) After 40.0 mL of titrant has been added -

moles of acid:

nacid = 0.0500L × 0.100 M = 5.00 x 10-3 mol

moles of base (NaOH) added = moles of base (acetate ion)

formed:

nbase = 0.0400L × 0.100 M = 4.00 x 10-3 mol CH3COO-

moles of acid left:

nacid = 5.00 x 10-3 – 4.00 x 10-3 = 1.00 x 10-3 mol CH3COOH

A BUFFER!

slide44

1.00 x 10-3mol 4.00 x 10-3

Ka for acetic acid = 1.8 × 10-5, pKa = 4.74

slide45
A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M CH3COOH. Calculate the pH:

c.) After 50.0 mL of titrant has been added

moles of acid:

nacid = 0.0500L × 0.100 M = 0.00500 mol

moles of base (acetate ion)

formed:

nbase = 0.0500L × 0.100 M = 0.00500 mol

We are at the equivalence point now.

There is no original acetic acid left. The acetate ion will hydrolize:

We need to calculate the concentration of the hydroxide ion to

calculate the pH.

slide46

-x +x +x

0.0500-x x x

Remember to use the

total volume.

M = 0.00500mol/0.100L = 0.0500M

x = [OH-] = 5.3 × 10-6

pH = 14.00 – pOH = 14.00 – [-log(5.3 × 10-6)] = 14.00 – 5.28 = 8.72

Notice that the pH is not 7 since CH3COONa is a basic salt.

slide47
A 0.100 M NaOH solution is used to titrate 50.0 mL of 0.100M CH3COOH. Calculate the pH:

d.) After 50.2mL of titrant has been added

Calculate moles of acid:

nacid = 0.0500L × 0.100 M = 5.00 x 10-3 mol

Calculate moles of base:

nbase = 0.05020L × 0.100 M = 5.02 x 10-3 mol

5.00 x 10-3 mol of base react with 5.00 x 10-3 mol of acid

5.02 x 10-3 molbase- 5.00 x 10-3 molacid = 2 x 10-5 mol base

pH = 14.00 – pOH = 14.00 -log[OH-] =

-log(2.00 x 10-4) = 10.3

titration of a weak acid by a strong base
Titration of a weak acid by a strong base.
  • Before the titration the pH is higher for a weak acid; the acid and its conjugate base are at equilibrium
  • At the equivalence point pH is higher because of the hydrolysis of a basic salt formed.

pH=pKa

A slighter change in pH requires careful choice of indicator.

example titration of a weak base by a strong acid
Example - titration of a weak base by a strong acid.

What is the pH at the eqivalence point when 50 mL of 0.100M ammonia is titrated with 0.100 M HCl?

NH3(aq) + H3O+ (aq) NH4+(aq) + H2O(l)

from HCl

At the equivalence point:

moles of acid = moles of base

moles of base:

nbase = 0.0500L × 0.100 M = 5.00 x 10-3 mol

The ammonium ion will hydrolize

We need to calculate the concentration of the hydronium ion to

calculate the pH.

slide52

-x +x +x

0.0500-x x x

Volume of acid (HCl)= Volume of base (the same molarity and 1:1

mole ratio)

slide53

x = [H3O+ ] = 5.3 × 10-6

pH = -log(5.3 × 10-6) = 5.28

Notice that the pH is not 7 since NH4Cl is an acidic salt

titration solution stoichiometry
Titration: Solution Stoichiometry

An aqueous solution of sodium hydroxide is standardized by titration with a 0.381 M solution of hydrochloric acid. If 19.8 mL of base are required to neutralize 22.4 mL of the acid, what is the molarity of the sodium hydroxide solution?

titration solution stoichiometry1
Titration: Solution Stoichiometry

HCl + NaOH NaCl + H2O

0.381M

22.4 mL 19.8 mL

Calculate the number of moles of acid:

moleacid = M x V = 0.381M x 0.0224L = 8.53 x 10-3 mol

molebase = moleacid = 8.53 x 10-3 mol

Molarity of base:

solubility equilibria k sp
Solubility Equilibria - Ksp

Consider the following reaction:

AbBa(s) aA+(aq) + bB-(aq)

where AB is a moderately or slightly soluble salt

We can write an equilibrium constant expression:

Ksp = solubility product constant

(Notice that the concentration of the solid does not enter the Ksp

expression since it is constant. The aqueous concentrations are given in molarity and are concentrations at equilibrium.)

example solubility product constant
Example: Solubility Product Constant

Write the Ksp expression for the following salts:

CuBr

HgI2

Ag2SO4

(s) Cu+(aq) + Br-(aq)

Ksp = [Cu+][Br-]

(s) Hg2+(aq) + 2I-(aq)

Ksp = [Hg2+][I-]2

(s) 2Ag+(aq) + SO42-(aq)

Ksp = [Ag+]2[SO42-]

k sp and solubility
Ksp and Solubility

Solubility(s) is defined as amount of solute in moles per unit volume of solution that dissolves to form a saturated solution (i.e. one in equilibrium).

