Part 2: Reactions & Problems involvingweak electrolytes

2

Review: Strong Electrolytes

Remember: Strong electrolytes ionize or dissociatecompletely.

____________ electrolytes only dissociate to a small extent. To solve problems that involve weak acids and weak bases, we will revisit Equilibrium.

Three classes of strong electrolytes

1. ____________ ____________

2. ____________ ____________

3. ____________ ____________

3

Review: Strong ElectrolytesCalculation of concentrations of ions in solution of

strong electrolytes is fairly easy

~ Calculate the concentrations of ions in 0.070 Mstrontium hydroxide, Sr(OH)2

Not as simplistic with weak acids and weak bases.

ACID-BASE THEORIES ___________________________________

acid--donates a hydrogen ion (H+) in water

base--donates a hydroxide ion in water (OH−) This theory iswas limited

___________________________________ acid--donates a proton in water

base--accepts a proton in water This theory is better; it explains ammonia as a base! This is the main

theory that we will use for our acid/base discussion.

___________________________________ acid--accepts an electron pair

base--donates an electron pair This theory explains most traditional acids and bases plus a host of

coordination compounds and is used widely in organic chemistry. Uses coordinate covalent bonds.

ACIDS DONATE ONLY ONE PROTON AT A TIME

____________—acids that can donate one H+

ex.

____________—acids that can donate two H+'s

ex.

____________—acids that can donate many H+'s

ex.

Acid Dissociation (Ionization, Protonation) ReactionsEx. 1) Write the simple dissociation (ionization) reaction

(omitting water) for each of the following acids.

a. Hydrochloric acid (HCl)

b. Acetic acid (CH3COOH or HC2H3O2)

c. The ammonium ion (NH4+)

d. The anilinium ion (C6H5NH3+)

7

The water concentration in dilute aqueous solutions is very high.

1 L of water contains 55.5 moles of water.

Thus in dilute aqueous solutions:

M5.55OH 2 • The water concentration is many orders of magnitude greater than the ion concentrations.• Thus the water concentration is essentially _________. Which is why we can omit water from our equilibrium expressions.

8

The Auto-Ionization of WaterPure water ionizes ________________________

less than one-millionth molar

Because the activity of pure water is 1, the equilibrium constant for this reaction is

H O + H O H O + OH2 2 3+ -

K H O OHc 3 +

9

Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7 M at 250C

This particular equilibrium constant is called the

ion-product for water, Kw so Kw = ____________

K H O OH

1.0 x 10 1.0 x 10

1.0 x10

c 3+

-7 -7

14

10

Remember: A convenient way to express ____________and ____________ is through pH. pH is defined as

If we know either [H3O+] or [OH-], then we can calculate

____________and vise versa.

[H3O+] = 10^-pH [OH-] = 10^-pOH

pH = -log H O3 +

The pH and pOH scales

14.00pOHpH

100.1OHOH 143

11

Ex. 2) Calculate the number of H3O+ and OH- ions in

one liter of pure water at 250C.

pH Scale Notice:

When [H+] = 10-3 pH = 3

When [H+] = 10-9 pH = 9

13

Develop familiarity with pH scale by looking at a series of solutions in which [H3O

+] varies between 1.0 Mand 1.0 x 10-14 M.

[H3O+] [OH

-] pH pOH

1.0 M 1.0 x 10-14

M 0.00 14.00

1.0 x 10-3

M 1.0 x 10-11

M 3.00 11.00

1.0 x 10-7

M 1.0 x 10-7

M 7.00 7.00

2.0 x 10-12

M 5.0 x 10-3

M 11.70 2.30

1.0 x 10-14

M 1.0 M 14.00 0.00

WEAK ACIDS AND BASES: The vast majority of acids/bases are weak. They are

considered weak b/c they do not ionize very much.

This means an equilibrium is established and it lies far to the left (reactant favored).

The equilibrium expression for acids is known as the Ka(the acid dissociation constant).

It is set up the same way as any other equilibrium expression.

15

Ionization Constants for Weak Monoprotic Acids and Bases

Let’s look at the dissolution of acetic acid, a weak acid, in water as an example.

