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Acid Base Equilibria maybe do packet 2 before packet 3? Combine concepts from 1 and 3?. Arrhenius Acids & Bases (1 st year chem definition!). An Arrhenius acid is a substance that, when dissolved in water, increases the concentration of H + (needs H + in it)

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arrhenius acids bases 1 st year chem definition
Arrhenius Acids & Bases(1st year chem definition!)

An Arrhenius acid is a substance that, when dissolved in water, increases the concentration of H+ (needs H+ in it)

Example: HCl (monoprotic)

H2SO4 (diprotic)

An Arrhenius base is a substance that, when dissolved in water, increases the concentration of OH–

Example: NaOH(monobasic)

Ca(OH)2 (dibasic)

br nsted lowry acids and bases
Brønsted-Lowry Acids and Bases

Brønsted-Lowry acid is a species that donates H+ (proton)

Brønsted-Lowry base is a species that accepts H+(proton)

Brønsted-Lowry definition of a base does not mention OH–and the reaction does not need to be aqueous.

the h ion in water
The H+ Ion in Water
  • The H+(aq) ion is simply a proton with no surrounding valence electrons.
  • In water, clusters of hydrated H+(aq) ions form.
slide5
The simplest cluster is H3O+(aq)

We call this a hydronium ion.

  • Larger clusters are also possible (such as H5O2+ and H9O4+).
  • Generally we use H+(aq) and H3O+(aq) interchangeably.
proton transfer reactions
Proton-Transfer Reactions
  • Consider

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

  • H2O donates a proton to ammonia.
    • Therefore, water is acting as an acid.
  • NH3 accepts a proton from water.
    • Therefore, ammonia is acting as a base.
amphoteric substances
Amphotericsubstances
  • Can behave as acids and bases.
  • Water is an example of an amphoteric species.

Water as an acid (proton donor)

H2O + NH3NH4+ + OH-

Water as a base (proton acceptor)

H2O + HNO2H30+ + NO2-

conjugate acid base pairs
Conjugate Acid-Base Pairs
  • A conjugate acid is the substance formed by adding a proton to the base.
  • A conjugate base is the substance left over after the acid donates a proton.

Within a pair the acid has more hydrogen!

strong acids bases
Strong Acids & Bases

All other acids and bases are weak!

slide10
Ions

Most anions are weak bases

Most cations are weak acids

Anions of strong acids and cations of strong bases are neutral

strengths of acids and bases
Strengths of Acids and Bases

Strong acids completely ionize in water.

HCl + H2O  H3O+ + Cl-

HCl H+ + Cl-

  • Essentially no un-ionized molecules remain in solution so the equation usually does not contain equilibrium arrows. Keq >>1
  • Their conjugate bases have negligible tendencies to become protonated

Cl- + H+  HCl.

The conjugate base of a strong acid is a neutral anion.

strengths of acids and bases1
Strengths of Acids and Bases

Strong bases completely dissociate in water

NaOH(aq)  Na+(aq) + OH-(aq)

  • Essentially no undissociated compound remains in solution so the equation usually does not contain equilibrium arrows. Keq >>1
  • The ions have negligible tendencies to attract OH- in solution.

Na+(aq) + OH-(aq)  NaOH(aq)

The cation of a strong base is a neutral cation.

slide13
All other acids are Weak acids. They only partially dissociate in aqueous solution.

HC2H3O2(aq) + H2O(l)  H3O+(aq) + C2H3O2-(aq)

  • They exist in solution as a mixture of molecules and component ions. (usually mostly molecules in equilibrium)
  • Their conjugate bases are weak bases.
    • Besides non-neutral anions, weak bases tend to be nitrogen containing organic compounds
  • Example: Acetic acid is a weak acid; acetate ion (conjugate base) is a weak base.

conjugateacid

conjugatebase

acid

base

non acid compounds with hydrogen
Non-acid compounds with hydrogen

Not all compounds containing hydrogen are acidic.

