Honors ChemistryName______

Acid/Base Equilibria

16.1 Acids and Bases: A Brief Review

"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)

Acids and bases are extremely important in many everyday applications: our own bloodstream, our

environment, cleaning materials, and industry.

ACID-BASE THEORIES

  • Arrehnius definition

Acid - a substance that, when dissolved in water, increases the concentration of H+ ions

Base - substance that, when dissolved in water, increases the concentration of OH- ions

This theory has the limitations of being restricted to aqueous solutions.

  • Lewis Definition

Acid - accepts an electron pair

Base - donates an electron pair

This theory explains all traditional acids and bases plus a host of coordination compounds and is used

widely in organic chemistry. Uses coordinate covalent bonds.

  • Bronsted-Lowry Model

Acids: are proton donors HA + H2O  A- + H3O+

Bases: are proton acceptors B- + H2O  BH + OH-

This theory is better. This is the main theory that we will use for our acid/base discussion.

16.2 Bronsted-Lowry Acids and Bases

  • Hydronium Ion: an ion form when an acid donate H+ ion combines with a H2O molecule to form a H3O+

-essentially has the same function as a H+ion, but H3O+denotes that we are using the Brønsted-Lowry model.

  • Conjugate Acid-Base Pairs: A pair of compounds that differ by the presence of one H+ unit

Ex) H2O + HCl → H3O+ + Cl-

base acid conjugate acidconjugate base

→

Ex) HNO3+ H2O → H3O++ NO3−neutral compound as an acid

acidbase CA CB

NH4+ + H2O→ H3O++NH3cation as an acid

acidbase CA CB

H2PO4− + H2O → H3O++ HPO42− anion as an acid

Acid base CA CB

  • Amphiprotic or amphoteric - molecules or ions that can behave as EITHER acids or bases; water, anions of weak acids (look at the examples above—sometimes water was an acid, sometimes it acted as a base)

Relative Strength of Acids and Bases

  • The stronger the acid, the weaker its conjugate base.
  • The stronger the base, the weaker its conjugate acid.
  • Strong acids: Substances that dissociate completely into ions in water. Completely transfer all their protons.

Ex. Hydrochloric acid: HCl HCl + H2O → Cl- + H3O+

  • Weak acid: substance that does not dissociate completely in water. They exist in solution as a mixture of acid molecules and their constituent ions.

Ex. Acetic Acid (Vinegar) CH3COOH + H2O ⇌ CH3COO- + H3O+

  • Strong base: substance that dissociates completely increasing [OH-]

Ex. Sodium Hydroxide: NaOH NaOH → Na+ + OH-

  • Weak Base: substance that when dissolved remove protons from water and to form OH- ions

Ex. Ammonia: NH3 NH3 + H2O ⇌ NH4+ + OH-

  • In every acid base reaction, the position of the equilibrium favors the formation of the weaker acid.

16.3 AutoIonization of water

  • the process where one water molecule donates a proton to another to form hydronium and hydroxide ions
  • Since the water molecule is amphoteric, it may dissociate with itself to a slight extent.

H2O + H2O(l) ↔ H3O+(aq) + OH-(aq)

↔

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

Kw = [H3O+][ OH-] = 1.0 ×10−14*since Kw is so small, the equilibrium lies far to the left

[H3O+] = [ OH-] = 1.0 ×10−7 in neutral solution

  • water is amphoteric, it can act as an acid or a base

16.4 Acid Strength and the pH Scale

  • Acid Strength: determined by the relative concentration of H+ and measured using the pH scale.
  • Acidic, basic, and neutral solutions can be distinguished as shown below:

Type of Solution / pH / [H+] / Color of litmus
Acidic / < 7.00 / / pink
Neutral / = 7.00 / = / in between
Basic / > 7.00 / / blue
  • Equations:pH = -log[H+] , [H+] = 10-pH , pOH = -log[OH-], [OH-] = 10-pOH, pH + pOH = 14

