Arrhenius Definition

Arrhenius Definition

An acid is a substance which forms H+ ions as the only positive ion in aqueous solution.

An alkali is a substance which forms OH- ions as the only negative ion in aqueous solution.A base is an insoluble hydroxide.

The Brønsted-Lowry Concept of Acids and Bases

An acid is a proton donor.

A base is a proton acceptor.

Johannes Nicolaus Brønsted Thomas Martin Lowry

http://en.wikipedia.org/wiki/File:Johannes_Br%C3%B8nsted.jpg

An acid is an electron acceptor.

The Lewis Concept of Acids and Bases

A base is an electron donor.

Gilbert Newton Lewis (October 23, 1875 – March 23, 1946)

http://en.wikipedia.org/wiki/File:GilbertLewis.jpg

pH = - log [H+]

The pH of distilled water is 7, this is neutral. Any solution with a pH below 7 (i.e. pH 1.0 to pH 6.9) is an acid and any solution with a pH above 7 (i.e. pH 7.1 to pH 14) is an alkali.Acidic solutions have a pH between 1 and 6.9

Alkaline solutions have a pH between 7.1 and 14.

Neutral solutions are neither acidic nor alkaline so their pH is 7.

Acids all produce Hydrogen ions (H+). Acids like Hydrochloric acid produce lots of Hydrogen ions; this is because when Hydrogen Chloride gas dissolves in water the molecules of Hydrogen Chloride dissociate into Hydrogen ions and Chloride ions.

HCl(aq) = H+ + Cl- Vant Hoff factor =2

The important fact here is that every HCl that splits up produces 1 H+ so 2 moles of HCL will produce 2 moles of H+

Water also dissociates to produce ions, this time it is Hydrogen ions and Hydroxyl ions.H2O(l) = H+ + OH-

Sodium Hydroxide also dissociates to produce ions when it is dissolved in water, this time it is Sodium ions and Hydroxyl ions.NaOH(aq) = Na+ + OH-

In each case we can measure or calculate the concentration of Hydrogen ions present. We use the symbol [H+], we use square brackets to mean that it is the concentration of Hydrogen ions.

In a 1 molar HCl(aq) Hydrogen Chloride solution known as Hydrochloric acid[H+] = 0.01 = pH 2

In H2O(l) water [H+] = 0.0000001 = pH 7

In NaOH(aq) Sodium Hydroxide solution [H+] = 0.00000000000001 = pH 14

The Dissociation of Water

2H2O(l) H3O+(aq) + OH-(aq)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Strong acid weak acid Strong alkaliweak alkali

Crossword

pH + pOH must always = 14

Brønsted-LowryA base is a proton

acceptor.An acid is a proton

donor.

Arrhenius DefinitionAn acid is a substance which

forms H+ ions as the only positive ion in aqueous solution

Neutralization needs equal amounts of

H+ & OH-

Always makes water and a salt.

Using Titration allows you to work out the Concentration

of a an acid or alkaliProviding you know 3 of the

items belowM1V1 = M2V2

pH = -log [H3O+]

pH + pOH = 14

[H3O+] x [OH-]=1 x 10-14

Acids

Acid Alkali Salt Water

Neutralization

To makeNeutral

Neutral

Aqueous

Adding Acid to Alkali

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Oiy! come over here and I’ll rip your hydroxide

off!

You come here it’s just as far!

You acid breathed heathen!

Acid - A substance that has the potential to donate a proton (H+) or accept an electron pair. Base - A substance that can accept a proton, release OH-, or donate an electron pair.Arrhenius Model - Arrhenius proposed that acids are substances that produce protons, H+, in aqueous solution, whereas bases produce hydroxide ions, OH-, in aqueous solution. Bronsted-Lowry Definition - Bronsted and Lowry define an acid as a proton (H+) donor and a base as a proton acceptor. Compare this model with the Arrhenius Model and the Lewis definition. Lewis Definition - Lewis defined an acid as an electron pair acceptor and a base as an electron pair donor. Compare his model with the Arrhenius model and the Bronsted-Lowry definition.

