ACIDS & BASES

Models of Acids & Bases

Arrhenius Acids produce H+ in aqueous solutions Bases produce OH- in aqueous solutions Limited approach

Brønsted-Lowry Acids are proton donors (H+) Bases are proton acceptors

H3O+ - Hydronium ion

Terminology Monoprotic – one acidic proton Diprotic – two acidic protons Triprotic – three acidic protons Oxyacids – acids in which the acidic

proton is attached to an oxygen atom

Organic acid – contains the mildly acidic carboxyl group Generally weak acids Equilibrium lies far to the left

Conjugate Pairs Conjugate base – what remains after an

acid has donated a proton Cl- is the conjugate base of HCl

Conjugate acid – what is formed when a base accepts a proton H3O+ is the conjugate acid of H2O

H2O + HCl H3O+ + Cl-

HCl is a stronger acid than H3O+ so equilibrium lies far to the right

General reaction : H2O + HA H3O+ + A-

Conjugate Pairs

Strong acids (almost) completely dissociate Produce weak conjugate bases

Weak acids (almost) completely stay together Produce strong conjugate bases

Strong bases Produce weak conjugate acids

Weak bases Produce strong conjugate acids

Acid Dissociation Constant (Ka) Ka is the equilibrium constant for the

reaction in which a proton is removed from the HA to form the conjugate base A-

General reaction : H2O + HA H3O+ + A- OR HA H+ + A-

Ka = [H3O+][A-] = [H+][A-] [HA] [HA]

Water is not included In dilute solutions the [H2O] is high and changes

very little

Practice

Write the dissociation reaction for the following acids

Acetic acid HC2H3O2

Hydrofluoric acid HF

Acid Strength

Strong Acids

Acids for which the equilibrium lies far to the right Almost completely dissociated Yield weak conjugate bases

Common strong acids Strong six (or seven) HCl, H2SO4, HBr, HI, HClO4, HNO3, (HClO3)

Weak Acids

Equilibrium lies far to the left Mostly together, few ions Yield relatively strong conjugate bases

Examples – acetic (HC2H3O2) oxalic (H2C2O4)

Using Ka and pKa

• pKa = -log(Ka)

• Can also be used to identify acid strength

• A strong acid has a large Ka value

• Small pKa

• A weak acid has a small Ka value

• Large pKa

Water is Amphoteric

Water can act as both an acid and a base

Autoionization of water H2O + H2O H3O+ + OH-

Ion-product constant Kw

At 25°C [H+] = [OH-] = 1.0 x 10-7

Kw = [H+] [OH-]

Kw = (1.0 x 10-7) (1.0 x 10-7) = (1.0 x 10-

14)

Solution Characteristics

Neutral solution [H+] = [OH-]

Acidic solution [H+] > [OH-]

Basic solution [H+] < [OH-]

Basic – pH > 7 Neutral – pH = 7Acidic – pH < 7

pH Scale

pH and pOH

pH = -log[H+] The number of decimal places in the

log is equal to the number of significant figures in the original number [H+] = 1.0 x 10-7

pH = 7.00 Keep the two decimal places

pOH = -log[OH-] pH + pOH = 14

Calculating the pH of Strong Acid Solutions Determine what species are present Determine which components are

significant Those present in large amounts

1.0 M solution HCl Solution components – H+, Cl-, H2O (not

HCl) Major species – H+, Cl-, H2O

OH- is present in only small amounts - ignored

Calculate the pH of 1.0 M HCl solution Consider the H+ from HCl

H+ from H2O is negligible The H+ from HCl will drive the

autoionization back according to LeChatelier’s Principle

pH =-log [H+] pH = -log(1.0) pH = 0

Calculating the pH of Weak Acid Solns1. List the major species in the solution

2. Choose the species that can produce H+ and write the balanced equations for the reactions producing H+

3. Using the values of the equilibrium constants for the reactions you have written decide which equilibrium will dominate in producing H+

4. Write the equilibrium expression for the dominant equilibrium

5. List the initial concentrations of the species participating in the dominant equilibrium

6. Define the change needed to achieve equilibrium; that is define x

7. Write the equilibrium concentrations in terms of x

8. Substitute the equilibrium concentrations into the equilibrium expression

9. Solve for x the “easy” way (by assuming [HA]0 – x ≈ [HA]0)

10. Use the 5% rule to verify whether the approximation is valid

11. Calculate [H+] and pH

Example

Calculate the pH of 1.0 M solution of HF. Ka = 7.2 x 10-4

Percent Dissociation

Percent dissociation = amount dissociated x 100% initial concentration

Percent dissociation = percent ionization Specifies how much of the weak acid has

dissociated For a given weak acid, the percent dissociation

increases as the acid becomes more dilute More dilute (water) = more dissociation Percent dissociation of acetic acid is significantly

greater in 0.10 M solution than in a 1.00 M solution

Example

Calculate the percent dissociation of acetic acid (Ka = 1.8 x 10-5) in each of the following 1.00 M Solution 0.10 M Solution

Mixture of Weak Acids

In a mixture of weak acids, if one acid has a relatively higher Ka value, it will be the focus of the solution in order to calculate pH

Same process as before

Example

Calculate the pH of a solution that contains 1.00 M HCN (Ka = 6.2 x 10-

10) and 5.00 M HNO2 (Ka = 4.0 x 10-4). Also calculate the concentration of CN- in this solution at equilibrium.

