daniel l. reger scott r. goode david w. ball chapter 16 reactions between acids and bases

Daniel L. RegerScott R. GoodeDavid W. Ball

www.cengage.com/chemistry/reger

Chapter 16Reactions Between

Acids and Bases

Titration of Strong Acids and Bases

• Titration: a method used to determine the concentration of a substance known as the analyte by adding another substance, the titrant, which reacts in a known manner with the analyte.

• analyte + titrant → products

Laboratory Titrations

(a) A known volume of acid is measured into a flask.

(b) Standard base is added from a buret.

(c) The endpoint is indicated by a color change.

(d) The volume of base is recorded.

• Titration Curve: a graph of pH of a solution as titrant is added.• For a titration of a strong acid with a

strong base, the pH will start at a very low value and stay low as long as strong acid is still present.

Titration: Strong Acid and Base

• The pH will rise sharply to 7 at the equivalence point, where the acid and base are present in stoichiometrically equivalent amounts.• After excess strong base has been

added, the pH levels off at a high value.

Titration: Strong Acid and Base

Titration Curve for a Strong Acid with a Strong Base

• Calculate the equivalence point in the titration of 20.00 mL of 0.1252 M HCl with 0.1008 M NaOH.

HCl + NaOH H2O + NaCl

The Equivalence Point in a Titration

• Calculate the equivalence point in the titration of 40.00 mL of 0.2387 M HNO3 with 0.3255 M NaOH.

Test Your Skill

• Millimole: one thousandth of a mole.• If molarity is expressed in moles/liter (M)

and volume in milliliters (mL), n will be in millimoles (mmol).

•

. Liters cancel but the milli- multiplier remains.

millimoles smilliliterliter

molesn

Units of Millimoles

Calculating a Titration Curve

• Calculate the pH in the titration of 20.0 mL of 0.125 M HCl with 0.250 M NaOH after 0, 2.00, 10.00, and 20.00 mL base are added.

• Calculate the pH in the titration of 20.0 mL of 0.125 M HCl with 0.250 M NaOH after 5.00 mL and 12.00 mL base are added.

Test Your Skill

Calculating a Titration Curve

• Titration curve of 50.00 mL 0.500 M KOH with 1.00 M HCl.

Strong Base + Strong Acid Curve

• 10 mL of two different 0.100 M acids titrated with 0.100 M NaOH.

Stoichiometry and Titration Curves

Estimating the pH of Mixtures• Fill in first 3 three lines of sRf table.• Look at the final solution (f-line).• If a strong acid is present, the solution will be

strongly acidic.• If a strong base is present the solution will be

strongly basic.

• If only water is present, the solution will be neutral.

Solution Estimate of pH

Strongly acidic 1

Neutral 7

Strongly basic 13

Buffers

• Buffer: a solution that resists changes in pH.• A buffer is a mixture of a weak acid or

base and its conjugate partner.

• HA + OH- → H2O + A-

Weak acid reacts with any added OH-.

• A- + H3O → HA + H2O

Weak base reacts with any added H3O+.

The pH of a Buffer System

• For the chemical reaction

HA + H2O → H3O+ + A-

• [HA]

][Alogp pH

obtain to function)-(p log- the take

][A

[HA] ]O[H or

[HA]

]][AO[H

-

a

-a

3

-3

a

K

KK

The pH of a Buffer System

• Calculate the pH of a solution of 0.50 M HCN and 0.20 M NaCN, Ka = 4.9 x 10-10.

• Calculate the pH of a solution of 0.40 M NH3 and 0.10 M NH4Cl. For NH3 Kb = 1.8 x 10-5.

The pH of a Buffer System

• Calculate the pH of a buffer that is 0.25 M HCN and 0.15 M NaCN, Ka = 4.9 x 10-10 .

Test Your Skill

• Calculate the amount of sodium acetate that must be added to 250 mL of 0.16 M acetic acid in order to prepare a pH 4.68 buffer. Ka = 1.8 x 10-5

The Composition of a pH Buffer

Test Your Skill

• How many moles of NaCN should be added to 100 mL of 0.25 M HCN to prepare a buffer with pH = 9.40?

