Acids & BasesAcids & BasesAcids:Acids:

acids are sour tastingArrhenius acidArrhenius acid: Any substance that, when dissolved

in water, increases the concentration of hydronium ion (H3O+)

Bronsted-Lowry acidBronsted-Lowry acid: A proton donorLewis acidLewis acid: An electron acceptor

Bases:Bases:bases are bitter tasting and slipperyArrhenius baseArrhenius base: Any substance that, when dissolved

in water, increases the concentration of hydroxide ion (OH-)

Bronsted-Lowery baseBronsted-Lowery base: A proton acceptorLewis acidLewis acid: An electron donor

Lone Hydrogen ions do not exist by themselves in solution. H+ is always bound to a water molecule to form a hydronium ion

Brønsted-Lowry Theory of Acids & BasesConjugate Acid-Base Pairs

General Equation

Brønsted-Lowry Theory of Acids & Bases

Brønsted-Lowry Theory of Acids & Bases

Brønsted-Lowry Theory of Acids & BasesNotice that water is both an acid & a base = amphoteric

Reversible reaction

ELECTROLYTESELECTROLYTESElectrolytes are species which conducts electricity when

dissolved in water. Acids, Bases, and Salts are all electrolytes.

Salts and strong Acids or Bases form Strong Electrolytes. Salt and strong acids (and bases) are fully dissociated therefore all of the ions present are available to conduct electricity.

HCl(s) + H2O H3O+ + Cl-

Weak Acids and Weak Bases for Weak Electrolytes. Weaks electrolytes are partially dissociated therefore not all species in solution are ions, some of the molecular form is present. Weak electrolytes have less ions avalible to conduct electricity.

NH3 + H2O NH4+ + OH-

Acids & BasesAcids & Bases STRONG vs WEAKSTRONG vs WEAK_ _ completely ionizedcompletely ionized _ partially ionized_ partially ionized_ strong electrolyte_ strong electrolyte _ weak electrolyte_ weak electrolyte_ ionic/very polar bonds_ ionic/very polar bonds _ some covalent bonds_ some covalent bonds

Strong AcidsStrong Acids:: Strong Bases:Strong Bases:HClOHClO44 LiOHLiOHHH22SOSO44 NaOHNaOHHIHI KOHKOHHBrHBr Ca(OH)Ca(OH)22

HClHCl Sr(OH)Sr(OH)22

HNOHNO33Ba(OH)Ba(OH)22

Acids & BasesAcids & Bases• One ionizable proton:

HCl → H+ + Cl-

• Two ionizable protons:H2SO4 → H+ + HSO4

-

HSO4- → H+ + SO4

2-

• Three ionizable protons:H3PO4 → H+ + H2PO4

–

H2PO4- → H+ + HPO4

2-

HPO42- → H+ + PO4

-3

Combined:H2SO4 → 2H+ + SO4

2-

Combined: H3PO4 → 3H+ + PO4

3-

Acids & BasesAcids & BasesFor the following identify the acid and the base as strong or weak .

a. Al(OH)3 + HCl

b. Ba(OH)2 + HC2H3O2

c. KOH + H2SO4

d. NH3 + H2O

Weak base Strong acid

Weak acid

Strong acid

Strong base

Strong base

Weak base Weak acid

Acids & BasesAcids & BasesFor the following predict the product. To check your answer left click

on the mouse. Draw a mechanism detailing the proton movement.

a. Al(OH)3 + HCl

b. Ba(OH)2 + HC2H3O2

c. KOH + H2SO4

d. NH3 + H2O

AlCl3 + 3 H2O

Ba(C2H3O2)2 + 2 H2O

K2SO4 + 2 H2O

NH4+ + OH-

2

2

3

Conjugate Acid-Base Pairs

Conjugate Acid-Base Pairs

Acids & BasesAcids & BasesFor the following Identify the conjugate acid and the conjugate base.

The conjugate refers to the acid or base produced in an acid/base reaction. The acid reactant produces its conjugate base (CBCB).

a. Al(OH)3 + 3 HCl AlCl3 + 3 H2O

b. Ba(OH)2 + 2 HC2H3O2 Ba(C2H3O2)2 + 2 H2O

c. 2 KOH + H2SO4 K2SO4 + 2 H2O

d. NH3 + H2O NH4+ + OH-

CB CACB CA

CB CACB CA

CB CACB CA

CA CBCA CB

TITRATIONTITRATIONTitration of a strong acid with a strong base

ENDPOINT = POINT OF NEUTRALIZATION = ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINTEQUIVALENCE POINT

At the end point for the titration of a strong acid with a strong base, the moles of acid (H+) equals the moles of base (OH-) to produce the neutral species water (H2O). If the mole ratio in the balanced chemical equation is 1:1 then the following equation can be used.

MOLES OF ACID = MOLES OF BASE

nnacidacid = n = nbasebase

Since M=n/V

MMAAVVAA = M = MBBVVBB

TITRATIONTITRATIONMMAAVVAA = M = MBBVVBB

1. Suppose 75.00 mL of hydrochloric acid was required to neutralize 22.50 mLof 0.52 M NaOH. What is the molarity of the acid?