Exercise – Complete the following table
pH / [H+] / pOH / [OH–] / Acidic, basic, or neutral?
(a) / 5.4 x 10–4
(b) / 7.8 x 10-10
(c) / 10.75
(d) / 5.00
Answers:
(a)pH = 3.27; pOH = 10.73; [OH–] = 1.85 x 10–11 = 1.9 x 10–11, acidic (since pH < 7).
(b)pH = 4.89, [H+] = 1.3 x 10–5, pOH = 9.11, acidic (since pH < 7).
(c)[H+] = 1.8 x 10-11, pOH = 3.25, [OH–] = 5.6 x 10–4, basic (since pH > 7).
(d)pH = 9.00, [H+] = 1.0 x 10–9, [OH–] = 1.0 x 10–5, basic (since pH > 7).

16.5 Strong acids and Bases

  • Know the seven strong acids

1)HCl - hydrochloric acid

2)HBr - hydrobromic acid

3)HI - hydroiodic acid

4)HNO3 - nitric acid

5)HClO3 - chloric acid

6)HClO4 - perchloric acid

7)H2SO4 - sulfuric acid

  • By definition, strong acids and bases are 100% ionized in water solution.
  • Ionization of a strong acid gives rise to H+ ions, and ionization of a strong base produces OH– ions.
  • The equilibrium constant for a strong acid or strong base is undefined, since the reaction the ionization is complete. There is no equilibrium!
  • The [H+] for a strong acid (or the [OH–] for a strong base) is determined completely by the stoichiometry of the reaction

16.6 Weak Acids

  • Weak acids only partially ionized in aqueous solution.
  • Equilibrium constant is used to express the extent of ionization for a weak acid.

HA (aq) + H2O(l) ↔ H3O+(aq) + A-(aq)

orHA (aq) ↔ H+(aq) + A-(aq)

pKa = -logKa, pKb = -logKb↑Ka = ↓pKa = ↑acid strength

  • The subscript a on Ka denotes that it is an equilibrium constant for the ionization of an acid, and Ka is called the acid-dissociation constant.
  • The magnitude of Ka indicates the tendency of the hydrogen atom to ionize: The larger the value of Ka, the stronger is the acid.

  • Percent Ionization – one wat to measure acid strength.

Using Ka to Calculate pH

  1. write the balanced equation
  2. setting up the equilibrium expression (Ka),
  3. Set up RICE Box defining initial concentrations, changes, and final concentrations in terms of x, substituting values and variables into the Ka expression and solving for x.

Ex) Calculate (a) the pH and (b) the percent ionization of a 0.250 M HC2H3O2 solution. Ka(HC2H3O2)=1.8x10-5.

Reaction / HC2H3O2 ↔ H+ + C2H3O2–
Initial Concentration (M) / 0.250 / 0 / 0
Change (M) / -x / +x / +x
Equilibrium Concentration (M) / 0.250 - x / x / x

(a)

x2 = 4.5 x 10-6

x = 2.12 x 10-3 = [H+].

pH = -log[H+] = -log(2.12 x 10-3) = 2.67.

(b)

Calculating Ka from pH

When given the pH value of a solution, solving for Ka requires the following steps:

1)Write balance equation

2)Set up an RICE box and write the equilibrium expression

3)Solve for the concentration of H3O+ (denoted x) using the equation for pH: [H3O+ ]=10−pH

4)Use the concentration of H3O+ to solve for the concentrations of the other products and reactants.

5)Plug all concentrations into the equation for Ka and solve.

Ex) Calculate the Ka value of a 0.2 M aqueous solution of propionic acid, CH3CH2CO2H, with a pH of 4.88.

CH3CH2CO2H + H2O ⇋ H3O+ + CH3CH2CO− 2

R / HA ↔ H+ + A–
I / 0.2 / 0 / 0
C / -x / +x / +x
E / 0.2 - x / x / x

x=[H+] pH=−log[H+] [H+] = 10-pH

[H+] = 10−4.88 = 1.32×10−5 = x

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.

Example: Consider the dissociation of phosphoric acid.