Ka = 4.9 x 10-10.

Determining the Response of a Buffer to Added Acid or Base

• Calculate the initial and final pH when 10 mL of 0.100 M HCl is added to (a) 100 mL of water, and (b) 100 mL of a buffer which is 1.50 M CH3COOH and 1.20 M CH3COONa.

Test Your Skill

• Calculate the final pH when 10 mL of 0.100 M NaOH is added to 100 mL of a buffer which is 1.50 M CH3COOH and 1.25 M CH3COONa.

Before any base added.

(a) Part way to equivalence point.

(b) Equivalence point.

(c) Beyond equivalence point.

Qualitative Aspects: Titration: Weak Acid + Strong Base

Titration: Weak Acid + Strong Base

• HA + OH- A- + H2O

• (a) Before any base is added the

solution is a weak acid has a low pH.• Estimated pH = 3

pH 2-4 is typical–Depends on the concentration of the acid.

–Depends on the value of Ka.

Titration: Weak Acid + Strong Base

• HA + OH- A- + H2O

• (b) After some base is added, but before the equivalence point is reached.• The solution is a mixture of the weak

acid HA and its conjugate base A-; therefore, the solution is a buffer.

• The estimated pH is equal to pKa.

Titration: Weak Acid + Strong Base

• HA + OH- A- + H2O

• (c) At the equivalence point the solution is salt of A-, all the HA having been consumed by the stoichiometric amount of OH-.• A- is the weak conjugate base of HA.• The estimated pH is 10.

Titration: Weak Acid + Strong Base

• HA + OH- A- + H2O

• (d) After excess strong base is added OH- is in excess.• The estimated pH is 13.

pH Estimates

Solution Estimate of pH

Strongly acidic 1

Weakly acidic 3

Neutral 7

Weakly basic 11

Strongly basic 13

Buffer(acidic to basic)

pKa

4-10 typical

Titration: Weak Acid + Strong Base

Titration Curves for Acids of Different Strengths

• Calculate the pH in the titration of 20.00 mL of 0.500 M formic acid (HCOOH Ka=1.8 x 10-4) with 0.500 M NaOH after 0, 10.00, 20.00, and 30.00 mL of base have been added.• The titration reaction is

HCOOH + OH- HCOO- + H2O

Calculating the Titration Curve for a Weak Acid

Titration of 25.00 mL of 0.500 M Formic Acid with 0.500 M NaOH

Test Your Skill

• Calculate the pH in the titration of 12.00 mL of 0.100 M HOCl with 0.200 M NaOH after 0, 3.00, 6.00, and 9.00 mL of base have been added.

Titration of 20.00 mL of 0.500 M Methylamine with 0.500 M HCl

pH Indicators

• Indicator: a substance that changes color at the endpoint of a titration.• pH indicators are weak acids or bases

whose conjugate species are a different color.

• HIn + H2O ⇌ H3O+ + In-

•

• pKIn = -log(KIn)

[HIn]

]][InO[H3In

K

pH Indicators

• When pH is lower than pKIn, the indicator will be in the acid form.

• When pH is greater than pKIn, the indicator will be in the base form.• An indicator should be chosen which

changes at or just beyond the equivalence point.

pH Indicators

* Thymol blue is polyprotic and has three color forms.

Name Acid Color

Base Color

pH Range

pKIn

Thymol blue* Red Yellow 1.2–2.8 1.6Methyl orange Red Yellow 3.1–4.4 3.5Methyl red Red Yellow 4.2–6.3 5.0Bromthymol blue

Yellow Blue 6.2–7.6 7.3

Phenolphthalein

Clear Pink 8.3–10.0 8.7

Thymol blue* Yellow Blue 8.0–9.6 9.2

Properties of Indicators

Titration Curves for Strong and Weak Acids

• Polyprotic acids provide more than one proton when they ionize.• Polyprotic acids ionize in a stepwise

manner.

