chapter 16 aqueous ionic equilibrium. 2 buffers buffers are solutions that resist changes in ph...

Chapter 16

Aqueous IonicEquilibrium

Chapter 16

Aqueous IonicEquilibrium

2

BuffersBuffers

buffers are solutions that resist changes in pH when an acid or base is added

they act by neutralizing the added acid or base

but just like everything else, there is a limit to what they can do, eventually the pH changes

many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion

buffers are solutions that resist changes in pH when an acid or base is added

they act by neutralizing the added acid or base

but just like everything else, there is a limit to what they can do, eventually the pH changes

many buffers are made by mixing a solution of a weak acid with a solution of soluble salt containing its conjugate base anion

Demo of a pH BufferDemo of a pH Buffer

Assume 1 drop = 0.05mL When this 1 drop is dissolved in 30 mL of H2O

What should the pH be when we add 1 drop of 0.1M HNO3 to 30mL of deionized water ?

Why are pH buffers important?

Why are pH buffers important?

Life on Earth is water-basedHuman 48-75% waterPlants as high as 95% waterH+ and OH- are chemically and structurally reactive in cellsThe functioning of the cell is very pH dependentBlood plasma pH = 7.4pH < 6.9 fatal (acidosis)pH > 7.9 fatal (alkalosis)Fish die if the pH of the water goes below 5 or above 9

Life on Earth is water-basedHuman 48-75% waterPlants as high as 95% waterH+ and OH- are chemically and structurally reactive in cellsThe functioning of the cell is very pH dependentBlood plasma pH = 7.4pH < 6.9 fatal (acidosis)pH > 7.9 fatal (alkalosis)Fish die if the pH of the water goes below 5 or above 9

5

Making an Acid BufferMaking an Acid Buffer

6

How Acid Buffers Work

HA(aq) + H2O(l) A−(aq) + H3O+

(aq)

How Acid Buffers Work


(aq)

buffers work by applying Le Châtelier’s Principle to weak acid equilibrium

buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize it

the buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more HA and keep the H3O+ constant

buffers work by applying Le Châtelier’s Principle to weak acid equilibrium

buffer solutions contain significant amounts of the weak acid molecules, HA – these molecules react with added base to neutralize it

the buffer solutions also contain significant amounts of the conjugate base anion, A− - these ions combine with added acid to make more HA and keep the H3O+ constant

7

H2O

How Buffers WorkHow Buffers Work

HA + H3O+A−A−

AddedH3O+

newHA

HA

8

H2O

HA

How Buffers WorkHow Buffers Work

HA + H3O+

A−

AddedHO−

newA−

A−

9

Common Ion Effect


(aq)

Common Ion Effect


(aq)

adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left

this causes the pH to be higher than the pH of the acid solution lowering the H3O+ ion concentration

adding a salt containing the anion, NaA, that is the conjugate base of the acid (the common ion) shifts the position of equilibrium to the left

this causes the pH to be higher than the pH of the acid solution lowering the H3O+ ion concentration

10

Common Ion EffectCommon Ion Effect

11

Write the reaction for the acid with water

Construct an ICE table for the reaction

Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HC2H3O2 + H2O C2H3O2- + H3O+

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change

equilibrium

What is the pH of a buffer that is 0.100 M HC2H3O2 and 0.100 M NaC2H3O2 given Ka for

HC2H3O2 = 1.8 x 10-5


HC2H3O2 = 1.8 x 10-5

12

[HA] [A-] [H3O+]

initial 0.100 0.100 0

change

equilibrium

represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+x-x

0.100 -x 0.100 + x x


HC2H3O2 = 1.8 x 10-5


HC2H3O2 = 1.8 x 10-5

HC2H3O2 + H2O C2H3O2- + H3O+

13

determine the value of Ka

since Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.100 0.100 x0.100 -x 0.100 +x


HC2H3O2 = 1.8 x 10-5


HC2H3O2 = 1.8 x 10-5

14

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

the approximation is valid

x = 1.8 x 10-5

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.100 0.100 x


HC2H3O2 = 1.8 x 10-5


HC2H3O2 = 1.8 x 10-5

Tro, Chemistry: A Molecular Approach

15

x = 1.8 x 10-5

substitute x into the equilibrium concentration definitions and solve

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.100 0.100 1.8E-50.100 + x x 0.100 -x


