A c i d s a n d B a s e s C h 1 6 P a g e | Ch 16: Acids...

A c i d s a n d B a s e s C h 1 6 P a g e | 1

Ch 16: Acids and Bases

Homework:

Read Chapter 16 Work out sample/practice exercises in the sections,

Chapter problems: 39, 41, 49, 63, 67, 83, 91, 95, 99, 107, 111, 115, 117, 123, 139

Check for the MasteringChemistry.com assignment and complete before due date

Acids, Bases, Salts:

Acids, bases and salts are very important and perform many essential functions.

Digestive juices (0.1 M HCl; needed to kill bacteria, break down food and

activate enzymes)

pH buffers; Households and Industrial uses

Drain cleaner (NaOH)

fertilizer (NH4NO3)

Car battery acid (40% H2SO4)

Table salt as a food preservative or for flavor (NaCl)

Example 1: Come up with more common acids, bases, or salts and their uses.


A few common acids, their uses and relative strength:

A few common bases, their uses and relative strength:


Review Electrolytes and Double Displacement Reactions:

Electrolytes:

Nonelectrolyte: A molecule or substance that remains whole in aqueous

solutions, it cannot split apart. Examples: any insoluble solid, gas (CO2, O2,

SO2), and molecules (sugar, CH4, H2O)

Weak Electrolyte: An ionic substance that will partially ionize into its separate

ions in aqueous solution. Examples: Weak acids (HF, HC2H3O2) and Weak

bases (NH4OH, CH3NH2 ), and slightly soluble solids (PbCl2) . Partial ionization

is an equilibrium reaction in which the reactant is favored, K<1 ; HF (aq)

H+1 (aq) + F-1 (aq).

Strong Electrolyte: An ionic substance that completely dissociates into its ions

in aqueous solution. Examples: Strong Acids (HCl), Strong Bases (NaOH),

Soluble Salts (KBr); K>>1; HCl (aq) H+1 (aq) +Cl-1 (aq).

Titration: Reacting a solution of unknown concentration with a known (standard)

concentration, stopping the titration when an indicator (phenolphthalein) changes

color.

Example 2:

Write the dissociation reactions in the solvent water for the following

substances. Strong electrolytes will have (), weak electrolytes have (), and

nonelectrolytes are no reaction (NR).

FeCl3 (aq)

HNO3 (aq)

Sr(OH)2 (aq)

CH3OH (aq)

HF(aq)


Double Displacement Reactions:

Double Displacement reactions have two ionic reactants. Reactants will

exchange ions in making products… AB + CD AD + CB

A reaction occurs if a nonelectrolyte (Solid, Liquid (H2O), Gas), or Weak

Electrolyte is formed as one or more of the products. If all the reactants and the

products are strong electrolytes, then no reaction takes place. Review the

Solubility Rules.

a) Whole or Molecular equation

2 AgNO3 (aq) + CaCl2 (aq) Ca(NO3)2 (aq) + 2 AgCl (s)

b) Total Ionic Equation with Spectator Ions

2 Ag+1 (aq) + 2 NO3-1 (aq) + Ca+2 (aq) +2 Cl-1 (aq)

Ca+2 (aq) + 2 NO3-1 (aq) + 2 AgCl (s)

c) Net Ionic Reaction

Ag+1 (aq) + Cl-1 (aq) AgCl (s)

Example 3:

For the Double Displacement reactions, write the whole, total ionic (circling

the spectator ions) and net ionic equations given the reactants. Identify the type of

reaction as Precipitation, Neutralization, or No Reaction.

a) HCl (aq) + NaOH (aq)

b) NH4OH, same as NH3 (aq) + H2S (aq)

c) Al(NO3)3 (aq) + KCl (aq)

d) K2S (aq) + Zn(ClO3)2 (aq)


Review Acids, Bases and Salts :

Acids:

Properties:

Taste Sour

Reacts with “active” metals to liberate H2:

Zn (s) + 2 HCl(aq) H2(g)+ ZnCl2(aq)

Reacts with carbonates to liberate CO2:

CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

React with bases to form ionic salts (neutralize):

NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l)

Conduct electrical current

Certain dyes change color with acids (litmus-red)

Acids ionize in water to increase the H+1 ion concentration

Nomenclature:

a) Binary acids, those anions ending in ide: Hydro root ic acid;

(HCl), Hydrochloric Acid; (H2S), hydrosulfuric acid

b) Ternary oxyacids: if anion ends with ate, root ic acid; (HNO3),

nitric acid, if anion ends with ite, root ous acid, (HNO2), nitrous

acid

Bases (also known as alkalis):

Properties:

Taste bitter

Slimy to the touch

Conducts electricity

Certain dyes change color with bases (litmus-blue)

React with acids to form ionic salts (neutralize):

Ionize in water to increase the OH-1 ion concentration

Nomenclature: metal name + hydroxide; (NaOH), Sodium Hydroxide; or

common name (NH3), ammonia, organic bases with N are amines,

(CH3NH2), methyl amine

Salts:

The ionic substances not readily identified as an acid or base, some are very

soluble, others only slightly soluble in water.

Nomenclature: cation name + anion name; (CuCl2) copper (II) chloride,

(NH4)2SO4 , ammonium sulfate, (NaCl), sodium chloride.


Example 4: Fill in the Table with names or formulas.

Name Formula

Hydrosulfuric acid

Ethyl amine

Potassium fluoride

MgHCO3

H2C2O4

Ba(OH)2

Acids and Bases Defined :

Arrhenius Definition (1884): Most limited definition requires water.

Acid: Substance that will increase the H+1 ion concentration in an aqueous

solution. (HF)

Base: Substance that will increase the OH-1 ion concentration in an aqueous

solution. (KOH)

Neutralization is the combination of an acid with a base to form water and a

salt.

Bronsted Lowry Definitions (1923): broader definition, more base possibilities

Acid: Donates one H+1 ion. (HA), NH4+1

Base: Accepts one H+1 ion. (A-1), NH3

Conjugate Acid/Base Pairs: These are different by only a single H+1. The

acid has one more H+1 compared to the base in a conjugate pair. NH4+1 is

the conjugate acid for NH3 the conjugate base. HF is the conjugate acid

for F-1 the conjugate base.

Amphiprotic Substance: One that can both accept and donate H+1 and can be

either acid or base dependent on the environment. H2O can accept H+1

and become H3O+1 or donate H+1 and become OH-1. HCO3

-1 is also

amphiprotic.


Amphoteric Substance is another term used. Although all amphiprotic species

must be amphoteric, not all amphoteric substances are amphiprotic. Amphoteric

substances are able to react with an acid or a base. For example, the amphoteric

metal oxide, ZnO, contains no hydrogen and cannot donate a proton. ZnO acts as a

Lewis acid or Lewis base which accepts or donates electron pairs.

Amphoteric ZnO reacts with both acids and with bases:

In acid: ZnO + 2H+ → Zn2+ + H2O

In base: ZnO + H2O + 2OH- → [Zn(OH)4]2-

Example 5:

a) Write the formulas for the conjugate bases given the acids:

Acid NH4+1 HF HNO2 H2SO3 H2O

Conjugate

base

b) Write the formulas for the conjugate acids given the bases:

Base C2H3O2-1 (CH3)2NH ClO-1 SO3

-2 H2O

Conjugate

acid

c) Which of the species in (a) and (b) above are amphiprotic?

http://en.wikipedia.org/wiki/Lewis_acid


Lewis Definitions: This definition does not require water or aqueous reactions.

Acid: Accepts a share of a nonbonding electron pair. (BH3)

Base: Donates a share of a nonbonding electron pair. (NH3)

Generally results in a covalent bond forming, the product is called an adduct

Example 6:

Identify the Lewis acid and base for the reactants below.

a) (CH3)3N + BF3 (CH3)3NBF3

b) FeBr3 + Br-1 FeBr4 -1

c) Zn+2 + 4 NH3 Zn(NH3)4+2

d) SO2 + H2O H2SO3


Structure of Acids:

Binary acids have the hydrogen attached to a nonmetal atom: HF

Oxy acids or ternary acids have the hydrogen attached to an oxygen atom:

Carboxylic acids have COOH group the hydrogen

attached to the COO group is acidic:

*ADD THE LONE PAIR ELECTRONS TO THE PICTURES OF

SULFURIC ACID, NITRIC ACID AND CARBOXYLIC ACID GROUP.

