Unit 9 Student Notes Packet Page 1

Unit 9 – Acids & Bases "ACID"--Latin word acidus, meaning sour. (lemon) "ALKALI"--Arabic word for the ashes that come from burning certain plants; water solutions feel

slippery and taste bitter. (soap)

ACID-BASE THEORIES

I. Arrhenius definition Acid – ______________________________________________________ Base – ______________________________________________________

This theory is limited to substances with those “parts”. Ammonia is a MAJOR exception!

II. Bronsted-Lowry definition Acid – ______________________________________________________ Base – ______________________________________________________

This theory is better; it explains ammonia as a base! This is the main theory that we will use for our acid/base discussion.

III. Lewis definition Acid – ______________________________________________________ Base – ______________________________________________________

This terminology is not as widely used as it used to because it is too general. Instead we focus on coordination complexes, which are products of Lewis Acid-Base reactions.

THE BRONSTED-LOWRY CONCEPT OF ACIDS AND BASES Using this theory, you should be able to write weak acid/base dissociation equations and identify acid, base, conjugate acid and conjugate base.

conjugate acid-base pair – A pair of compounds that differ by the presence of one H+ unit Label the acid, base, conjugate acid, and conjugate base in the following reactions.

Acidic SolutionsNeutral compound as an acid: HNO3 + H2O H3O+ + NO3 −

Cation as an acid: NH4+ + H2O H3O+ + NH3

Anion as an acid: H2PO4− + H2O H3O+ + HPO4

2−

In each of the acid examples notice the formation of H3O+ – this species is named the _____________It lets you know that the solution is acidic!

Unit 9 Student Notes Packet Page 2

Basic Solutions: Label the acid, base, conjugate acid, and conjugate base in the following reactions.

Neutral compounds: NH3 + H2O NH4+ + OH−

Anion as a base: CO32- + H2O HCO3

- + OH-

Anion as a base: PO43- + H2O OH- + HPO4

2−

Notice the formation of OH− in each of the alkaline examples. This species is named the ___________________. It lets you know that the resulting solution is basic!

Amphoteric --molecules or ions that can behave as EITHER acids or bases; water, anions of weak acids.

Notice how water could act as either the acid or the base in our previous examples.

Sometimes water was an acid [donated H+], sometimes it acted as a base [accepted H+]

For more information on how water can act as an acid OR a base, watch the following video:

https://www.youtube.com/watch?v=HdmCagtasYg

pH SCALEUse the following video to help fill in the notes about the pH scale: https://www.youtube.com/watch?v=pFK16GsU1e4

0 7 14

Why does [H+] = [OH-] at pH 7?

Unit 9 Student Notes Packet Page 3

pH and pOH pH [H+] (M) [OH-] (M) pOH

7

6

3

12

• What is the pattern between pH and [H+]? • What is the relationship between [H+] and [OH-]?

• How is pH related to pOH?

CALCULATING pHhttps://www.youtube.com/watch?v=aw-Y1d2IIJcAdd notes to this page using the video link above. Include examples given in the video.

The pH of a solution is 3.75, what is the Hydrogen ion concentration, [H+]?

Examples:1. What is the pH of a solution with a hydrogen ion concentration of 2.0 x 10 -11?

2. What is the pOH of a solution with a hydroxide ion concentration of 1.8 x 10 -4?

3. The pH of human blood was measured to be 7.41 at 25°C. Calculate the pOH, [H+], and [OH-] for the sample.

CONCENTRATED VS STRONG Concentrated –

pH = – log[H+] pOH = – log[OH–]

Answers: 1) 10.70; 2) 3.74; 3) pOH = 6.59, [H+] = 3.9 x 10-8 M, [OH-] = 2.6 x 10-7 M

Unit 9 Student Notes Packet Page 4

Strong –

Describe how to make a concentrated solution

Describe how to make a strong solution

ACID STRENGTH A strong acid completely __________________ in solution A weak acid only partially ionizes. The strength of an acid depends on the equilibrium:

HA(aq) + H2O(l) ⇄ H3O+(aq) + A– (aq)

If the equilibrium lies far to the left, the acid is weak (only a small percentage of the acid molecules ionize)

If the equilibrium lies far to the right, the acid is strong (completely ionizes)

STRONG ACIDSMemorize these 6 strong acids!