HC2H3O2 = 1.8 x 10-5


HC2H3O2 = 1.8 x 10-5

16

substitute [H3O+] into the formula for pH and

solve

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x

equilibrium 0.100 0.100 1.8E-5


HC2H3O2 = 1.8 x 10-5


HC2H3O2 = 1.8 x 10-5

17

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

the values match

[HA] [A-] [H3O+]

initial 0.100 0.100 ≈ 0

change -x +x +x



HC2H3O2 = 1.8 x 10-5


HC2H3O2 = 1.8 x 10-5


18

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


19

Practice - What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and

0.071 M KF?


0.071 M KF?

Write the reaction for the acid with water

Construct an ICE table for the reaction

Enter the initial concentrations – assuming the [H3O+] from water is ≈ 0

HF + H2O F- + H3O+

[HA] [A-] [H3O+]

initial 0.14 0.071 ≈ 0

change

equilibrium

20

[HA] [A-] [H3O+]

initial 0.14 0.071 0

change

equilibrium



represent the change in the concentrations in terms of x

sum the columns to find the equilibrium concentrations in terms of x

substitute into the equilibrium constant expression

+x+x-x

0.14 -x 0.071 + x x

HF + H2O F- + H3O+

21

determine the value of Ka

since 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 +x

equilibrium 0.012 0.100 x0.14 x



0.071 +x

Ka for HF = 7.0 x 10-4

22

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

KF?


KF?

Ka for HF = 7.0 x 10-4

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

the approximation is valid

x = 1.4 x 10-3

[HA] [A2-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x

equilibrium 0.14 0.071 x


23


0.071 M KF?


0.071 M KF?

x = 1.4 x 10-3

substitute x into the equilibrium concentration definitions and solve

[HA] [A2-] [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

24


KF?


KF?

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

[HA] [A-] [H3O+]

initial 0.14 0.071 ≈ 0

change -x +x +x


25



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


Ka for HF = 7.0 x 10-4

26

Henderson-Hasselbalch EquationHenderson-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 Ka and initial concentrations of the weak acid and salt of the conjugate base as long as the “x is small” approximation is valid

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 Ka and initial concentrations of the weak acid and salt of the conjugate base as long as the “x is small” approximation is valid

27

Deriving the Henderson-Hasselbalch Equation

Deriving the Henderson-Hasselbalch Equation

28

What is the pH of a buffer that is 0.050 M HC7H5O2 and 0.150 M NaC7H5O2 where

Ka=6.5x10-5 for HC7H5O2?



29





Assume the [HA] and [A-] equilibrium concentrations are the same as the initial

Substitute into the Henderson-Hasselbalch Equation

Check the “x is small” approximation

HC7H5O2 + H2O C7H5O2- + H3O+





HC7H5O2 + H2O C7H5O2- + H3O+

[HA] [A-] [H3O+]

initial 0.050 0.150 ≈ 0

change -x +x +x

equilibrium 0.050-x 0.150-x x

31

Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?

Do I Use the Full Equilibrium Analysis or the Henderson-Hasselbalch Equation?

It is not clear that the Henderson-Hasselbalch Equation is an approximation, but the way we use it, (by placing the initial concentration of [HA]o and [A-]o into the equation assuming they are equilibrium values), makes it an approximation

For this reason 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:

§ the initial concentrations of acid and salt are not very dilute

§ the Ka is fairly small

for most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of Ka

It is not clear that the Henderson-Hasselbalch Equation is an approximation, but the way we use it, (by placing the initial concentration of [HA]o and [A-]o into the equation assuming they are equilibrium values), makes it an approximation

For this reason 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:

§ the initial concentrations of acid and salt are not very dilute

§ the Ka is fairly small

for most problems, this means that the initial acid and salt concentrations should be over 1000x larger than the value of Ka

chapter 16 aqueous ionic equilibrium. 2 buffers buffers are solutions that resist changes in ph...

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