pH + pOH always equals

1 14

I’m afraid you’re wrong this time

Try Again

pH + pOH always equals

1 14

pH + pOH always equals 14

“A base is a proton acceptor” is a definition by

Bronsted - Lowry Lewis


Try Again

0.001 M NaOH has a pH of

11 3


“A base is a proton acceptor” is a definition by


Try Again

0.001 M NaOH has a pH of 11


“A base is a proton acceptor” is a definition by Bronsted Lowry

The Proton in Water: Arrhenius TheoryAn acid as a substance that ionizes in water to give hydrogen ions, and a base as a substance that ionizes in water to give hydroxide ions.

Hydrochloric acid, HCl, is a strong acid, and is very soluble in water. It dissociates into its component ions in the following manner: HCl (g) H+ (aq) + Cl- (aq) The hydrogen ion interacts strongly with a lone pair of electrons on the oxygen of a water molecule. The resulting ion, H3O+ is called the hydronium ion.

Strong acids are so named because they react completely with water, leaving no undissociated molecules in solution.

Consequently, [H+] normally equals the original concentration of the acid x VHF

For example, in a 0.10 M aqueous solution of the strong acid, nitric acid,

[H+] = 0.10 = 0.1M pH = 1

HNO3 H+ NO3-

Dissociating

HNO3 H+ NO3-

DissociatingX

X

The Brønsted-Lowry Concept of Acids and Bases

An acid is a proton donor.

A base is a proton acceptor.

Johannes Nicolaus Brønsted Thomas Martin Lowry

http://en.wikipedia.org/wiki/File:Johannes_Br%C3%B8nsted.jpg

Conjugate Acid-Base PairsConjugate means having a common

derivation, linked.Let's look at the reaction of NH3 and H2O

again: NH3 + H2O NH4

+ + OH- The reverse of this reaction is:

NH4+ + OH- NH3 + H2O

In this case, NH4+ acts as an acid which

donates a proton to OH-.OH- acts as a base.

Vinegar is a weak acid

CH3COOH + H2O CH3COO-- + H3O+

Write the formula of each of the following

The Acid

The base

The conjugate acid

The conjugate base

HA + H2O

H3O+ + A-

If HA is a strong acid because it gives up its proton readily, then A- is a weak base because it has little affinity for the proton. If HA is a weak acid because it donates very few protons to the water, then A- has a high affinity for a proton, and A- is a stronger base than water.

H3O+ + A-

HA + H2O

Writing acid equilibrium equations

The Ka for any acid solution in equilibrium:

HX (aq) H+ (aq) + X- (aq)is defined as:

ACIDIC solutions are formed when an acid transfers a proton to water. The reaction of HCl with water can be written in either of the following ways:

HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq) HCl (aq) H+ (aq) + Cl- (aq)

acid-base reactions are written as HX (aq) X- (aq) + H+ (aq). and that others have been written as

HX (aq) + H2O (l) X- (aq) + H3O+ (aq)

The exclusion of pure solids and liquids from equilibrium constant expressions such as Ka is especially important here because it makes the equilibrium expressions for these two reactions the same, provided that H+ and H3O+ are considered to be the same. Thus, either form of the acid-base equilibrium equation can be used.

. Common Strong Acids and Bases

Acids Bases

HCl - hydrochloric acid VHF 2NaOH - sodium hydroxide HNO3 - nitric acid KOH - potassium hydroxide H2SO4 - sulfuric acid Ca(OH)2 - calcium hydroxide

Most substances that are acidic in water are actually weak acids. Because weak acids

dissociate only partially in aqueous solution, an equilibrium is formed between the acid and its ions. The ionization equilibrium is given by:

HA(aq) H+(aq) + A-(aq)

where X- is the conjugate base. For calculation purposes, the following is

assumed:

The equilibrium constant is then:

Ka= { [H+][A-] / [HA] }

Products/reactants

Calculating the pH for Solutions of Weak Acids

Calculate the pH of a 0.10 M solution of acetic acid. If the Ka = 1.8 x 10-5.