Polyprotic Acids

Polyprotic Acids Dissociates in a stepwise manner

One proton at a time Most are weak acids

H2CO3 H+ + HCO3- Ka1 = 4.3 x

10-7

HCO3- H+ + CO3

2- Ka2 = 5.6 x 10-11

Conjugate base becomes the acid in the 2nd step

Ka1 > Ka2 > Ka3

Example – Phosphoric Acid

Calculate the pH of 5.0 M H3PO4solution, and equilibrium concentrations of H3PO4, H2PO4

-, HPO4

2-, PO43-

Ka1 = 7.5 x 10 -3

Ka2 = 6.2 x 10-8

Ka3 = 4.8 x 10-13

Only the first dissociation makes a significant contribution of H+ based on Ka values

Sulfuric Acid

Strong acid in its first dissociation Weak acid in its second

Second step cannot be ignored for dilute solutions

Calculate the pH of a 1.0 M H2SO4 solution.

Bases

Strong Bases Weak Bases

Group 1A metal hydroxides LiOH, NaOH, KOH,

RbOH, CsOH

Group 2A metal hydroxides Ca(OH)2, Ba(OH)2,

Sr(OH)2

Less soluble than 1A hydroxides Used as antacids

Ammonia (NH3) and other covalent bases

Bases don’t have to contain OH-

React with water to increase [OH-]

Proton acceptors Compounds with low

values of Kb

Bases

Reaction with waterB + H2O BH+ + OH-

Equilibrium constant = Kb

Kb = [BH+ ][OH-]

[B] Kb is the equilibrium constant for the

reaction of a base to form conjugate acid and OH-

Calculating the pH of Strong Bases Kw = [H+][OH-] = 1.0 x 10-14

If [OH-] is known, [H+] can be calculated From [H+], can calculate pH

Calculate the pH of 0.05 M NaOH

Calculation of pOH

pKw = 14 = pH + pOH pOH = 14-pH

Same process as a weak acid

Calculating pH of Weak Bases

Acid-Base Properties of Salts

Salts that Produce Neutral Solutions Salts that consist of the cations of strong

bases and the anions of strong acids have no effect on pH, ([H+]), when dissolved in water

Cations of strong bases Na+, K+ (Group 1A) No affinity for protons Can’t produce H+

Anions of strong acids Cl-, NO3

-

Have no affinity for protons

Solutions of NaCl, KCl, NaNO3, and KNO3 are neutral

Salts that Produce Basic Solutions For any salt whose cation has neutral

properties and whose anion is the conjugate base of a weak acid, the aqueous solution will be basic

Salts that Produce Basic SolutionsC2H3O2

-(aq) + H2O(l) HC2H3O2(aq) + OH-

(aq)

Acetic acid is a weak acid – acetate is the conjugate base

Significant affinity for protonsWill react with the best proton donor available

Water is the only source of protons

Produces hydroxide ions – basic solution

Salts as Weak Bases

For any weak acid and its conjugate base Ka x Kb = Kw

Calculate the pH of a 0.30 M NaF solution. Ka = 7.2 x 10-4

Solve for Kb

ICE Chart Solve for x Answer: pH = 8.31

Salts that Produce Acidic Solutions Salts in which the anion is not a base

and the cation is the conjugate acid of a weak base produce acidic solutions NH4Cl

Anion (Cl-) has no affinity for a proton Cation (NH4

+) behaves as a weak acid

Salts that Produce Acidic Solutions Calculate the pH of a 0.10 M NH4Cl

solution. Kb value for NH3 = 1.8 x 10-5

NH4+ and H2O can produce H+

Solve for Ka (NH4+)

ICE Chart Solve for x Find pH

Answer: pH

Salts that Produce Acidic Solutions Salt that contains a highly charged

metal ion AlCl3 Hydrated Al ion [Al(H2O)6

3+] is a weak acid

High metallic charge polarized the O-H bond in water Hydrogen in water becomes acidic

Typically – the higher the charge on the metal ion, the stronger the acidity of the hydrated ion

Salts that Produce Acidic Solutions Calculate the pH of 0.010 M AlCl3

solution Ka = 1.4 x 10-5 Al(H2O)6

3+

Solve like a typical weak acid

Answer: pH = 3.43

Solutions of Salts

Many salts - both ions can affect the pH of the solution

Predict whether the solution will be acidic, basic, or neutral by comparing Ka value of acidic ion and Kb of basic ion Ka > Kb – solution is acidic

Kb > Ka – solution is basic

Kb = Ka – solution is neutral

Example

Determine whether an aqueous solution will be acidic, basic, or neutral NH4C2H3O2

Ions in solution are NH4+ and C2H3O2

-

Ka = 5.6 x 10-10 NH4+

Kb = 5.6 x 10-10 C2H3O2-

Ka = Kb, solution is neutral

Effect of Structure on Acid-Base Properties Factors determining acid

characteristics of molecules with X-H bond Strength of bonds

Strong bonds are reluctant to break Polarity of bonds

High bond polarity tends to increase the acidity of hydrogen

Halogens – electronegativity decreases down a group HF has an unusually strong bond – weak acid

Effect of Structure

Molecules of form H-O-X If X has high electronegativity, the

hydrogen tends to be acidic If X has low electronegativity, the

compound tends to be basic -OH comes off

The more oxygen atoms around X, the more acidic the compound Oxygen pulls the electrons

Acid-Base Properties of Oxides

Acidic Oxides (acid anhydrides) Nonmetal oxides that react with water to form

acidic solutions SO3(g) + H2O(l) H2SO4(aq)

2 NO2(g) + H2O(l) HNO3(aq) + HNO2(aq)

Covalent bonds Basic Oxides (basic anhydrides)

Metallic oxides of Group 1A and 2A metals react with water to form basic solutions K2O(s) +H2O(l) 2 KOH(aq)

CaO(s) + H2O(l) Ca(OH)2(aq)

Ionic bonds

Lewis Acid-Base Model

Lewis acid – electron pair acceptor H+ can accept an electron pair

NH3 + H+ NH4+

Lewis base – electron pair donor OH- can donate an electron pair

OH- + H+ H2O