H3PO4(aq) + H2O(l) ⇌ H3O+(aq) + H2PO4−(aq) Ka1 = 7.5 × 10-3

H2PO4-(aq) + H2O(l)⇌ H3O+(aq) + HPO42−(aq) Ka2 = 6.2 × 10-8

HPO42-(aq) + H2O(l) ⇌ H3O+(aq) + PO43−(aq) Ka3 = 4.8 × 10-13

  • Most of the H+ in the solution comes from the first ionization reaction

As long as successive Ka values differ by a factor of 103 or more, it is possible to obtain a satisfactory estimate of the pH of polyprotic acid solutions by treating them as if they were monoprotic acids, considering only Ka1

16.7 Weak Bases

  • Weak base reacts with water, forming the conjugate acid and OH- ions

B(aq) + H2O(l) ⇌ HB+(aq) + OH-(aq)

  • The constant Kb always refers to the equilibrium in which a base reacts with H2O to form the corresponding conjugate acid and OH-
  • Weak bases fall into two categories

1)Neutral substances that have an atom with a lone pair that can serve as proton acceptor

2)Weak base that consists of anions of weak acids (basically, the conjugate base)

Calculating pH or pOH from Kb

Ex)Calculate the pH of a 0.600 M solution of methylamine CH3NH2. Kb = 4.4 x 10–4.

R / CH3NH2(aq) + H2O(l) ⇌ CH3NH3+ (aq) + OH–(aq)
I / 0.600 / - / 0 / 0
C / -x / - / +x / +x
E / 0.600 - x / - / x / x

X = 1.62 x 10-2M = [OH–]

pOH = -log[OH-] = 1.79

pH + pOH = 14

pH = 12.21.

Calculating Kb from pH

Ex)The pH of a 0.10 M solution of a weak base is 9.67. What is the Kb of the base?

R / B (aq) + H2O(l) ⇌ BH+ (aq) + OH–(aq)
I / 0.10 / - / 0 / 0
C / -x / - / +x / +x
E / 0.10 - x / - / x / x

At equilibrium, [OH–] = [BH+] = x.

Use the pH to calculate the [OH–] at equilibrium (which is the value of x).

Here pOH = 14.00 – pH = 14.00 – 9.67 = 4.33.

Thus [OH-] = 10-pOH = 10-4.33 = 4.68x10-5 = x

16.8: Relationship between Ka and Kb

  • The product of the acid-dissociation constant for an acid and the base-dissociation constant for its conjugate base equals the ion-product constant for water

Ka x Kb = Kw = 1.0 x 10-14 (at 25oC)

pKa = -log(Ka) and pKb = -log Kb pKa + pKb = 14

16.9: Acid-Base Properties of Salt Solutions and Hydrolysis (splitting of water)

  • Salts are ionic compounds produced when an acid and base react. Salts are not always neutral.

HA + BOH ⇌ H2O + BA

(acid) (base) (water) (salt)

How to Determine if Salts Will Undergo Hydrolysis

(*You Need to be able to identify strong and weak acids and bases.*)

1)Neutral Salts: Salts resulting from the reaction of a strong acid with a strong base will not undergo hydrolysis and thus the solution will remain neutral.

Ex) Salt produced by the reaction of a Strong Acid with a Strong Base

HCl + NaOH→NaCl + H2O

Will not undergo hydrolysis (reaction with water)

NaCl+ H2O → NaCl + H2O

2)Acidic Salts: Salts resulting from the reactions of a strong acid with a weak base (the salt will have a cation from a weak base) will undergo hydrolysis and make the solution acidic.