Structure of Bases:

Most ionic bases contain OH-1 ions

Some contain CO3-2 ions

Molecular bases contain structures that would like to add an H+1 ion,

mostly NH3 and amine groups.

Relative strengths:

A stronger acid will have a weaker conjugate base and vice versa. Strong

acids have a negligible conjugate base. Stronger bases have weaker conjugate

acids. Reactions always favor having a greater amount of the weaker acid

and base at equilibrium.


Example 7:

a) Which acid (HI or HF) has the weaker conjugate base?

b) Which base (C2H3O2-1 or OH-1) has a weaker conjugate acid?


Autoionization of Water:

Pure water will ionize just slightly; it is generally considered a nonelectrolyte since

the amount it ionizes is so small.

Kw = 1.0 x 10-14 at 25°C for the reaction…

H2O (l) + H2O (l) H3O+1 (aq) + OH-1 (aq)

The hydronium ion (H3O+1) is often written as a proton in water, H+1 (aq), even

though the H+1 is so reactive it cannot exist alone in water. H+1 is chemically

bonded to one or more water molecules in an aqueous solution connected by

hydrogen bonding. H(H2O)n+1

H2O (l) + H2O (l) H3O+1 (aq) + OH-1 (aq)

or

H2O (l) H+1 (aq) + OH-1 (aq);

Kw = [H+1][OH-1]

Temperature: As the temperature changes, so will Kw

0°C Kw = 1.1 x 10-15

25°C Kw = 1.0 x 10-14

37°C Kw = 2.5 x 10-14

60°C Kw = 9.6 x 10-14

pH Scale:

pH is a value that helps in determining the acidity of a solution. pH can be

determined using pH meter, pH paper, or indicators (intense colored organic dyes

that change color at different pH values).

pH = -log[H+1] [H+1] = 10-pH

Kw = 1.0 x 10-14 = [H+1][OH-1] at 25°C

pOH = -log[OH-1] [OH-1] = 10-pOH

pKw = -log[Kw] pKw = pH + pOH = 14 at 25°C

Acid has a pH < 7

Neutral has pH = 7

Base has a pH > 7

pX = -log[X], pKa = -logKa, pKb = -logKb; Ka = 10-Ka , Kb = 10-Kb


Significant Figures and Logs:

• When you take the log of a number written in scientific notation, the digit(s) before

the decimal point come from the exponent on 10, and the digits after the decimal

point come from the decimal part of the number

log(2.0 x 106) = log(106) + log(2.0)

= 6 + 0.30303… = 6.30303...

• Because the part of the scientific notation number that determines the significant

figures is the decimal part, the sig figs are the digits after the decimal point.

Example 8:

Given that the pH of a solution is 4.88, solve for the [H+1], [OH-1], and pOH using

appropriate significant digits.

Strong Acids and Bases:

Strong Acids: HCl, HBr, HI, HNO3, HClO4, HClO3, H2SO4

Strong Soluble Bases: LiOH, NaOH, KOH, RbOH, CsOH, FrOH, Ca(OH)2,

Sr(OH)2, Ba(OH)2, Ra(OH)2

Strong Acids and Strong Bases dissociate into ions nearly completely, so the

dissociation reaction equilibrium constant K is very, very large.


Example 9:

Calculate the concentration of hydrogen and hydroxide ions, and the pH and pOH.

a) 0.060 M HCl; [H+1] = [Acid] for monoprotic strong acids.

b) 0.012 M Ba(OH)2 ; [OH-1] = (number OH-1)x[Base] for strong soluble bases

Equilibrium Involving Weak Acids: Weak acids are much more common and numerous compared to strong acids.