Hydrochloric acid Nitric acid

Hydrobromic acid Perchloric acid

Hydriodic acid Sulfuric acid

CALCULATING THE pH OF STRONG ACID SOLUTIONS

Unit 9 Student Notes Packet Page 5

Because a strong acid completely ionizes, or dissociates ions, [HA] = [H+] for the first hydrogen ion released by the acid.

Monoprotic acids only release one hydrogen ion. Polyprotic acids (di- and tri-protic) completely dissociate from their first hydrogen ion, but the second

hydrogen ion is much less likely to dissociate.

Examples:1. What is the pH of 1.0 M hydrochloric acid?

2. Calculate the pH of 0.10 M HNO3.

3. What is the pH of 1.0 x 10-10 M HCl?

CALCULATING THE pOH OF STRONG BASIC SOLUTIONS Strong bases (soluble in water): NaOH, KOH, LiOH, RbOH, CsOH Strong bases (not very soluble in water): Ca(OH)2, Ba(OH)2, Sr(OH)2 Strong bases are OH- combined with group I or II cations. (NEED TO KNOW!!) When a strong base completely ionizes, or dissociates ions, [MOH] = [OH-] for the hydroxide ion released

by the base. (Note: M is for metal)

Examples:1. What is the pOH of a 0.250 M solution of potassium hydroxide?

2. Calculate the pOH of 1.1 x 10-5 M Mg(OH)2?

3. What is the pH of a solution of KOH that has a concentration of 3.24 x 10-6 M KOH?

RELATIVE STRENGTHS OF ACIDS AND BASES The more readily an acid gives up a proton, the less readily its conjugate base accepts a proton.

Unit 9 Student Notes Packet Page 6

o In other words, the stronger the acid, the _____________ its conjugate base.

The more readily a base accepts a proton, the less readily its conjugate acid gives up a proton.o In other words, the stronger the base, the _____________ its conjugate acid.

WEAK ACIDS If it isn’t one of the 6 strong acids, it’s weak!

HF(aq) + H2O(l) ⇄ H3O+(aq) + F – (aq)

• An HF solution contains a large number of intact (or un-ionized) HF molecules.• It also contains some H3O+(aq) and F –(aq).

Describing concentrations:A 1.0 M HF solution has an [H3O+] that is much less than 1.0 M because

________________________ ______________________________________________________________________________

____.

Some weak acids

Hydrofluoric acid Sulfurous acid

Acetic acid Carbonic acid

Formic acid Phosphoric acid

Unit 9 Student Notes Packet Page 7

ACID IONIZATION CONSTANT (Ka) A way to quantify the relative strength of a weak acid The equilibrium constant of the weak acid

HA(aq) ⇄ H+(aq) + A–(aq)

The smaller the constant, the less the acid ionizes

BASE IONIZIATION CONSTANT (Kb) The equilibrium expression for bases is known as the Kb. For weak base reactions:

B (aq) + H2O ⇄ HB+ (aq) + OH−(aq)

Notice that Ka and Kb expressions look very similar. The difference is that a base produces the hydroxide ion in solution, while the acid produces the hydronium ion in solution.

Another note on this point: H+ and H3O+ are both equivalent terms here. Often water is left completely out of the equation since it does not appear in the equilibrium. This has become an accepted practice. (* However, water is very important in causing the acid to dissociate.)

AUTOIONIZATION OF WATER Use this video as further reference for the autoionization of water. https://www.youtube.com/watch?

v=enq6e81VvZA

Fredrich Kohlrausch, around 1900, found that no matter how pure water is, it still conducts a minute amount of

electric current. This proves that water self-ionizes.