H2A + H2O H⇌ 3O+ + HA- Step 1

HA- + H2O H⇌ 3O+ + A2- Step 2

Polyprotic Acids

• There is a separate acid ionization constant for each step

H2A + H2O H⇌ 3O+ + HA- Step 1

HA- + H2O H⇌ 3O+ + A2- Step 2

A][H

]][HAO[H

2

3a1

K

][HA

]][AO[H-

-23

a2

K

Polyprotic Acids

• HA- is the conjugate base of H2A, so it is a weaker acid than H2A.

• Ka1 is always larger than Ka2.

• For triprotic acids (such as H3PO4), Ka2 is always larger than Ka3.

Polyprotic Acids

• When successive Ka values differ by a factor of 1000 or more, each step can be assumed to be essentially unaffected by the occurrence of the subsequent step.

Calculating Concentrations of Species in Polyprotic Acid Solutions

• Calculate the concentrations of all species in 0.250 M malonic acid, Ka = 1.6 x 10-2.

• Consider the first ionization and solve by usual approach.

H2C3H2O4 + H2O HC⇌ 3H2O4- + H3O+

Concentrations of Species in Polyprotic Acid Solutions

• The second step is needed only to calculate the concentration of C3H2O4

2-

because the concentration of H3O+ is determined by the first step.• You can ignore the effect of the second

step on the pH because the Ka1 is so much larger than Ka2.

Concentrations of Species in Polyprotic Acid Solutions

Test Your Skill

• Calculate the pH of a 0.040 M solution of ascorbic acid. (Ka1 = 8.0 x 10-5, Ka2 = 1.6 x 10-12)

Amphoteric Species

• Amphoteric: having both acidic and basic properties.• Conjugate bases of weak polyprotic

acids are amphoteric.

• The hydrogen oxalate ion, HC2O4-, is a

weak acid (Ka2 = 1.6 ×10-4).

• HC2O4- + H2O C⇌ 2O4

2- + H3O+

Amphoteric Species• Weak Acid• The hydrogen oxalate ion,

HC2O4-, is a weak acid.

HC2O4- + H2O C⇌ 2O4

2- + H3O+

• Ka2 = 1.6 x 10-4

• Weak Base• The hydrogen oxalate ion,

HC2O4-, can also act as a

weak base.

HC2O4- + H2O → H2C2O4 + OH-

• Kb = Kw/Ka1 = 1.0 x 10-14 / 5.6 x 10-2

• Kb = 1.9 x 10-13

• Since Ka > Kb, the ion will act as a weak acid in water.• When comparing Ka to Kb note that Ka is Ka2 and Kb is Kw/Ka1.

Test Your Skill

• Ka for the hydrogen malonate ion, HC3H2O4

-, is 2.1 x 10-6. Is a solution of sodium hydrogen malonate acidic or basic?

Factors That Influence Solubility

• The pH affects the solubility of salts of weak acids.• Complex ion formation affects the

solubility of salts of transition metal cations.

Salts of Anions of Weak Acids• The solubility of salts of anions of weak

acids is enhanced by lowering the pH.

• Cd(CN)2(s) Cd⇌ 2+(aq) + 2CN-(aq)

Ksp = 1.0 x 10-8

• Adding acid reduces [CN-] in solution, by the reaction

• H3O+(aq) + CN-(aq) HCN⇌ (aq) + H2O(l)

Salts of Transition Metal Cations

• Transition metal cations form complexes with Lewis bases such as H2O, NH3, or OH-.• Formation of a complex reduces the

concentration of metal ion and increases the solubility of the salt.

Solubility of Amphoteric Species

• Amphoteric species, such as Be(OH)2, Al(OH)3, Sn(OH)2, Pb(OH)2, Cr(OH)3, Ni(OH)2, Cu(OH)2, Zn(OH)2, and Cd(OH)2 react with acid or base to form the soluble metal ion or complex ions

M(OH)x + xH+ Mx+ + xH2O x = 2,3

M(OH)x + yOH- M(OH)xy+ x = 2,3,

y = 1,2