Weak acids (WA or HA) only will partially ionize. We use the equilibrium

constant, Ka, in which Ka<1, the smaller the Ka, the weaker the acid. The book

appendix has Ka values.

The weak acid equilibrium reaction is generally of the form…

HA H+1 + A-1 or HA + H2O H3O +1 + A-1

Approximation in Calculations: When using RICE you may avoid the quadratic

equation if x is added or subtracted from a relatively large number compared to the

value of Ka. The addition or subtraction of the x can be considered negligible when

Ka is 1000 times smaller than the concentration. Approximation values can be used

if Ka is between 100 to 1000 times smaller than the concentrations.


Example 10:

In 0.120 M solution, a weak monoprotic acid (HA) is 5.00% ionized. Calculate

using the RICE equation the equilibrium concentrations, pH and Ka.

Example 11:

The pH of a 0.100 M solution of a monoprotic acid (HA) is 2.97. Using RICE,

calculate Ka.

Example 12:

Calculate the equilibrium concentrations and pH for the following solutions. (a)

0.150 M HC2H3O2 (b) 0.150 M HCN


Equilibrium Involving Weak Bases:

Weak soluble bases are generally going to be amines, ammonia, and conjugate base

ions of acids. Weak bases (B or A-1) need to have water as an additional reactant to

provide the H+1 and only will partially ionize. We use the equilibrium constant, Kb,

in which Kb<1, the smaller the Kb, the weaker the base. Appendix D has K values

The weak base equilibrium reaction is generally of the form…

B + H2O BH+1 + OH-1

Or A-1 + H2O HA + OH-1

every anion can potentially act as a base and accept an H+1.

Example 13:

Calculate the % ionization, pOH, pH and the equilibrium concentrations in 0.15 M

NH3 (aq). Kb = 1.8 x 10-5


Example 14:

The pH is 11.37 for a NH3 (aq) solution. Calculate the Molarity of NH3.

Polyprotic Acid Equilibria:

Ionization of polyprotic acids occur stepwise. Ka is different for each step,

decreasing as each H+1 is lost (Ka1 > Ka2 > Ka3).

Generally, the difference in Ka values is enough so that the second ionization

does not happen to a large extent… except for extremely dilute solutions, the

[H+1] can be assumed to come from the first step alone.

[A-2] = Ka2 as long as the second ionization is negligible.


Example 15:

Calculate the concentrations of all species in 0.25 M H2SO4 solution.

Given: Ka1 >>1 (assume complete ionization), Ka2 = 1.2 x 10-2

Example 16:

Calculate the concentrations of all species in 0.40 M H3AsO4 solution.

Given: Ka1 = 5.5 x 10-3, Ka2 = 1.7 x 10-7, Ka1 = 5.1 x 10-12


Relating Ka and Kb for Conjugate Acid/Base Pairs:

Ka HF H+1 + F-1 Ka = 3.5 x 10-4

Kb F-1 + H2O HF + OH-1 Kb = ?

Kw H2O H+1 + OH-1 Kw = 1.0 x 10-14

Kw = Ka x Kb

Example 17:

Demonstrate how Kw = Ka x Kb and calculate the Kb value for F-1

Salt Solutions (Acid-Base Properties): Salts are ionic compounds with cation and anion that may be conjugates of a

base and acid.

The reaction…

Acid + Base Water + Soluble Salt

…has several General Categories forming soluble salts:

SA + SB soluble neutral salt + water

WA + SB soluble basic salt + water

SA + WB soluble acidic salt + water

WA + WB salt of unknown acidity + water

The reaction… Acid + Base Water + Salt

Can be reversed to create the reaction for the hydrolysis of a salt…

Salt + Water Acid + Base

Writing the whole, total ionic equation and net ionic equations:

Balance all atoms and charges

SA, SB and SS (Soluble salts) are written as charged ions

WA, WB, gases, solids, liquids are written whole

Include phases


Example 18:

Write the whole, total ionic, and net ionic reactions for the hydrolysis of the

following salts. Identify spectator ions. Predict acidity of the salt, (Is it neutral,

acidic or basic)

a) CaBr2

b) NaNO2

c) NH4NO3

d) NH4C2H3O2

e) CH3NH2F

Example 19:

Calculate [OH-1], pH, % hydrolysis for 0.10 M NaClO solution found in Clorox

bleach. Ka of HClO = 3.5 x 10-8


Example 20:

Write the whole, total and net ionic hydrolysis reactions for NH4Br. Calculate

[H+1], pH, % hydrolysis for 0.20 M NH4Br solution. Look up the appropriate

equilibrium constant you will require in the calculations.