• Since the water molecule is amphoteric, it may dissociate with itself to a slight extent.

• Only about 2 in a billion water molecules are ionized at any instant!

• The equilibrium expression used here is referred to as the autoionization constant for water,

Kw

• In pure water or dilute aqueous solutions, the concentration of water can be considered to be

a

constant, so we include that with the equilibrium constant and write the expression as:

Ka =

Kb =

Unit 9 Student Notes Packet Page 8

♦ Knowing this value allows us to calculate the OH− and H+ concentration for various situations.

♦ [OH− ] = [H+ ] solution is neutral (in pure water, each of these is 1.0 × 10−7 )

♦ [OH− ] > [H+ ] solution is basic

♦ [OH− ] < [H+ ] solution is acidic

It is important to recognize the meaning of Kw. In an aqueous solution at 25°C, no matter what it contains, [H+] x [OH-] = 1.0 x 10-14.

Examples:1. Calculate either the [H+] or [OH-] as required for each of the following solutions at 25°C, and state

whether the solution is neutral, acidic, or basic.

a. 1.0 x 10-5 M OH-

b. 1.0 x 10-7 M OH-

c. 10.0 M H+

2. At 10°C, the autoionization of water equals 2.9 x 10-15. Calculate [H+] and [OH-] as necessary under each of the following conditions:

a. Calculate [H+] if [OH-] = 9.3 x 10-4 M. Is the solution acidic or basic?

b. Calculate [H+] and [OH-] for a neutral solution.

c. Calculate [OH-] if [H+] = 6.7 x 10-11 M. Is the solution acidic or basic?

Kw =

Unit 9 Student Notes Packet Page 9

CALCULATING THE pH OF WEAK ACID SOLUTIONS(ICE box method)Steps

1. Choose the species in the solution that can produce H+, and write balanced equations for the reactions producing H+ (usually dissociation or ionization).

2. Write the equilibrium expression for the equation that dominates production of H+. 3. Write equilibrium concentrations of each reactant and product (row E from the ICE box). 4. Substitute equilibrium concentrations into the equilibrium expression, setting equal to Ka.5. Assume the change (x) is much smaller than the initial acid concentration, and assume equilibrium

concentration of acid is approximately equal to the initial concentration of acid.6. Calculate [H+] and pH.

Ex. 1 Calculate the pH of a 1.00 × 10−4 M solution of acetic acid. The Ka of acetic acid is 1.8 × 10−5 .

Ex. 2 The hypochlorite ion (OCl-) is a strong oxidizing agent found in household bleaches and disinfectants. It also has a relatively high affinity for protons (it is a much stronger base than Cl - for example) and forms the weakly acidic hypochlorous acid (HOCl, Ka = 3.5 x 10-8). Calculate the pH of a 0.100 M aqueous solution of hypochlorous acid.

Unit 9 Student Notes Packet Page 10

CALCULATING THE pH OF WEAK BASE SOLUTIONS The same steps are used to calculate weak bases and weak acids. (Make an ICE box) Remember, that are now calculating pOH and [OH-] instead of pH!

Ex. 3 Calculate the pH for a 15.0 M solution of NH3. (Kb = 1.8 x 10-5)

PERCENT IONIZATION

% ionization=¿¿¿

Examples:

1. What is the percent ionization in a 0.200 M solution nitrous acid if the [H3O+] is 0.0096 M?

2. If a 2.5 M solution of nitrous acid is 1.4% ionized, what is the hydronium concentration in the solution? What is the pH of the solution?

Finding the percent ionization of a weak acid

Unit 9 Student Notes Packet Page 11

1. Determine equilibrium concentrations for all species in the ionization reaction (Use ICE box if needed).2. Use Ka and the equilibrium constant expression to find [H+].3. Use [H+]eq and the initial acid concentration, [HA]0, to solve for percent ionization.