The equation for the ionization of acetic acid is:

The equilibrium constant for this ionization is expressed as:

COOCHOHOH COOHCH -3323

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Values for several ionization constants

Other values are found on Table B.8

ACID FORMULA IONIZATION

CONSTANT

Acetic CH3COOH 1.8 x 10-5

Nitrous HNO2 5.1 x 10-4

Hydrofluoric HF 7.2 x 10-4

Hypochlorous HClO 2.9 x 10-8

Hydrocyanic HCN 6.0 x 10-10

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From the above table we see that the order of increasing acid strength for these weak acids is:

HF > HNO2 > CH3COOH >HClO >HCN

The order of increasing base strength of the anions (conjugate bases) of these acids is:

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Ex. 3) Write the equation for the ionization of the weak acid HCN and the expression, Ka for its ionization constant. Also label the conjugate acid/base pairs

Calculations involving Weak Acids

Use the RICE TABLE method you learned in general equilibrium

Set up an equilibrium reaction.

Start by writing the balanced equation

Define initial concentrations, changes, and final concentrations in terms of x

Set up the acid equilibrium expression (Ka) and solve for KaOR

Substitute values and variables into the Ka expression and solve for x.

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Calculation of Ionization Constants

Ex. 4) In 0.12 M solution, a weak monoprotic acid, HX, is 5.0% ionized.

a) Calculate the concentrations of all species in solution.

b) Calculate the ionization constant for the weak acid.

Since the weak acid is 5.0% ionized, it is also 95% unionized.

21

Ex. 5) The pH of a 0.100 M solution of a weak monoprotic acid, HA, is found to be 2.970. What is the value for its ionization constant?

pH = 2.970 so [H+]= 10-pH

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Calculations Based on Ionization ConstantsEx. 6) Calculate the concentrations of the various

species in 0.15 M acetic acid, CH3COOH, solution. Ka = 1.8 x 10

-5

Always write down the ionization reaction and the ionization constant expression.

5

3

-

33a

-

3323

108.1COOHCH

COOCHOHK

COOCHOH OHCOOHCH

23

Combine the basic chemical concepts with some algebra to solve the problem

Substitute the algebraic quantities into the ionization expression.

Solve the algebraic equation. If Ka is less than x 10-4

you can cancel out – x or + x , using the simplifying assumption (happy face rule)

Complete the algebra and solve for concentrations.

Reminder:

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Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does.

2a

4acbb

c b a

0107.2108.1

108.115.0

2

652

5

x

xx

X

xx

3-3

6255

101.6- and 106.1

12

107.214108.1108.1

x

x

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Ex. 7) Now calculate the percent ionization for the 0.15 M acetic acid. From Ex. 6, we know the concentration of CH3COOH that ionizes in this solution is 1.6 x 10

-3

M. The percent ionization of acetic acid is

% ionization = [CH3COOH] ionized x 100%[CH3COOH] original

26

Ex. 8) Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. Then find the % ionization. Ka=4.0 x 10

-10 for HCN

27

Let’s look at the percent ionization of two weak acids as a function of their ionization constants for Ex. 7 & 8

The [H+] in 0.15 M acetic acid is ____________ times greater than for 0.15 M HCN.

Solution Ka [H+] pH % ionization

0.15 M

CH3COOH

1.8 x 10-5 1.6 x 10-3 2.80 1.1

0.15 M

HCN

4.0 x 10-10

7.7 x 10-6

5.11 0.0051

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Weak bases work the same way as weak acids.

However, the x you solve for is the [OH−] and taking the negative log of x will give you the pOH and not the pH!

Ex. 9) Calculate the concentrations of the various species in 0.25 M aqueous ammonia and the percent ionization. Kb for ammonia = 1.8 x 10

-5

Calculations involving Weak Bases

29

Calculations Based on Ionization ConstantsEx. 10) The pH of an aqueous ammonia solution is 11.37.

Calculate the molarity (original concentration) of the aqueous ammonia solution.

Use the ionization expression and some algebra to get the equilibrium concentration.

Determination of the pH of a Mixture of Weak Acids Only the acid with the largest Ka value will contribute

an appreciable [H+]. Determine the pH based on this acid and ignore any others.

Determination of the pH of a Mixture of Weak Acids

Ex. 11) a. Calculate the pH of a solution that contains 5.00 M HCN (Ka = 6.2 × 10

−10) and 5.00 M HNO2 (Ka = 4.0 × 10−4).

b. Calculate the concentration of cyanide ion (CN−) in this solution at equilibrium.

Polyprotic Acids Many weak acids contain two or more acidic hydrogens.

Acids with more than one ionizable hydrogen Ionize in steps

Each dissociation has its own Ka value.