STEP 1: Write the ionization equilibrium for acetic acid: HC2H3O2 (aq) H+ (aq) + C2H3O2

- (aq)

STEP 2: Create an I.C.E. grid, and determine the concentration from information provided in the problem: HC2H3O2 (aq) H+ (aq) + C2H3O2

- (aq)

HC2H3O2 H+ C2H3O2-

INITIAL 0.10M 0 0

CHANGE -xM+xM

+xM

EQUILIBRIUM

(0.10 - x) M

xM xM

HC2H3O2 H+ C2H3O2- +

STEP 3: Substitute the equilibrium concentrations into the equilibrium constant expression: The equilibrium constant expression is: Ka = { [H+][C2H3O2

-] / [HC2H3O2] }

Ka = { (x)(x) / (0.10 - x) } = 1.8 x 10-5 This equation has only one unknown and can be solved using the quadratic formula. However, we can make things easier:

STEP 4: Remove the denominator x as it is very small compared to the initial conc of the acid The equation becomes: 1.8 x 10-5 = {(x)(x) / 0.10}1.8 x 10-5 = {X2/ 0.10}

0.1 x 1.8 x 10-5 = {X2/ 0.10} x 0.1

1.8 x 10-6 = X2

x = 1.3 x 10-3 M = [H+].

STEP 5: Now we find pH: pH = -log(1.3 x 10-3M)pH = 2.89

If you need to find the pH of a weak base its exactly the same process only you would calculate pOH and then subtract that value from 14 to find the pH

2.

Identify a conjugate acid-base pair in the reaction NH3 + H2O ‹–› NH4

+ + OH-

a. NH3 and OH-

b. H2O and OH-

c. H2O and NH4+

d. NH3 and H2O

3.

Compared to strong acids, weak acids produce __________ ions and conduct electricity __________ efficiently. a. fewer, more b. more, more c. fewer, less d. more, less

4. A 0.001M solution of HCl is ____________. a. a dilute acid solution b. neutral c. a weak acid solution d. a concentrated acid solution

Calculate the pH of a solution that has a [OH-] = 2.50 x 10-4M. a. -3.6 b. 10.4 c. 0.4 d. 3.6

Calculate the H+ ion and OH- ion concentrations in a 0.50M solution of HBr. a. [H+] = 0.50M and [OH-] = 2.0M b. [H+] = 1.0 x 10-7M and [OH-] = 1.0 ´ 10-7M c. [H+] = 0.50M and [OH-] = 0.50M d. [H+] = 0.50M and [OH-] = 2.0 x 10-14M

If ka for the reaction H2A 2H+ + A- is equal to 0.002. If 1 mole of H2A is dissolved into a 1 liter of water. What will be the expression for Ka at equilibrium?

If ka for the reaction H2A 2H+ + A- is equal to 0.002. If 1 mole of H2A is dissolved into a 1 liter of water. What will be the expression for Ka at equilibrium?

Buffers Key points

Buffers can be thought of as two reactions occurring simultaneouslyFirst is a weak acid or base in equilibrium with waterAnd second a fully dissociated salt of that acid or base Both of these produce the same conjugate ion (A-)

Buffers

• Buffers are made up of a weak acid or weak base mixed with a salt of that acid or base.

• Eg ethanoic acid and sodium ethanoate

Lets see how that works

HA A- H+

NaA A- Na+

The Salt completely dissociates Increasing the Conc of the conjugate base

HA H+ A--

HA A- H+

A-

At first this will reduce the amount of H+ so that the Equilibrium balance is restored

HA A- H+

So what happens to this balance when more acid is added

HA A- H+

HA A- H+

HA A- H+

So what happens to this balance when base is added

So the base reacts with the protons. Then the HA dissociates to re-establish the equilibrium

But the first reaction is in equilibrium and has a balance between HA and A-And the second once it happens is irreversible So this gives the following balance When H+ is added it reacts with A- and forms HA When OH is added it reacts with H+ and produced water

This means the H+ is used up so more HA dissociates to replace the H+ restoring the pHHow do we calculate the pH of a buffer solution?