Find the percent ionization of a 2.5 M HNO2 solution. The Ka of nitrous acid is 4.6 x 10-4.

HNO2 ⇄ H+ + NO2-

[HNO2]eq = __________________

[H+]eq = _________

[NO2-]eq = _________

ACID-BASE PROPERTIES OF SALTS – SALT HYDROLYSIS Salts are ionic compounds (cation + anion) When dissolved in water, some salts are pH-neutral, some are acidic, and some are basic In order to determine acidity, take the salt and add water. The water will dissociate into the H+ and OH-; this process is hydrolysis.

Compare the ions separately.o Add H+ to the anion (to make its conjugate acid) and OH- to the cation (to make its conjugate

base).o Strong acids/bases completely dissociate.o The weak acid/base does NOT complete dissociate.

Ex. NaNO3

Determine whether the following salts will produce acidic, basic, or neutral solutions.1. KNO2

2. NH4Br3. CaCl2

Determining the pH of a salt solution The same equilibrium expression is used Add in the hydrolysis of water in order to complete the expression Remember: solve for Kb if basic and solve for ka for acidic solutions.

Example: Calculate the pH of a 1.0 M solution of sodium acetate. The ka for acetic acid is 1.8 x 10 -5.

Buffers

Unit 9 Student Notes Packet Page 12

A buffer is ___________________________________________________________________________ A solution can be buffered at an acidic or basic pH.

o Components of an acidic buffer: ____________________________________________________o Components of a basic buffer: _____________________________________________________

Example 1: Calculate the pH of a solution containing 0.20 M of HC2H3O2 and 0.10 M NaC2H3O2. (Ka = 1.8 x 10-5)

Next, calculate the pH of the solution after 0.20 mol of HCl is added into 1 L of the solution.

Example 2. Consider a solution containing 0.50 M HA (Ka = 1.0 x 10-6) and 0.50 M NaA.

A. Calculate the pH of this solution.

B. Calculate the pH after 0.10 mol of NaOH is added to 1 L of the solution.

Henderson-Hasselbalch Equation

pH = 6

pH = 6.17

Unit 9 Student Notes Packet Page 13

Write the equation for the dissociation of a weak acid, HA.

Deriving the H-H equation

Titrations

Standard Solution______________________________________________________________________

Analyte_______________________________________________________________________________

Equivalence point______________________________________________________________________

Endpoint______________________________________________________________________________

Strong Acid –Strong Base TitrationExample – Consider the titration of 100 mL of 1.00 M HCl with 0.500 M NaOH. Calculate the [H+] in the solution after 50.0 mL of 0.50 M NaOH has been added.

Titration curve (with class data)

Titration of a weak acid and strong base

Unit 9 Student Notes Packet Page 14

Example 1. 30 mL of 0.10 M NaOH solution is added to 50 mL of 0.10 M HF. Calculate the pH of the mixed solution. (Ka = 7.2 x 10-4)

Titration of a weak Base-Strong Acid

Unit 9 Student Notes Packet Page 15

Calculating pH of polyprotic acids

Acids with more than one ionizable hydrogen will ionize in steps. Each dissociation has its own Ka value. The first dissociation will be the greatest and subsequent dissociations will have much smaller

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

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

Except for H2SO4, polyprotic acids have Ka2 and Ka3 values so much weaker than their Ka1 value that the 2nd and 3rd (if applicable) dissociation can be ignored. The [H+ ] obtained from this 2nd and 3rd dissociation is negligible compared to the [H+ ] from the 1st dissociation. Because H2SO4 is a strong acid in its first dissociation and a weak acid in its second, we need to consider both if the concentration is more dilute than 1.0 M. The quadratic equation is needed to work this type of problem.

***Following are all of the types of titration curves you should be familiar with. Watch the video and take notes as you go along. The password for the video is: linuspauling