HCl + NaOH HCl + NaOH H H22O + NaClO + NaCl

MMaa V Vaa = M = Mbb V Vbb rearranges to M rearranges to Maa = M = Mbb V Vbb / V / Vaa

so Mso Maa = (0.52 M) (22.50 mL) / (75.00 mL) = (0.52 M) (22.50 mL) / (75.00 mL)

= 0.16 M= 0.16 M

Now you try:2. If 37.12 mL of 0.843 M HNO3 neutralized 40.50 mL of KOH,

what is the molarity of the base?Mb = 0.773 mol/L

Molarity and Titration

TITRATIONTITRATIONTitration of a strong acid with a strong base

ENDPOINT = POINT OF NEUTRALIZATION = ENDPOINT = POINT OF NEUTRALIZATION = EQUIVALENCE POINTEQUIVALENCE POINT

At the end point for the titration of a strong acid with a strong base, the moles of acid (H+) equals the moles of base (OH-) to produce the neutral species water (H2O). If the mole ratio in the balanced chemical equation is NOT 1:1 then you must rely on the mole relationship and handle the problem like any other stoichiometry problem.

MOLES OF ACID = MOLES OF BASEMOLES OF ACID = MOLES OF BASE

nnacidacid = n = nbasebase

TITRATIONTITRATION1. If 37.12 mL of 0.543 M LiOH neutralized 40.50 mL

of H2SO4, what is the molarity of the acid?2 LiOH + H2 LiOH + H22SOSO44 Li Li22SOSO44 + 2 H + 2 H22OO

First calculate the moles of base:0.03712 L LiOH (0.543 mol/1 L) = 0.0202 mol LiOH0.03712 L LiOH (0.543 mol/1 L) = 0.0202 mol LiOHNext calculate the moles of acid:0.0202 mol LiOH (1 mol H0.0202 mol LiOH (1 mol H22SOSO4 4 / 2 mol LiOH)= 0.0101 mol H/ 2 mol LiOH)= 0.0101 mol H22SOSO44

Last calculate the Molarity:: MMa a = n/V = 0.010 mol H= n/V = 0.010 mol H22SOSO44 / 0.4050 L = 0.248 M / 0.4050 L = 0.248 M

2. If 20.42 mL of Ba(OH)2 solution was used to titrate29.26 mL of 0.430 M HCl, what is the molarity of the barium hydroxide solution?

Mb = 0.308 mol/L

Molarity and Titration• A student finds that 23.54 mL of a 0.122 M

NaOH solution is required to titrate a 30.00-mL sample of hydr acid solution. What is the molarity of the acid?

• A student finds that 37.80 mL of a 0.4052 M NaHCO3 solution is required to titrate a 20.00-mL sample of sulfuric acid solution. What is the molarity of the acid?

• The reaction equation is:H2SO4 + 2 NaHCO3 → Na2SO4 + 2 H2O + 2 CO2

Water Equilibrium

Water Equilibrium

Kw = [H+] [OH-] = 1.0 x 10-14

Equilibrium constant for water

Water or water solutions in which [H+] = [OH-] = 10-7 M are neutral solutions.

A solution in which [H+] > [OH-] is acidic

A solution in which [H+] < [OH-] is basic

pHpHA measure of the hydronium ionA measure of the hydronium ion

• The scale for measuring the hydronium ion concentration [H3O+] in any solution must be able to cover a large range. A logarithmic scale covers factors of 10. The “p” in pH stands for log.

• A solution with a pH of 1 has [H3O+] of 0.1 mol/L or 10-1

• A solution with a pH of 3 has [H3O+] of 0.001 mol/L or 10-3

• A solution with a pH of 7 has [H3O+] of 0.0000001 mol/L or 10-7

pH = - log [HpH = - log [H33OO++]]

The pH scaleThe pH scaleThe pH scale ranges from 1 to 10The pH scale ranges from 1 to 10-14-14 mol/L or from 1 mol/L or from 1

to 14.to 14. pH = - log [H3O+]

1 2 3 4 5 6 7 8 9 10 11 12 13 14acidacid neutral neutral basebase

Manipulating pHManipulating pHAlgebraic manipulation of:Algebraic manipulation of:

pH = - log [H3O+] allows for:allows for:

[H3O+] = 10-pH

If pH is a measure of the hydronium ion If pH is a measure of the hydronium ion concentration then the same equations could be concentration then the same equations could be used to describe the hydroxide (base) concentration. used to describe the hydroxide (base) concentration.

[OH-] = 10-pOH pOH = - log [OH-] thus:thus:

pH + pOH = 14 ; the entire pH range!

PRACTICE PROBLEM #251. How many milliliters of 1.25 M LiOH must be added to neutralize

34.7 mL of 0.389 M HNO3?

2. What mass of Sr(OH)2 will be required to neutralize 19.54 mL of 0.00850 M HBr solution?

3. How many mL of 0.998 M H2SO4 must be added to neutralize 47.9 mL of 1.233 M KOH?

4. What is the molar concentration of hydronium ion in a solution of pH 8.25?

5. What is the pH of a solution that has a molar concentration of hydronium ion of 9.15 x 10-5?

6. What is the pOH of a solution that has a molar concentration of hydronium ion of 8.55 x 10-10?

10.8 mL

0.0101 g

29.6 mL

5.623 x 10-9 M

pH = 4.0

pOH = 4.9

GROUP STUDY PROBLEM #25______1. How many milliliters of 0.75 M KOH must be added to neutralize

50.0 mL of 2.50 M HCl?

______2. What mass of Ca(OH)2 will be required to neutralize 100 mL of 0.170 M HCl solution?

______3. How many mL of 0.554 M H2SO4 must be added to neutralize 25.0 mL of 0.9855 M NaOH?

______ 4. What is the molar concentration of hydronium ion in a solution of pH 2.45?

______ 5. What is the pH of a solution that has a molar concentration of hydronium ion of 3.75 x 10-9?

______ 6. What is the pOH of a solution that has a molar concentration of hydronium ion of 4.99 x 10-4?