These are extremely weak acids…so much that we don’t consider them acids at all.

Their conjugate bases are strong bases!

Negligible acidity: OH- H2 CH4

Strong bases: O2- H- CH3-

slide16
The stronger an acid is, the weaker its conjugate base will be.

In acid-base reactions, the reaction favors the transfer of a proton from the stronger acid to the stronger baseto form a weaker acid and weaker base.

We need a more specific way to determine acid strength!

slide17

Where does that acid/base ranking come from?

Keq!!!

Since weak acids and bases are in equilibrium…we can write equilibrium constant expressions!

When looking at the reaction of a weak acid with water we label the equilibrium constant Ka:

HF(aq) + H2O(l)  H3O+(aq) + F-(aq)

When looking at the reaction of a weak base with water we label the equilibrium constant Kb:

NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)

k a and k b
Ka and Kb
  • The value of Ka or Kb indication the extent to which the weak acid or base ionizes/dissociates.
    • Larger Kaor Kb means more products!
  • Weak acids with larger Ka values are stronger.
  • Weak bases with larger Kb values are stronger.

(Values in Appendix D)

slide19
Percent Ionization

• Percent ionization is another method to assess acid strength.

• For the reaction: HA(aq)  H+(aq) + A–(aq)

• The higher the percent ionization,

the stronger the acid.

polyprotic acids
Polyprotic Acids

Polyprotic acids have more than one ionizable proton.

  • The protons are removed in successive steps.

Consider the weak acid, H2SO3 (sulfurous acid):

H2SO3(aq)  H+(aq) + HSO3–(aq) Ka1= 1.7 x 10–2

HSO3–(aq)  H+(aq) + SO32–(aq) Ka2= 6.4 x 10–8

  • It is always easier to remove the first proton in a polyprotic acid than the second.

Ka1 > Ka2 > Ka3, etc

the autoionization of water
The Autoionization of Water

In pure water the following equilibrium is established:

H2O(l)+ H2O(l)H3O+(aq) + OH–(aq)

acid base acid base

  • This process is called the autoionizationof water.

We can write an equilibrium constant expression for the autoionization of water:

Kw = [H3O+] [OH–] = 1.0*10-14 @25°C

Kwis called the “ion-product constant”

Kw = Ka*Kb for conjugate pair

the ion product constant
The Ion Product Constant
  • This applies to pure water as well as to aqueous solutions. (for our purposes…)
  • A solution is neutral if [OH–] = [H3O+].
  • If the [H3O+] > [OH–], the solution is acidic.
  • If the [H3O+] < [OH–], the solution is basic.

In a neutral solution at 25oC,

[H+] = [OH-] = 1.0 x 10-7 M

slide23
Give the conjugate base of the following Bronsted-Lowry acids:

(a) HIO3 (b) NH4+1 (c) H2PO4-1 (d) HC7H5O2

remove H+

(a) IO3-1

(b) NH3

(c) HPO4-2

(d) C7H5O2-

slide24
Designate the Bronsted-Lowry acid and the Bronsted-Lowry base on the left side of each of the following equations, and also designate the conjucate acid and base on the right side:

(a) NH4+1(aq) + CN-1(aq)  HCN(aq) + NH3(aq)

acid base acid base

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

base acid base acid

(c) HCHO2(aq) + PO4-3(aq)  HPO4-2(aq) + CHO2-1(aq)

acid base acid base

slide25
(a) The hydrogen oxalate ion (HC2O4-1) is amphoteric. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base towards water.

(b) What is the conjugate acid and base of HC2O4-1?