If we know the solubility, we can calculate the Ksp.

If we know the Ksp, we can calculate solubility.

The results of such calculations give us only the estimated values in cases when ions produced have charges of -1 or +1.

example2
Example

Calculate the solubility of AgBr at 100ºC. AgBr, Ksp = 5.0×10-5

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

Ksp = [Ag+][Br-] = 5.0×10-5

The Ag+ ions and Br- are formed in equal amounts.

Lets use S for solubility of AgBr. The concentration of each ion

will be also S.

[Ag+][Br-] = 5.0×10-5

(S)(S) = 5.0×10-5

S = 7.1 × 10-3mol/L

practice1
Practice
  • Calculate the solubility of BaF2, Ksp=1.8x10-7.

s=3.6x10-3 M

  • A saturated solution of silver sulfate contains 1.2g of solid dissolved in 250.0 mL. What is the Ksp for silver sulfate?

Ksp=1.5x10-5

  • Solubility and LeChatelier 17.9

How does the formation of bicarbonate from carbonate affect the solubility of CaCO3?

effect of ph
Effect of pH

CaCO3(s) Ca+2(aq) + CO3-2(aq)

CO3-2(aq) +H3O+(aq) HCO3-1(aq) + H2O (l)

HCO3-1(aq) H3O+ (aq) H2CO3(aq) + H2O (l)

H2CO3(aq) CO2(aq) + H2O (l)

Ksp = 8.7 x 10-9

Large Kc

K~105

Insoluble salts containing anions that are B-L bases dissolve in acidic solutions.

solubility and common ion effect
Solubility and Common Ion Effect

It is often desirable to remove an ion from a solution by forming a precipitate of its insoluble compound.

The common-ion effect is the shift in an ionic equilibrium caused by the addition of a solute that provides an ion common to the equilibrium.

BaSO4(s) Ba2+(aq) + SO42-

What would happen if we add Na2SO4 to this equilibrium?

solubility and common ion effect1
Solubility and Common Ion Effect

Adding Na2SO4 to this equilibrium will shift the equilibrium

to the left by removing Ba2+ ions.

practice2
Practice

Calculate the solubility (in mol/L) of Mg(OH)2 in the following: (Ksp = 6 x 10-12)

a) pure water

b) 0.041 M Ba(OH)2

c) 0.0050 M MgCl2

1 x 10-4 mol/L

9 x 10-10 mol/L

2 x 10-5 mol/L

Masterton, Ch. 18, #16

k sp and precipitation
Ksp and Precipitation

The solubility rules (chapter 5) apply to solutions with concentrations of 0.1 M or higher.

If we are dealing with lower concentrations we need to compare ion product Q value with Ksp value to determine if the precipitation will take place.

Ion product, Q , expression has the same form as Ksp expression. The concentrations used in Q expression, however, are not the equilibrium but the original concentrations.

k sp and precipitation1
Ksp and Precipitation

We can consider the following cases when comparing the values of Ksp and Q.

– The solution contains ions at a concentration lower than required for equilibrium with the solid. The solution is unsaturated and no precipitate will form.

– The solution contains ions at a concentration higher than required for equilibrium with the solid. The solution is supersaturated and to reach an equilibrium the precipitate will form.

– The solution is saturated and at the point of precipitation.

Q < Ksp

Q > Ksp

Q = Ksp

example3
Example

20.0 mL of 4.5 × 10-3 M AgNO3is mixed with 10.0 mL of 7.5 × 10-2 M NaBrO3. Will the precipitate form? Ksp for AgBrO3 = 6.7 × 10-5

AgBrO3(s) Ag+(aq) + BrO3-(aq)

moles of Ag+:

M × V = 4.5 × 10-3M × 0.0200L = 9.0 × 10-5mol

moles of BrO3- :

M × V = 7.5 × 10-2M × 0.0100L = 7.5 × 10-4mol

Q = (3.0 × 10-3)(2.5 × 10-2) = 7.5 × 10-5

Q > Ksp - the precipitate will form

example4
Example

Calculate the solubility of AgBr at 100ºC. The Ksp = 5.0×10-5

Based on your result will a precipitate form if you add 0.25 mol of AgBr to 500 mL of water?

[Ag+][Br-] = 5.0×10-5

(S)(S) = 5.0×10-5

S = 7.1 × 10-3mol/L

0.25 mol in 0.500 L = 0.50 mol/L

This value is much larger than the value for solubility.

The precipitate will form.

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