Hydrated Metal Cations can act as Weak Acids:

Alkali metal and alkaline earth metal cations are pH neutral, negligible

counter-ions of strong bases. They do not act as weak acids.

Small highly charged metals can coordinate with water and release H+1 ion

from water to reduce the charge.

Cu(H2O)6+2 (aq) H+1 (aq) + Cu(H2O)5(OH)+1(aq) Ka = 3 x 10-8

Smaller metals with higher positive charges more acidic

Acid Strength:

The stronger an acid is at donating H+1, the weaker the conjugate base is

at accepting H+1

Cation makes a stronger acid than neutral molecule which is more acidic

than anion… H3O+1 > H2O > OH-1 or NH4

+ > NH3 > NH2−

Larger Ka = stronger acid


Chemical Structure Influences the Acid/Base Strength:

BinaryAcids:

Bond Strength (Greatest factor in same vertical column),

weaker bond (larger anion) more acidic

Bond strengths: HF >> HCl > HBr > HI

Electronegativity difference: (factor when comparing anions in the same

period, horizontal row)

greater difference more acidic.

H2S is a weaker acid than HCl

F-1 ion (special case) causes an increased ordering in water molecules

creating the unfavorable lowering of entropy. This helps to explain why

HF is a weak acid.

Ternary Oxyacids:

Electronegativity of Center Atom, higher electronegativity more acidic

H2SO4 > H2SeO4 > H2TeO4

Oxidation State of Center,

Larger oxidation number (more oxygens) more acidic.

HClO4 > HClO3 > HClO2 > HClO

Carboxylic Acids: The ability for the conjugate base to have resonance

structures will stabilize the base and it is more likely to have the

hydrogen ion lost. R-COOH

Polyprotic Acids:

The fewer H+1 a species has, the weaker the acid becomes.

H3PO4 is a stronger acid than H2PO4-1 and both are stronger than the

HPO4-2 acid since each successive Ka gets much smaller.

Example 21:

Explain the following observations:

a) H3PO4 is a stronger acid than H3AsO4

b) H2SO3 is a stronger acid than HSO3-1

Acid Rain:

Over 85% of U.S. fuel is from fossil fuels producing CO2, SO2, and NO2 which are linked

to acid rain and damages to ecosystems and structures. Natural processes as volcanoes

also add to it.

Nonmetal oxides and water create acids.


More Practice:

1. Give the conjugate base of the following Bronsted-Lowry acids…

a) NH4+1 b) H2PO4

-1 c) HC7H5O2

2. Give the conjugate acid of the following Bronsted-Lowry bases…

a) CN-1 b) H2PO4-1 c) C2H5NH2

3. Designate the Bronsted-Lowry acid and base on the reactant side of each equation

and the conjugate acid and base on the product side: (all aqueous)

a) NH4+1 + CN-1 HCN + NH3

b) (CH3)3N + H2O (l) (CH3)3NH+1 + OH-1

c) HCHO2 + PO4-3 CHO2

-1 + HPO4-2

4. The hydrogen oxalate ion, HC2O4-1, is amphiprotic. Write the balanced chemical

equation showing how it acts as an acid and how it acts as a base in water.