(a) Acid: HC2O4-1(aq) + H2O(l)  C2O4-2(aq) + H3O+1(aq)

acid base conj base conj acid

Base: HC2O4-1(aq) + H2O(l)  H2C2O4(aq) + OH-1(aq)

base acid conj acid conj base

(b)

slide26
Label each of the following as being a strong acid, a weak acid, or a species with negligible acidity. In each case write the formula of its conjugate base, and indicate whether the conjugate base is a strong base, a weak base or a species with negligible basicity: (a) HNO2 (b) H2SO4 (c) HPO4-2 (d) CH4 (e) CH3NH3+1

(a) HNO2,

(b) H2SO4,

(c) HPO4-2 ,

(d) CH4 ,

(e) CH3NH3+1 ,

weak acid

NO2-1 ,

weak base

strong acid

HSO4-1 ,

negligible base

weak base

weak acid

PO4-3 ,

negligible acid

strong base

CH3-1 ,

weak base

weak acid

CH3NH2 ,

ACID BASE

strong negligible

weak weak

negligible strong

IMPORTANT

slide27
Which of the following is the stronger acid, HBrO or HBr?
  • Which is the stronger base, F-1 or Cl-1?

Briefly explain your choices.

(a) HBr - It is one of the seven strong acids

  • F-1

HF is a weak acid so F- is a weak base

HCl is a strong acid so Cl- is neutral

slide28
Predict the products of the following acid-base reactions & determine whether equilibrium lies to the right or left:

(a) O-2(aq) + H2O(l) 

(b) CH3COOH(aq) + HS-1(aq) 

(c) NO3-1(aq) + H2O(l) 

* You will need a chart like this or Ka/Kb values to determine equilibrium!

slide29

O-2(aq) + H2O(l) OH-1(aq) + OH-1(aq)

base acid acid base

*OH- is the weaker acid so equilibrium lies to the right

  • CH3COOH(aq) = HC2H3O2 (aq)
    • HC2H3O2 (aq) + HS-1(aq)  C2H3O2-1(aq) + H2S(aq)

acid base base acid

*H2S is the weaker acid so equilibrium lies to the right

  • NO3-1(aq) + H2O(l) HNO3(aq) + OH-1(aq)
  • base acid acid base
  • *H2O is the weaker acid so equilibrium lies (far!) to the left
the p function
The “p” Function
  • p is short for “– log10”

pH = -log10[H+] = -log[H+]

pOH = -log10[OH-] = -log [OH-]

  • Note that this is a logarithmic scale.
  • Thus a change in [H+] by a factor of 10 causes the pH to change by 1 unit.
  • Most pH values fall between 0 and 14.
slide31
In neutral solutions at 25oC, [H+] = 1.0 x 10-7

pH = -log[1.0 x 10-7] = 7.00

Lower pH = more acidic

Higher pH = more basic

another one pk
Another one: pK
  • We can use a similar system to describe the equilibrium constant

pK= - log[K]

  • The value of Kw at 25oC is 1.0 x 10–14

pKw= - log (1.0 x 10–14)= 14.0

Kw=[H+][OH-]

pKw= pH + pOH = 14.0

the ph loop
The pH Loop

[H+]=10-pH

pH

[H+]

pH = -log[H+]

[H+] [OH-]=10-14

pH + pOH=14.0

[OH-]=10-pOH

[OH-]

pOH

pOH = -log[OH-]

measuring ph
Measuring pH

The most accurate method to measure pH is to use a pH meter.

Acid Base Indicators:

  • certain dyes change color as pH changes.
  • Indicators are less precise than pH meters.
    • Many indicators do not have a sharp color change as a function of pH.
  • Most acid-base indicators can exist as either an acid or a base.
    • These two forms have different colors.
    • The relative concentration of the two different forms is sensitive to the pH of the solution.
  • Thus, if we know the pH at which the indicator turns color, we can use this color change to determine whether a solution has a higher or lower pH than this value.
slide37
Example 1: Calculate [H+1] for each of the following solutions and indicate whether the solution is acidic, basic or neutral. (a) [OH-1] = 0.0007 M (b) a solution where [OH-1] is 100 times greater than [H+1]
slide38
Example 2:

(a) If NaOH is added to water, how does the [H+1] change? How does pH change?

