Need to Know

What is a Net Ionic Reaction

simply shows the chemical species that undergo reaction while excluding spectator ions.

What is a spectator ion

Ionic species that show up on both sides of the reaction and do not participate in the chemical change.

Multiply charged species are generally

Insoluble

Sulfates are generally

Soluble

Nitrates (No3^-), chlorates (ClO3^-), and ammonium (NH4⁺) salts

Soluble

Alkali metals and halogens are generally

Soluble

Silver (Ag) salts

Insoluble

A measure of how much they dissolve and ionize is the

molar solubility and Ksp that results

Ksp is

an equilibrium constant that represents the equilibrium between a solid salt and its dissolved ions

Molar solubility is the

actual amount (concentration) of solute that can exist in a saturated solution at equilibrium

If Ksp is small (a large negative exponent)

the solubility of that compound is low

2 AB Ksp = [A][B]

Ksp = x²

3 AB₂ or A₂B Ksp = [A][B]²

Ksp = 4x³

4 AB₃ or A₃B Ksp = [A][B]³

Ksp = 27x⁴

5 A₂B₃ or A₃B₂ Ksp = [A]² [B]³

Ksp = 108x⁵

As a rule of thumb, to estimate the solubility in ranking salts

Take the root of the Ksp where the root is the number of ions in the salt

A soluble salt is added to an insoluble salt. They have an ion in common. The result is that, by Le Chatelier’s principle

The dissolution of the soluble salt shifts left, making it less soluble

Quick steps to solve:

1. Identify the common ion and the insoluble salt (IS).

• IS = the one given with a Ksp = Mg(OH)2

• Common ion shows up in both compounds = OH–

2. Write the Ksp expression for the insoluble salt.

• Ksp = [Mg2+][OH–]

2 = 1.2 x 10-11

3. Plug in the concentration of the common ion directly.

• Ksp = [Mg2+](0.1)2 = 1.2 x 10-11

4. Solve for the concentration of the other ion. This is its

new molar solubility.

• 1.2 x 10-11 = [Mg2+](0.1)2

[Mg2+] = 1.2 x 10-9

Arrhenius Theory

States that acids and bases are the dissociation products for water

Brønsted–Lowry Theory

We use this theory most often in CH 302. It is built around the proton, H donor and acceptor concept, defining acids as proton donors and bases as proton acceptors.

Lewis Theory

build around the unbonded electron pair, defining acids as electron pair acceptors and bases as electron pair donors.

Water dissociation is

Endothermic

From Le Chatelier’s Principle and the van’t Hoff equation

as T↑, Kw↑ and as T↓, Kw↓

100 C, pH =

6

25 C, pH =

7

0 C, pH =

8

pH =

-log[H+]

pOH =

-log[OH– ]

Kw =

[H+ ][OH– ] = 10^-14 at 25°C

pKa =

-log(Ka)

pKb =

-log(Kb)

Kw

KaKb = 10-14 at 25°C

pKw

pKa + pKb = 14 at 25°C

The larger the Ka (= smaller pKa)

the stronger the acid

The larger the Kb (= smaller pKb)

the stronger the base

Kw = KaKb, thus

the stronger the acid, the weaker the conjugate base (and vice versa)

HCl

Hydrochloric acid

HBr

Hydrobromic acid

HI

Hydroiodic acid

HNO3

Nitric acid

H2SO4

Sulfuric acid

HClO3

Chloric acid

HClO4

Perchloric acid

LiOH

Lithium hydroxide

NaOH

Sodium hydroxide

KOH

Potassium hydroxide

RbOH

Rubidium hydroxide

Ba(OH)2

Barium hydroxide

Sr(OH)2

Strontium hydroxide

HCl

SA

Hydrobromic acid

SA

HI

SA

Nitric acid

SA

H2SO4

SA

Chloric acid

SA

HClO4

SA

LiOH

SB

Sodium hydroxide

SB

KOH

SB

Rubidium hydroxide

SB

Ba(OH)2

SB

Strontium hydroxide

SB

CrO4 2–

Chromate

Cr2O7 –

Dichromate

CH3COO–

Acetate

HCO3 –

Bicarbonate

BrO4 –

Perbromate

BrO3 –

Bromate

BrO2 –

Bromite

BrO–

Hypobromite

PO3 3–

Phosphite

O2 2–

Peroxide

IO4 –

Periodate

IO3 –

Iodate

IO2 –

Iodite

IO–

Hypoiodite

Simple A/B Calculation Steps

Step 1: Remove spectator ions.

Step 2: Identify compound as strong or weak acid or

base and its form: H+, OH–, HA, A–, BH+, or B.

Step 3: There is no neutralization happening in the

“simple” case, so just solve the appropriate calculation.

Weak Acid Calculation

[H+ ] = (KaCa)^1/2

Weak Base

[OH– ] = (KbCb)^1/2

Buffer Equation for Acid and Conjugate Base (When in Equilibrium)

[H+ ] = Ka(𝐶𝑎/𝐶𝑏)

Buffer Equation for Base and Conjugate Acid (When in Equilibrium)

[OH– ] = Kb(𝐶𝑏/𝐶𝑎)

Buffers are

solutions that resist changes in pH when small amounts of acid or base are added. They consist of a weak acid and its conjugate base or a weak base and its conjugate acid.

Neutralization reactions are

chemical reactions between an acid and a base that produce salt and water, typically resulting in a solution with a pH close to 7.

Neutralization Calculation Steps

Step 1: Remove spectator ions.

Step 2: Identify compounds as strong or weak acid or base

and its form: H+, OH–, HA, A–, BH+, or B. For buffers,

recognize that the compounds are conjugates.