5. Which of the following is the stronger acid, HBrO or HBr?

6. Which is the stronger base, F-1 or Cl-1?

7. Calculate [H+1], pH and pOH for the following and determine if acidic or basic…

a) [OH-] = 0.00040 M

b) 2.5 x 10-2 M HCl

c) solution where 100x[H+] =[OH-]

8. By what factor does [H+] change for a pH change of a) 2.00 units, b) 0.50 units ?

9. Predict the products of the following acid-base reactions, and also predict whether

the reactants or the products are preferred when at equilibrium.

a) NH3 (aq) + HBr (aq)

b) HCO3-1(aq) + F-1 (aq)

c) H2O (l) + NO3-1 (aq)

10. The average pH of normal arterial blood is 7.40 at body temperature (37˚C), at

which Kw = 2.4 x 10-14. Calculate [H+1], [OH-1] and pOH for this temperature.

11. A 0.100 M solution of lactic acid (HC3H5O3) has a pH of 2.44. Calculate Ka.

12. A 0.100 M solution of chloroacetic acid is 11.0% ionized. Solve for the equilibrium

concentrations of its ions and the Ka.


13. Saccharin (HNC7H4SO3) has a pKa of 2.32 at 25˚C. Solve for the pH of a 0.100 M

solution of saccharin.

14. Phosphoric acid is triprotic (H3PO4). Calculate the pH and equilibrium

concentrations involved in phosphoric acid for a 0.100 M solution.

Given: Ka1 =7.5 x 10-3, Ka2 = 6.2 x 10-8, Ka3 = 4.2 x 10-13

15. Calculate the pH of 0.075 M ethylamine (C2H5NH2), Kb = 6.4 x 10-4

16. Ephedrine (C10H15ON) is used in nasal sprays as a decongestant. A 0.035 M

solution has a pH of 11.33. Solve for the equilibrium concentrations and Kb.

17. Predict wheter the following aqueous salts will be acidic, basic, or neutral…

a) NH4Br b) NaC2H3O2 c) KClO4

18. An unknown salt is either NaOCl or NaF. When 0.050 moles of the salt is

dissolved in 0.500 L the pH of the solution is 8.08. What is the salt?


Answers:

1) a) NH3 b) HPO4

-2 c) C7H5O2-1

2) a) HCN b) H3PO4 c) C2H5NH3+1

3) a) NH4+1 + CN-1 HCN + NH3

acid base acid base

b) (CH3)3N + H2O (l) (CH3)3NH+1 + OH-1

base acid acid base

c) HCHO2 + PO4-3 CHO2

-1 + HPO4-2

acid base base acid

4) Acid: HC2O4-1 + H2O C2O4

-2 + H3O+1

Base: HC2O4-1 + H2O H2C2O4 + OH-1

5) HBr

6) F-1

7) a) [H+1]= 2.5 x 10-11, pH = 10.60, pOH = 3.40

b) [H+1]= 2.5 x 10-2 , pH = 1.60 , pOH = 12.40

c) [H+1]= 1.0 x 10-8 , pH = 8.00, pOH = 6.00

8) a) 100x b) 3.16x

9) a) Products: NH3 (aq) + HBr (aq) NH4+1 (aq) + Br-1 (aq)

b) Reactants: HCO3-1(aq) + F-1 (aq) CO3

-2(aq) + HF (aq)

c) Reactants: H2O (l) + NO3-1 (aq) OH-1

(aq) + HNO3 (aq)

10) [H+1]= 4.0 x 10-8, [OH-] = 6.0 x 10-7, pOH = 6.22

11) Ka = 1.4 x 10-4

12) [H+1] = [ClCH2COO-1] = 0.0110M, [ClCH2COOH] = 0.0890M, Ka = 1.4 x 10-3

13) [H+1]= 2.0 x 10-2, pH = 1.71

14) [H3PO4] = 0.073 M, [H+1]= 0.027 M, [H2PO4-1] = 0.027 M, [HPO4

-2] = 6.2 x 10-8,

[PO4-3] = 3x10-20 , [OH-] = 3.7 x 10-13 , pH=1.56

15) pH =11.82

16) [OH-1] = [C10H15ONH+1] = 0.0021M, [C10H15ON] = 0.033M, Ka = 1.4 x 10-4

17) a) acid, b) base, c) neutral

18) NaF

A c i d s a n d B a s e s C h 1 6 P a g e | Ch 16: Acids...

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