The first dissociation will be the greatest and subsequent dissociations will have much smaller Ka

As each H+ is removed, the remaining acid gets weaker and therefore has a smaller equilibrium constants.

As the negative charge on the acid increases it becomes more difficult to remove the positively charged proton.

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Polyprotic Acids

Consider arsenic acid, H3AsO4, which has three ionization constants

1 K1=2.5 x 10-4

2 K2=5.6 x 10-8

3 K3=3.0 x 10-13

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Polyprotic Acids The first ionization step is

35

Polyprotic Acids The second ionization step is

36

Polyprotic Acids The third ionization step is

37

Polyprotic Acids Notice that the ionization constants vary in the

following fashion:

This is a general relationship.

K K K1 2 3

38

Polyprotic AcidsEx. 12) Calculate the concentration of all species in

0.100 M arsenic acid, H3AsO4, solution.

1. Write the first ionization ionization step and represent the concentrations.

2. Substitute into the expression for K1.

3. Use the quadratic equation to solve for x, and obtain two values. Can’t use assumption. (too close for the 1st H+)

39

4. ionization and represent the concentrations.

5. Substitute into the second step ionization expression.

6. Now we repeat the procedure for the third ionization step.

7. Substitute into the third ionization expression.

8. Last Use Kw to calculate the [OH-] in the 0.100 M

H3AsO4 solution.

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A comparison of the various species in 0.100 M H3AsO4solution follows.

Species Concentration

H3AsO4 0.095 M

H+ 0.0049 M

H2AsO4- 0.0049 M

HAsO42- 5.6 x 10-8 M

AsO43- 3.4 x 10-18 M

OH- 2.0 x 10-12 M

Ka and pKa

pKa = - log Ka

The stronger the acid, the higher the Ka value and the lower the pKa value

Ka and pKaEx. 13) The pH of 0.10 M HClO solution was measured to

be 4.77 What is the value of the pKa for HClO?

43

Solvolysis ____________ is the reaction of a substance with the

solvent in which it is dissolved.

____________ refers to the reaction of a substance with water or its ions.

Combination of the anion of a weak acid with H3O+

ions from water to form nonionized weak acid molecules.

44

Hydrolysis

at 25oC

in ________solutions: [H3O+] = [OH-] = 1.0 x 10-7 M

in ________ solutions:[H3O+] < [OH-] > 1.0 x 10-7 M

in ________ solutions:[OH-] < [H3O+] > 1.0 x 10-7 M

for all conjugate acid/base pairs in aqueous solns.

Kw = Ka KbSo if we know the value of either Ka or Kb, the other

can be calculated.

45

Salts of Strong Soluble Bases and Weak Acids

A H O HA OH

K =HA OH

A

K

K

2

bW

a for HA

Note: This same method can be applied to the anion of any weak monoprotic acid.

46

Salts of Strong Soluble Bases and Weak AcidsEx. 14) Calculate the hydrolysis constants for the

following anions of weak acids.

a) F- , fluoride ion, the anion of hydrofluoric acid, HF. For HF Ka= 7.2 x 10

-4.

47

Salts of Strong Soluble Bases and Weak AcidsEx. 14) Calculate the hydrolysis constants for the

following anions of weak acids.

b) CN-, cyanide ion, the anion of hydrocyanic acid, HCN. For HCN, Ka= 4.0 x 10

-10.

48

Ex. 15) Calculate [OH-], pH and percent hydrolysis for the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO, solution. “Clorox”, “Purex”, etc., are 5% sodium hypochlorite solutions.

Na ClO Na ClO

0.10 0.10 0.10

~100%inH O2

M M M

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Salts of Acids and Bases Aqueous solutions of salts of strong acids

and strong soluble bases are ____________.

Aqueous solutions of salts of strong bases and weak acids are ____________.

Aqueous solutions of salts of weak bases and strong acids are ____________.

Aqueous solutions of salts of weak bases and weak acids can be ____________ ____________ ____________.

50

Rain water is slightly acidic because it absorbs carbon dioxide from the atmosphere as it falls from the clouds. (Acid rain is even more acidic because it absorbs acidic anhydride pollutants like NO2 and SO3 as it falls to earth.) If the pH of a stream is 6.5 and all of the acidity comes from CO2, how many CO2 molecules did a drop of rain having a diameter of 6.0 mm absorb in its fall to earth? Ka for H2 CO3 = 4.2 x 10

-7

A Fun Chemistry Problem for you