HA and A– or BH+ and B

e.g. HF and F– or NH4 + and NH3

Step 3: Do you neutralize? Yes if there is an acid + base and

at least one is strong.

Step 4: Set up RICE table, converting to moles, and solve

the limiting reagent problem.

Step 5: Identify what type of solution remains and solve

the appropriate calculation.

Step 6: Convert to pH, pOH, [H+], or [OH–] as requested.

Buffer Capacity

how much of a strong acid or base can be added to a buffer solution before the buffer is lost (i.e. before neutralizing all of either the weak acid or weak base in the buffer)

Buffer Capacity Steps

Example 1: There are 3 moles HA and 4 moles A–.

What is the buffer capacity?

Answer: This means 3 moles of OH– would

perfectly neutralize the HA, or 4 moles H+ would perfectly neutralize the A–.

Example 2: A buffer is made with 100 mL of 0.1 M HCOOH and 200 mL of 0.1 M HCOONa. Is the buffer maintained if 0.015 moles

of HCl are added?

Answer: First, remove spectators and identify the species given.

HCOOH is a weak acid HA, and HCOO–

is its conjugate weak base, A–. HCl is a strong acid, H+. Convert to moles: 0.1 L x 0.1 M = 0.01 moles HA

0.2 L x 0.1 M = 0.02 moles A–

This buffer has capacity to add 0.01 moles OH–, to

completely neutralize the HA, and 0.02 moles H+, to

completely neutralize the A–. Therefore, the buffer is maintained if only 0.015 moles of H+

is added.

Analyte

The "unknown" solution for which you would like to know either the concentration or the equilibrium constant. This is the solution you start with

Titrant

The "known" solution which has a precise and accurate concentration, generally a strong acid or base. This is the solution you are adding in.

Equivalence point

is the point at which the number of moles of added base/acid is equal to the number of moles of acid/base in the analyte solution.

Strong with Strong Titration EQ Point =

pH 7

End Point

ideally around the equivalence point.

Buffer Region

Is the pH range where the pH of a solution remains static

Half-Equivalence Point

the point at which exactly half the number of moles of added base/acid is equal to the number of moles of acid/base in the analyte solution

pH = pKa at the

Half-Equivalence Point

Weak Acid with Strong Base

is a type of acid-base titration where a weak acid is titrated with a strong base, resulting in a pH change that is less dramatic compared to strong acid-strong base titrations

Weak Base with Strong Acid

is a type of acid-base titration where a weak base is titrated with a strong acid, leading to a gradual pH change and typically resulting in a more complex pH curve compared to strong acid-strong base titrations

Polyprotic Acid with Strong

Base is a type of acid-base titration involving an acid that can donate more than one proton per molecule, resulting in multiple equivalence points during titration.

Titration to Buffer Region

Example: What is the pH when 50 mL of 0.01 M hydroiodic acid is added to 100 mL of 0.1 M sodium formate? (For formic acid, Ka = 10-4 .)

Step 1: We are left with H+ from HI and A– from formate.

Step 2: H+ is strong acid, A– is weak base.

Step 3: Yes – neutralization occurs! H+ + A– → HA

Step 4: Converting to moles, we have 0.05 L x 0.01 M = 0.0005 mol H+ and

0.1 L x 0.1 M = 0.01 mol A– . The limiting reagent is H +. H+ (aq) + A– (aq) → HA (aq)

0.0005 0.01 0

-0.0005 -0.0005 +0.0005

0 0.0095 0.0005

Step 5: There are two species left in solution, A– and HA. They are conjugates,

so this is a buffer. The calculation is [H

+] = Ka(Ca/Cb). You can plug in and solve, or, on a multiple-choice test, realize that pH~pKa , and since there is more base (A–) compared to acid (HA), pH will be slightly greater than pKa.

Step 6: [H+] = 10⁻⁴ M(.0005/.0095) = 5.26 ∙ 10⁻⁴ M

pH = -log[H+] = -log(5.26∙10-4 M) = 5.28

Titration to Equivalence Point

Example: What is the pH when 80 mL of 0.125 M hydroiodic acid is added to 20 mL of 0.5 M sodium formate? (For formic acid, Ka = 10-4 .)

Step 1: We are left with H+ from HI and A– from formate.

Step 2: H + is strong acid, A– is weak base.

Step 3: Yes – neutralization occurs! H+ + A– → HA

Step 4: Converting to moles, we have 0.08 L x 0.125 M = 0.01 mol H+ and 0.02 L x 0.5 M = 0.01 mol A– . With equal moles, there is no limiting reagent, and this titration reaches the equivalence point. H + (aq) + A – (aq) → HA (aq)

0.01 0.01 0

-0.01 -0.01 +0.01

0 0 0.01

Step 5: There is only HA left in solution, so this is a weak acid solution where [H+ ] = (KaCa) 0.5 = (10-4 x 0.01 𝑚𝑜𝑙 0.1 𝐿 ) 0.5 = (10-4 x 10-1 ) 0.5 = 10-2.5 M

Step 6: pH = -log[H+ ] = -log(10-2.5) = 2.5

What makes a buffer

1) By mixing a weak acid and its conjugate base, or a weak

base and its conjugate acid.

2) By neutralizing a weak acid/base with a strong base/acid by

adding less strong than you have of weak.

Diprotic Acids (H₂A)

H2A ⇌ HA– + H+

𝐾𝑎1 = [𝐻𝐴 – ][𝐻+]/𝐻2𝐴

HA– ⇌ A 2– + H+

𝐾𝑎2 = [𝐴 2– ][𝐻+]/𝐻𝐴–