(b) If [H+1] = 0.005 M, what is the pH of the solution? Is the solution acidic or basic?

(c) If pH = 6.3, what are the molar concentrations of H+1(aq) and OH-1(aq) in the solution?

  • Kw= [H+] [OH-].

the [OH-] will increase and the [H+] will decrease. [H+] decreases, pH increases.

  • pH = - log [H+] = - log (0.005) = 2.3 acidic
  • pH = 6.3

[H+] = 10-pH = 10-6.3 = 5 x 10-7 M

pOH= 14 – pH = 14 – 6.3 = 7.7

[OH-] = 10-pOH = 10-7.7 = 2 x 10-8 M

strong acid calculations
Strong Acid Calculations
  • In solution the strong acid is usually the only source of H+
  • The pH of a solution of a monoprotic acid may usually be calculated directly from the initial molarity of the acid.

Caution: If the molarity of the acid is less than 10–6M then the autoionization of water needs to be taken into account.

slide40
Example 3: Calculate the pH of each of the following strong acid solutions: (a) 1.8  10-4 M HBr (b) 1.02 g HNO3 in 250 mL of solution (c) 2.00 mL of 0.500 M HClO4 diluted to 50.0 mL (d) a solution formed by mixing 10.0 mL of 0.0100 M HBr with 20.0 mL of 2.5  10-3 M HCl
  • 1.8 x 10-4M HBr = 1.8 x 10-4M H+ pH = -log(1.8 x 10-4) = 3.74

(b)

0.0647 M HNO3 = 0.0647 M H+ pH = -log (.0647) = 1.19

  • M1V1 = M2V2

(0.500 M)(.00200 L) = (x M)(0.0500 L)

x = .0200 M HCl

(d)

pH = -log(.0050) = 2.30

[H+] = .0200 M

pH = -log(.0200) = 1.70

strong base calculations
Strong Base Calculations
  • Strong bases are strong electrolytes and dissociate completely in solution.
  • For example:

NaOH(aq)  Na+(aq) + OH–(aq)

The pOH (and thus the pH) of a strong base may be calculated using the initial molarity of the base.

slide42
Example 4: Calculate [OH-1] and pH for (a) 3.5  10-4 M Sr(OH)2 (b) 1.50 g LiOH in 250 mL of solution (c) 1.00 mL of 0.095 M NaOH diluted to 2.00 L (d) a solution formed by adding 5.00 mL of 0.0105 M KOH to 15.0 mL of 3.5  10-3 M Ca(OH)2
  • [OH-] = 2[Sr(OH)2] = 2(0.00035 M) = .00070 M OH-

pOH = -log(.00070) = 3.15

pH = 14 – 3.15 = 10.85

(b)

pOH = -log (.251) = 0.601

pH = 14 - .601 = 13.399

(c) M1V1 = M2V2 (0.095 M)(.00100 L) = (x M)(2.00 L)

x = .000048 M NaOH = [OH-]

pOH = -log(.000048) = 4.32

pH = 14 – 4.32 = 9.68

(d)

pOH = -log(.0079) = 2.1

pH = 14 – 2.1 = 11.9

weak acid calculations
Weak Acid Calculations
  • Weak acids are only partially ionized in aqueous solution.
  • Therefore, weak acids are in equilibrium:

HA(aq) + H2O(l)  H3O+(aq) + A–(aq)

OR

HA(aq)  H+(aq) + A–(aq)

  • We can write Ka for this dissociation:
calculating k a from ph for weak acids
Calculating Ka from pH for weak acids
  • In order to find the value of Ka, we need to know all of the equilibrium concentrations. (ICE Chart)

• The pH gives the equilibrium concentration of H+.

  • We then substitute these equilibrium concentrations into the equilibrium constant expression and solve for Ka.
slide45

Example 4: A 0.20 M solution of niacin (a monoprotic weak acid) has a pH of 3.26. What is the Ka for niacin?

using k a to calculate ph for weak acids
Using Ka to Calculate pH for weak acids
  • Write the balanced chemical equation clearly showing the equilibrium.
  • Write the equilibrium expression. Look up the value for Ka (in a table).
  • Write down the initial and equilibrium concentrations for everything except pure water. (ICE table)
  • We usually assume that the equilibrium concentration of H+ is x.
  • Substitute into the equilibrium constant expression and solve.
  • Remember to convert x to pH if necessary.
polyprotic acids1
Polyprotic Acids
  • Polyprotic acids have more than one ionizable proton.

H2SO3(aq)  H+(aq) + HSO3–(aq) Ka1 = 1.7 x 10–2

HSO3–(aq)  H+(aq) + SO32–(aq) Ka2 = 6.4 x 10–8

  • The majority of the H+(aq) at equilibrium usually comes from the first ionization
  • If the successive Ka values differ by a factor of 103, we can usually get a good approximation of the pH of a solution of a polyprotic acid by considering the first ionization only.
  • If not, then we have to account for the successive ionizations
slide48
Example 5: The acid dissociation constant for benzoic acid, HC7H5O2 is 6.5 x 10-5. Calculate the equilibrium concentrations of H3O+, C7H5O2-, and HC7H5O2 . The initial concentration of HC7H5O2 is 0.050M.

HC7H5O2 (aq) H+(aq) + C7H5O2- (aq)

I 0.050 0 0

C -x x x

E (.050-x) x x

x2 + (6.5*10-5)x – (3.25*10-6) = 0

x= .0018M = [H+]=[C7H5O2- ]

[HC7H5O2] = .050 - .0018 = .048 M

slide49
What if I don’t have a quadratic equation program?

Algebra Shortcut:

Assume x is much smaller (less than 5% of .050)

To simplify: .050-x = .050

Now you don’t need the quadratic equation!

x= .0018M = [H+]=[C7H5O2- ]

[HC7H5O2] = .050 - .0018 = .048 M

If you made this assumption you need to check and make sure it’s valid – if this answer isn’t less than 5% use the quadratic equation!

slide50
Example 6: Calculate the pH of the following solution (Kaand Kb values are in Appendix D). 0.175 M hydrazoic acid, HN3
  • HN3(aq)  H+(aq) + N3-(aq)

I 0.175M 0 0

C -x x x

E 0.175-x x x

x2 + (1.9*10-5)x – (3.325*10-6) = 0

x=.0018M H+

pH = -log(0.0018) = 2.74

slide51
Example 7: Calculate the percent ionization of 0.400 M hydrazoic acid, HN3, solution.

HN3(aq)  H+(aq) + N3-(aq)

I 0.400M 0 0

C -x x x

E 0.400-x x x

x2 + (1.9*10-5)x – (7.6*10-6) = 0

x=.0028M H+

slide52
Example 8: Citric acid, which is present in citrus fruits, is a triprotic acid. Calculate the pH of a 0.050 M solution of citric acid.

H3C6H5O7(aq)  H+(aq) + H2C6H5O7-1(aq) Ka = 7.4*10-4

H2C6H5O7-1(aq)  H+(aq) + HC6H5O7-2(aq) Ka = 1.7*10-5

HC6H5O7-2(aq)  H+(aq) + C6H5O7-3(aq) Ka = 4.0*10-7

To calculate the pH of a .050M solution, assume initially that only the first ionization is important

H3C6H5O7(aq) H+(aq) + H2C6H5O7-1(aq)

I 0.050 0 0

C -x x x

E 0.050-x x x

x = 0.0057

0.0057=[H+] = [H2C6H5O7-1]

slide53
Does the second ionization have any effect?

H2C6H5O7-1(aq)  H+(aq) + HC6H5O7-2(aq) Ka = 1.7*10-5

I 0.0057 0.0057 0

C -y y y

E 0.0057-y 0.0057+y y

y = 0.000017

This value is small compared to 0.0057 (think SDs)

Total [H+] = 0.0057 + .000017 = 0.0057

which indicates the 2nd (and any subsequent) ionizations can be ignored

pH = -log(0.0057) = 2.24

weak base calculations
Weak Base Calculations
  • Weak bases remove protons from substances.
  • There is an equilibrium between the base and the resulting ions:

Example:

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq).

The base-dissociation constant, Kb, is

The larger Kb,the stronger the base.

slide55
16.8 Relationship Between Kaand Kb

• Generally only either Ka or Kb for a conjugate pair is reported in tables. If you know one you can find the other!

• Consider the following equilibria:

NH4+(aq)  NH3(aq) + H+(aq)

NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq)

• We can write equilibrium expressions for these reactions:

slide56
• If we add these equations together:

NH4+(aq)  NH3(aq) + H+(aq) Ka

NH3(aq) + H2O(l)  NH4+(aq) + OH–(aq) Kb

• The net reaction is the autoionization of water.

H2O(l)  H+(aq) + OH–(aq) Kw

• When we add equations we multiply K.

Kw =Kax Kb

Alternatively, we can express this as:

pKa + pKb = pKw= 14.00 (at 25oC)

slide57
Example 9: Calculate the molar concentration of OH-1 ions in a 0.050 M solution of hydrazine, H2NNH2,

Kb = 1.3  10-6. What is the pH of this solution?

slide58
Example 10: Although the acid dissociation constant for phenol, C6H5OH, is listed in Appendix D, the base dissociation constant for the phenolate ion, C6H5O-1 is not. (a) Explain why it is not necessary to list both. (b) Calculate the Kb for the phenolate ion. (c) Is the phenolate ion a weaker or stronger base than ammonia, NH3?

Appendix D

slide59
16.9 Acid-Base Properties of Salt Solutions

• All soluble salts are strong electrolytes.

- In solution, they exist nearly entirely of ions.

- Acid-base properties of salts are due to the reactions of their ions in solution.

• Many ions can react with water to form OH– or H+.

• This process is called hydrolysis.

remember this
Remember this?

Most anions are weak bases

Most cations are weak acids

Anions of strong acids and cations of strong bases are neutral

summary
Summary

Anions

  • Of strong acids are neutral.
    • Example: Cl- of HCl
      • Cl- + H2O  X
  • Of weak acids are basic.
    • Example: F- of HF
      • F- + H2O  HF + OH-
  • With ionizable protons are amphoteric.
    • Example: HSO4–
summary1
Summary

Cations

  • Of strong bases are neutral.
    • Example: Na+ of NaOH
      • Na+ + H2O  X
  • All other cations are weak acids
    • Example: Fe3+
slide63

Cation hydrolysis reaction:

  • Smaller and more highly charged ions = stronger (weak) acids
slide65
The pH of a solution may be qualitatively predicted:

Salts from a weak acid and weak base can be either acidic or basic.

Compare Ka of the cation and Kb of the anion

- If the Ka is larger the solution will be acidic

- If the Kb is larger the solution will be basic

For example, consider NH4CN.

The Kb of CN-1 is larger than the Ka of NH4+1

so the solution will be basic.

slide66

basic

Example 1: Predict whether aqueous solutions of the following compounds are acidic, basic or neutral.

  • NH4Br (b) FeCl3 (c) Na2CO3

(d) KClO4 (e) NaHC2O4

acidic

acidic

neutral

acidic

Ka for acid HC2O4-1 = 6.410-5

Kb for base HC2O4-1 = 1.710-13

slide67
Example 2: Using data from Appendix D, calculate [OH-1] and pH for the following solution 0.10 M NaCN

Ka for HCN (given in Appendix D) = 4.9x10-10

slide68
16.10 Acid-Base Behavior and Chemical Structure

Acidity is directly related to the strength of attraction for a pair of electrons to a central atom.

4 situations to consider:

  • Ions

Ionic Charge and Size

When comparing ions of similar structure:

More positive ions are stronger acids.

tie breaker: Smaller ion is stronger acid

Acid strength: Na+ < Ca2+ < Cu2+ < Al3+

PO43- < HPO42- < H2PO4- < H3PO4

slide69
Example 3: Predict which member of each pair produces the more acidic aqueous solution:

(a) K+1 or Cu+2 (b) Fe+2 or Fe+3 (c) Al+3 or Ga+3

  • Cu+2 has higher charge (and K+1 is neutral)

(b) Fe+3 has higher charge

(c) Al+3 has a smaller size

slide70
Binary Acids:

Bond Polarity (Electronegativity) & Strength

  • The H–X bond strength is important in determining relative acid strength in any groupin the periodic table.
    • The weaker the bond the easier it will break
    • The H–X bond strength tends to decrease down a group - acid strength increases down a group
  • H–X bond polarity is important in determining relative acid strength in any period of the periodic table.
    • The H-X bond polarity tends to increase across a period - acid strength increases (from left to right) across a period
slide71

larger atom

weaker bond

stronger acid

more polar bond – stronger acid

slide72
3. Oxyacids (Acids with oxygen) with different central atoms

Generally, the larger the electronegativity of the central atom the stronger the acid.

  • The stronger the pull on electrons the less tightly the H is held

Acid Strength: H3BO3 < H2CO3 < HNO3

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The higher EN of central atom means more electron density shift away from H - H is easier to remove – stronger acid

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4. Oxyacids with the same central atom

Generally, the more oxygens attached to the central atom the stronger the acid.

  • The more atoms pulling on electrons the less tightly the H is held

Acid Strength: HClO < HClO2 < HClO3 < HClO4

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More O atoms means more electron density shift away from H and H is easier to remove – stronger acid

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Example 4: Explain the following observations:

(a) HNO3 is a stronger acid than HNO2

(b) H2S is a stronger acid than H2O

(c) H2SO4 is a stronger acid than HSO4-1

(d) H2SO4 is a stronger acid than H2SeO4

(e) CCl3COOH is a stronger acid than CH3COOH

(a) more oxygen atoms - electron density shifts away from H bond - easier to remove H

(b) bond strength decreases down a group - H easier to remove from S

(c) More positive ion is stronger

(d) S has higher electronegativity than Se – pulls electrons from H – easier to remove H

(e) the more electronegative 3 Cl atoms (as opposed to the 3 H atoms) pull electron density more weakens the O-H bond and makes H easier to remove

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16.11 Lewis Acids and Bases

• A Brønsted-Lowry acid is a proton donor.

• Lewis proposed a new definition of acids and bases that emphasizes the shared electron pair.

• A Lewis acid is an electron pair acceptor.

• A Lewis base is an electron pair donor.

• Note: Lewis acids and bases do not need to contain protons.

• Therefore, the Lewis definition is the most general definition of acids and bases.

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• What types of compounds can act as Lewis acids?

• Lewis acids must have a vacant orbital (into which the electron pair can be donated).

• Lewis acids sometimes have an incomplete octet (e.g., BF3).

• Transition-metal ions can be Lewis acids (empty d orbitals)

• Compounds with multiple bonds can act as Lewis acids.

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Example 5: Identify the Lewis acid and Lewis base in each of the following reactions:

(a) Fe(ClO4)3 + 6 H2O  Fe(H2O)6+3 + 3 ClO4-1

(b) CN-1 + H2O  HCN + OH-1

(c) (CH3)3N + BF3 (CH3)NBF3

(d) HIO + NH2-1 NH3 + IO-1

Acid Base

H donor H acceptor e acceptor e donator empty orbitals/mult. bonds has lone pairs

incomplete octet/cation often contains N

(a) Fe(ClO4)3 or Fe+3 H2O

(b) H2O CN-1

(c) BF3 (CH3)3N

(d) HIO NH2-1

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