How to calculate pH of buffer solution containing both acid and conjugate base?

Or

(note that due to sign change [A-] was moved to nominator).

This is so called Henderson-Hasselbalch equation (or buffer equation). It can be used for pH calculation of solution containing pair of acid and conjugate base - like HA/A-, HA-/A2- or B+/BOH. For solutions of weak bases sometimes it s more convenient to use equation in the formBoth equations are perfectly equivalent and interchangeable. Example Problem Applying the Henderson-Hasselbalch EquationCalculate the pH of a buffer solution made from 0.20 M HC2H3O2 and 0.050 M C2H3O2

- that has an acid dissociation constant for HC2H3O2 of 1.8 x 10-5. Solve this problem by plugging the values into the Henderson-Hasselbalch equation for a weak acid and its conjugate base. pH = pKa + log ([A-]/[HA]) pH = pKa + log ([C2H3O2

-] / [HC2H3O2]) pH = -log (1.8 x 10-5) + log (0.50 M / 0.20 M) pH = -log (1.8 x 10-5) + log (2.5) pH = 4.7 + 0.40 pH = 5.1

This is so called Henderson-Hasselbalch equation (or buffer equation). It can be used for pH calculation of solution containing pair of acid and conjugate base - like HA/A-, HA-/A2- or B+/BOH. For solutions of weak bases sometimes it s more convenient to use equation in the formBoth equations are perfectly equivalent and interchangeable.

Example Problem Applying the Henderson-Hasselbalch EquationCalculate the pH of a buffer solution made from 0.20 M HC2H3O2 and 0.50 M C2H3O2

- that has an acid dissociation constant for HC2H3O2 of 1.8 x 10-5.

pH = pKa + log ([A-]/[HA]) pH = pKa + log ([C2H3O2

-] / [HC2H3O2]) pH = -log (1.8 x 10-5) + log (0.50 M / 0.20 M) pH = -log (1.8 x 10-5) + log (2.5) pH = 4.7 + 0.40 pH = 5.1

or

or

Concentration of added salt ( conjugate base)

Concentration of weak acid (undissociated)

buffer capacity: the amount of an acid or base that can be added to

a volume of a buffer solution before its pH changes

significantly.

Titrations

One of the most common types of question that occur from the topic of acids and bases are titration questions.. This is because it combined your knowledge of acids with your skills in balancing equations, calculating concentrations and finding which aqueous ions will form. So today we will review titration calculations and improve all the skill required to solve themSo what tools will we need Concentration = moles/volume@ neutralization [H+] = [OH-]So Mole H+ = Moles OH-

Volume H+ Volume OH-

This is commonly written as M1/V1 = M2/V2 ne of the most common types of question that occur from the topic of acids

10.0 cm3 of a solution of potassium hydroxide was titrated with a 0.10 M solution of hydrochloric acid. 13.5 cm3 of the acid was required for neutralization. Calculate the concentration of the potassium hydroxide solution.Step 1Write down everything you know. Placing the one you know most about on the left (you don’t have to do this, but its just a good idea). Hydrochloric Potassium Acid hydroxideV = 13.5 cm3 V = 10.0 cm3

C = 0.10 M C = TBD

10.0 cm3 of a solution of potassium hydroxide was titrated with a 0.10 M solution of hydrochloric acid. 13.5 cm3 of the acid was required for neutralization. Calculate the concentration of the potassium hydroxide solution.

Step 2Write the equation or as much as you know about it.HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

This is necessary to check out the ratios of reactants – in this case 1:1 so we don’t have to worry about any ratios.

Step 3 make a table

Step 4Find all missing values

HCl KOH

R

n

C 0.10 ?

V 13.5 10.0

HCl KOH

R 1 1

n 0.00135 0.00135

C 0.10 ?

V 13.5 10.0

Step 5Work out the concentration of the potassium hydroxide.Concentration = 0.135 M

How does a mole hide from an owl?

It uses gRound CoVer


It uses gRound CoVerR n C V

H+ OH-

R n C V


It uses gRound CoVerR n C V

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Arrhenius Definition