Ex) Salt produced by the reaction of a Strong Acid with a Weak Base

HCl + NH3⇌NH4Cl

Will react with water to yield a conjugate base and a hydronium ion (makes solution acidic)

NH4Cl + H2O ⇌ NH3 + H3O+ + Cl-

3)Basic Salts: Salts resulting from the reactions of a weak acid with a strong base (the salt will have an anion from a weak acid)

Ex) Salt produced by the reaction of a Weak Acid with a Strong Base

HF + NaOH ⇌ NaF + H2O

Will react with water to yield a conjugate acid and a hydroxide ion (makes solution basic)

NaF + H2O ⇌ HF + OH- + Na+

Common Ions and Relative pH

Ions of Neutral Salts
Cations
Na+ / K+ / Rb+ / Cs+
Mg2+ / Ca2+ / Sr2+ / Ba2+
Anions
Cl- / Br- / I-,
ClO4- / BrO4- / ClO3- / NO3-
Acidic Ions
NH4+ / Al3+ / Pb2+ / Sn2+
Transition metal ions
HSO4- / H2PO4-
Basic Ions
F- / C2H3O2- / NO2- / HCO3-
CN- / CO32- / S2- / SO42-
HPO42- / PO43-

Calculating pH of a salt solution

Ex) Calculate the pH of a 0.500 M solution of KCN. Ka for HCN is 5.8 x 10-10.

1)Write the equation for the dissolving process, and examine each ion formed to determine whether the salt is an acidic, basic, or neutral salt.

KCN(s) → K+(aq) + CN-(aq)K+ is a neutral ion and CN- is a basic ion. KCN is a basic salt.

2)Write the equation for the reaction of the ion with water and the related equilibrium expression.

CN-(aq) + H2O(l) ⇌ HCN(aq) + OH-(aq)

3)Use the given Ka for HCN to find the value of Kb for CN-.

(5.8 x 10-10)(Kb) = 1 x 10-14

Kb = 1.7 x 10-5

4)Make RICE box and solve for x.

R / CN- (aq) + H2O(l) ⇌ HCN (aq) + OH–(aq)
I / 0.500 / - / 0 / 0
C / -x / - / +x / +x
E / 0.500 - x / - / x / x

Ex) What would be the pH of a 0.20 M ammonium chloride (NH4Cl) solution? Kb ammonia = 1.8 x 10-5.

NH4Cl(s) → NH4+(aq) + Cl-(aq) NH4+ is an acidic ion and Cl- is a neutral ion; solution will be acidic.

R / NH4+ (aq) + H2O(l) ⇌ NH3 (aq) + H3O+(aq)
I / 0.20 / - / 0 / 0
C / -x / - / +x / +x
E / 0.20 - x / - / x / x

(Ka)(1.8 x 10-5) = 1 x 10-14

Ka = 5.6 x 10-10

Salts of weak acids and weak bases

  • A salt formed between a weak acid and a weak base can be neutral, acidic, or basic depending on the relative strengths of the acid and base.

If Ka(cation) > Kb(anion) the solution of the salt is acidic.

If Ka(cation) = Kb(anion) the solution of the salt is neutral.

If Ka(cation) < Kb(anion) the solution of the salt is basic.

Ex) Using the Ka and Kb values below, determine if the following salts will be acidic, basic or neutral.

Ka(acetic acid) = 1.85x10-5

Ka(hydrogen cyanide) = 6.2x10-10

Ka(oxalic acid) = 5.6x10-2

Kb(NH3) = 1.8x10-5

(a)NH4CH3COO (ammonium acetate) –neutral, Ka = Kb

(b)NH4CN (ammonium cyanide) - basic Ka(HCN) < Kb(NH3)

(c)NH4HC2O4 (ammonium oxalate) – acidic, Ka(H2C2O4) > Kb(NH3)

16.10 Acid-Base Behavior and Chemical Structure

  • In general, the stronger the A–H or B–H+ bond, the less likely the bond is to break to form H+ ions and thus the less acidic the substance.

Binary Acids

•Period Trend: For elements in the same period, electronegativity is the dominant effect.

–The higher the electronegativity, the more it is capable of stabilizing a negative charge

•Group Trend: For elements in the same column, the dominant effect is size.

–The larger the atom the weaker the H-X bond and the more polarizable it is, therefore the greater the acidity

–Larger atoms can better stabilize a negative charge over a larger volume of space.

OxyAcids (HXOn)

  • The acidity of an oxyacid increases with the number of oxygen atoms: