chapter fifteen 1 hall © 2005 prentice hall © 2005 general chemistry 4 th edition, hill, petrucci,...

Chapter Fifteen

1

Prentice Hall © 2005Hall © 2005General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

Acids, Bases, andAcid–Base Equilibria

Chapter FifteenChapter Fifteen

Chapter Fifteen

2


The Brønsted–Lowry Theory

• Arrhenius theory: an acid forms H+ in water; and a base forms OH– in water.

• But not all acid–base reactions involve water, and many bases (NH3, carbonates) do not contain any OH–.

• Brønsted–Lowry theory defines acids and bases in terms of proton (H+) transfer.

• A Brønsted–Lowry acid is a proton donor.

• A Brønsted–Lowry base is a proton acceptor.

• The conjugate base of an acid is the acid minus the proton it has donated.

• The conjugate acid of a base is the base plus the accepted proton.

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Ionization of HCl

HCl acts as an acid by donating

H+ to H2O

H2O is a base in this reaction because it

accepts the H+ Conjugate acid of H2O

Conjugate base of HCl

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Ionization of Ammonia

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Water Is Amphiprotic

H2O acts as an acid when it donates H+, forming the

conjugate base ___

H2O acts as a base when it accepts H+, forming the

conjugate acid ___

Amphiprotic: Can act as either an acid or as a base

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Example 15.1 Identify the Brønsted–Lowry acids and bases and their

conjugates in:

(a) H2S + NH3 NH4+ + HS–

(b) OH– + H2PO4– H2O + HPO4

2–

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Ka and Kb

The equilibrium constant for a Brønsted acid is represented by Ka, and that for a base is represented by Kb.

CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO–(aq)

[H3O+][CH3COO–]Ka = ––––––––––––––––– [CH3COOH]

[NH4+][OH–]

Kb = ––––––––––––– [NH3]

NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

Notice that H2O is not included in either

equilibrium expression.

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Strength of Conjugate Acid–Base Pairs

• A stronger acid can donate H+ more readily than a weaker acid.

• The stronger an acid, the weaker is its conjugate base.

• The stronger a base, the weaker is its conjugate acid.

• An acid–base reaction is favored in the direction from the stronger member to the weaker member of each conjugate acid–base pair.

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The stronger the acid …

… the weaker the conjugate

base.

And the stronger the base …

… the weaker the conjugate

acid.

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Acid/Base Strength and Direction of Equilibrium

• In Table 15.1, HBr lies above CH3COOH in the acid column.

• Since HBr is a stronger acid than CH3COOH, the equilibrium for the reaction:

lies to the left.

Weaker acid Stronger acid

• We reach the same conclusion by comparing the strengths of the bases (right column of Table 15.1).

• CH3COO– lies below Br– ; CH3COO– is the stronger base:

Weaker base Stronger base

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Strong Acids

• The “strong” acids—HCl, HBr, HI, HNO3, H2SO4, HClO4—are considered “strong” because they ionize completely in water.

• The “strong” acids all appear above H3O+ in Table 15.1.

• The strong acids are leveled to the same strength—to that of H3O+—when they are placed in water.

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Periodic Trends in Acid Strength

• The greater the tendency for HX (general acid) to transfer a proton to H2O, the more the forward reaction is favored and the stronger the acid.

• A factor that makes it easier for the H+ to leave will increase the strength of the acid.

• Acid strength is inversely proportional to H—X bond-dissociation energy. Weaker H—X bond => stronger acid.

• Acid strength is directly proportional to anion radius. Larger X radius => stronger acid.

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Periodic Trends in Acid Strength

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02.11.01 12:28 AM6 16.3 Acids - Base i. Salt Solution ii. Characteristics of Acid-Base

I. AcidI. Acid--Base Strength: Binary HydridesBase Strength: Binary Hydrides

Bond Strength; Bond Strength; Stronger the bond strength the weaker the acid character.

1. Size increase of X, leads to longer and weaker H-X bond, so H+ is release more easily. (prominent down PT)

2. EN of X: As X become more electronegative, the H-X becomes more polar, and H+ is more easily release. (prominent across PT)

HH--FF HH--ClCl HH--Br Br HH--II569 kJ/mol569 kJ/mol 431 kJ/mol431 kJ/mol 368 kJ/mol368 kJ/mol 297 kJ/mol297 kJ/mol

BondStrength from atomic radius (orbital overlap) Increasing atomic size, Increasing acid strengthIncreasing atomic size, Increasing acid strength

NaNa--HH CHCH44 NHNH33 HFHF1.01.0 2.52.5 3.13.1 4.14.1

Electro-negativity

Increasing EN, Increasing Acid StrengthIncreasing EN, Increasing Acid Strength

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AcidAcid--Base Strength: Binary Base Strength: Binary hydrideshydrides

Trend of Acidity for Binary compoundsBond Strength: Atomic size (Down PT)

Bond polarity: Electronegative (across PT)

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Strength of Oxoacids

Acid strength increases with the electronegativity of the central atom, and with the number of terminal oxygen atoms.

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Hypochlorous acid HOCl (OH)Cl pK a = 7.5

chlorous acid HClO2 (OH)ClO pKa= 2.0

chloric acid HClO3 (OH)ClO2 pKa= -3

perchloric acid HClO4 (OH)ClO3 pKa= - 8

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02.11.01 12:28 AM11 16.3 Acids -Base i. Salt Solution ii. Characteristics of Acid-Base

2. 2. AcidAcid--Base Strength: Base Strength: Oxyacids and EN (Example)Oxyacids and EN (Example)

Variation on the identity of third atomElectronegativity

Which is the stronger acid ?

H2SO4because EN (S) > EN (Se).

(HO)mXOn

(HO)2SO2

(HO)2SeO2

OO Se

HO

HO

OO S

HO

HO

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Strength of Carboxylic Acids

• Carboxylic acids all have the –COOH group in common.

• Differences in acid strength come from differences in the R group attached to the carboxyl group.

• In general, the more that electronegative atoms appear in the R group, the stronger is the acid.

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Aspirin: acetylsalicylic acid

Vitamin C: Ascorbic acid

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Respiratory System - Ventilation can affect carbon dioxide, and therefore carbonic acid

1. Action of Carbonic Anhydrase

CO2 + H2 O <===> H2 CO3

Carbon Dioxide Water Carbonic anhydrase Carbonic acid

2. Effect of Respiration on pH

resp up---->"'blows off" CO2 ---->CO2 down ----> H2CO3 down----> pH up

resp down----> CO2 accumulates ----> CO2 up----> H2CO3 up----> pH down

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Example 15.2Select the stronger acid in each pair:

(a) nitrous acid, HNO2, and nitric acid, HNO3

(b) Cl3CCOOH and BrCH2COOH

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Strengths of Amines as Bases

• Aromatic amines are much weaker bases than aliphatic amines.

• This is due in part to the fact that the electrons in the benzene ring of an aromatic molecule are delocalized and can involve the nitrogen atom’s lone-pair electrons in the resonance hybrid.

• As a result, the lone-pair electrons are much less likely to accept a proton.

• Electron-withdrawing groups on the ring further diminish the basicity of aromatic amines relative to aniline.

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Example 15.3Select the weaker base in each pair:

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25



AcidAcid--Base and Salt solutionsBase and Salt solutionsConjugates:

Strong Acid - Base CA C+ + A (100%)

Weak Acids - Base CA C+ + A - < (100%)A- is a base (proton acceptor) A- + H2O AH + OH-

C+ is an acid (proton donor) C+ + H2O *C + H3O+

C = NH4+ or polyatomic with H in formula i.e., H2PO3

-2, HSO4-, HCO3

-

(i) Salts from Strong base and strong acid:(i) Salts from Strong base and strong acid:

Consider NaOH and HCl Consider NaOH and HCl the salt NaClthe salt NaCl no pH = 7no pH = 7

(ii) Salts from strong base and weak acid:(ii) Salts from strong base and weak acid:

Consider NaOH and HClO Consider NaOH and HClO the salt NaClO the salt NaClO pH > 7pH > 7

(iii) Salts from weak base and a strong acid:(iii) Salts from weak base and a strong acid:

Consider NH4OHand HNOConsider NH4OHand HNO33 the salt NHthe salt NH44NONO3 3 pH < 7pH < 7

(iv) Salts from a weak base and a weak acid:(iv) Salts from a weak base and a weak acid:

Consider NHConsider NH44OH and HClOOH and HClO the salt NHthe salt NH44ClO ClO pH = ?pH = ?

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Self-Ionization of Water

• Even pure water conducts some electricity. This is due to the fact that water self-ionizes:

• The equilibrium constant for this process is called the ion product of water (Kw).

• At 25 °C, Kw = 1.0 x 10–14 = [H3O+][OH–]• This equilibrium constant is very important because it

applies to all aqueous solutions—acids, bases, salts, and nonelectrolytes—not just to pure water.

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The pH Scale

• Concentration of H3O+ can vary over a wide range in aqueous solution, from about 10 M to about 10–14 M.

• A more convenient expression for H3O+ is pH.

pH = –log [H3O+] and so [H3O+] = 10–pH

• The “negative logarithm” function of pH is so useful that it has been applied to other species and constants.

pOH = –log [OH–] and so [OH–] = 10–pOH

pKw = –log Kw

• At 25 °C, pKw = 14.00

• pKw = pH + pOH = 14.00

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The pH Scale

Since pH is a logarithmic scale, cola drinks (pH about 2.5) are about ____ times as acidic as

tomatoes (pH about 4.5)

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Example 15.4

By the method suggested in Figure 15.5, a student determines the pH of milk of magnesia, a suspension of solid magnesium hydroxide in its saturated aqueous solution, and obtains a value of 10.52. What is the molarity of Mg(OH)2 in its saturated aqueous solution? The suspended, undissolved Mg(OH)2(s) does not affect the measurement.

Example 15.5

A Conceptual Example

Is the solution 1.0 x 10–8 M HCl acidic, basic, or neutral?

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30


Equilibrium in Solutions of Weak Acids and Weak Bases

These calculations are similar to the equilibrium calculations performed in Chapter 14.

1. An equation is written for the reversible reaction.

2. Data are organized, often in an ICE format.

3. Changes that occur in establishing equilibrium are assessed.

4. Simplifying assumptions are examined (the “5% rule”).

5. Equilibrium concentrations, equilibrium constant, etc. are calculated.

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Chapter Fifteen

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Types of Weak Bases

• C2H3O2- + H2O = HC2H3O2 + OH-

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Kw = Ka Kb

Kw / Ka = Kb

1.0 X 10-14 / 1.8 X 10-5 = 5.6 X 10-10

What is the pH of a 1.0 X 10-2 M NaC2H3O2 solution?

C2H3O2- + H2O = HC2H3O2 + OH-

Initial 1.0 X 10-2 0 0*

Change -x +x +x

Equilibrium

1.0 X 10-2 – x X X

Since [ C2H3O2- ]initial > 400 X Kb, we can neglect x compared to 1.0 X 10-2

5.6 X 10-10 = x2 / 1.0 X 10-2

x = [OH-] = 2.4 X 10-6 M => [H+] = Kw / [OH-] = 4.2 X 10-9

pH = - Log 4.2 X 10-9 = 8.38 (you see, I told you it was a base).

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0 2 . 1 1 . 0 1 1 2 : 2 8 A M4 1 6 . 3 A c i d s - B a s e i . S a l t S o l u t i o n i i . C h a r a c t e r i s t i c s o f A c i d - B a s e

H y d r o l y s i sH y d r o l y s i s : : S a l t s f r o m S a l t s f r o m w e a k b a s ew e a k b a s e a n d a n d w e a k a c i d sw e a k a c i d s

W h a t i s t h e p H o f a s a l t s o l u t i o n w h e n 3 7 g o f N H 4 F i s d i s s o l v e d i n 1 . 0 0 L o f w a t e r .

H 3 O Calcu latio n :

NH 4 NH 3 H 3 O +

i 2 M 0 1 • 1 0 -7

- x + x + x

e 2 M - x + x 1 • 1 0 -7 + x

11-b

-

1 0-a

+334

4

10 1. 47 K OH HF F

10 5. 56 K OH NH NH

F NH NH4F

OH Calcu latio n :

F H 2 O HF OH

i 2 M Ex cess 0 1 • 1 0 -7

- x - x + x + x

e 2 M - x Ex cess x 1 • 1 0 -7 + x

K a =x 2

2 M 5 . 5 6 • 1 0 -1 0 . . . . x = [ H 3 O + ] = 3 . 3 3 • 1 0 -5 M

K b =x 2

2 M 1 . 4 7 • 1 0 -11 . . . . x = [ O H - ] = 5 . 4 2 • 1 0 -6 M

[ OH - ] + [ H 3 O + ] [ H 2 O] 3. 33 • 10 -5 M - 5. 42 • 10 -6 M

Ac i di c [ H 3 O + ] E x cess = 2. 79 • 10 -5 M

Chapter Fifteen

35


Example 15.6Ordinary vinegar is approximately 1 M CH3COOH and as shown in Figure 15.6, it has a pH of about 2.4. Calculate the expected pH of 1.00 M CH3COOH(aq), and show that the calculated and measured pH values are in good agreement.

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Example 15.7What is the pH of 0.00200 M ClCH2COOH(aq)?

Chapter Fifteen

37


Example 15.8What is the pH of 0.500 M NH3(aq)?

Example 15.9The pH of a 0.164 M aqueous solution of dimethylamine is 11.98. What are the values of Kb and pKb? The ionization equation is

(CH3)2NH + H2O (CH3)2NH2+ + OH– Kb = ?

Dimethylamine Dimethylammonium ion

Example 15.10 A Conceptual ExampleWithout doing detailed calculations, indicate which solution has the greater [H3O+], 0.030 M HCl or 0.050 M CH3COOH.

Chapter Fifteen

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Polyprotic Acids

• A monoprotic acid has one ionizable H atom per molecule.

• A polyprotic acid has more than one ionizable H atom per molecule.– Sulfuric acid, H2SO4 Diprotic

– Carbonic acid, H2CO3 Diprotic

– Phosphoric acid, H3PO4 Triprotic

• The protons of a polyprotic acid dissociate in steps, each step having a value of Ka.

• Values of Ka decrease successively for a given polyprotic acid. Ka1 > Ka2 > Ka3 , etc.

• Simplifying assumptions may be made in determining the concentration of various species from polyprotic acids.

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Example 15.11

Calculate the following concentrations in an aqueous solution that is 5.0 M H3PO4:

(a) [H3O+] (b) [H2PO4–] (c) [HPO4

2–] (d) [PO43–]

Example 15.12

What is the approximate pH of 0.71 M H2SO4?

Chapter Fifteen

40


Ions as Acids and Bases

• HCl is a strong acid, therefore Cl– is so weakly basic in water that a solution of chloride ions (such as NaCl) is virtually neutral.

• Acetic acid, CH3COOH, is a weak acid, so acetate ion, CH3COO–, is significantly basic in water.

• A solution of sodium acetate (which dissociates completely into sodium and acetate ions in water) is therefore slightly basic:

CH3COO– + H2O CH3COOH + OH–

Chapter Fifteen

41


Carbonate Ion as a Base

A carbonate ion accepts a proton from water, leaving behind an OH–

and making the solution basic.

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42


Ions as Acids and Bases (cont’d)

• Salts of strong acids and strong bases form neutral solutions: NaCl, KNO3

• Salts of weak acids and strong bases form basic solutions: KNO2, NaClO

• Salts of strong acids and weak bases form acidic solutions: NH4NO3

• Salts of weak acids and weak bases form solutions that may be acidic, neutral, or basic; it depends on the relative strengths of the cations and the anions: NH4NO2, CH3COONH4.

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Aqueous Cations

[Fe(H2O)6]+3 + H2O

[Fe(H2O)5OH]+2 + H3O+

Chapter Fifteen

44


Example 15.13 A Conceptual Example(a) Is NH4I(aq) acidic, basic,or neutral? (b) What conclusion can you draw from Figure 15.8d about the equilibrium constants for the hydrolysis reactions in CH3COONH4(aq)?

Chapter Fifteen

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Ions as Acids and Bases (cont’d)

• In order to make quantitative predictions of pH of a salt solution, we need an equilibrium constant for the hydrolysis reaction.

• The relationship between Ka and Kb of a conjugate acid–base pair is:

Ka x Kb = Kw

• If instead we have values of pKa or pKb:

pKa + pKb = pKw = 14.00 (at 25 °C)

Chapter Fifteen

46


Example 15.14Calculate the pH of a solution that is 0.25 M CH3COONa(aq).

Example 15.15What is the molarity of an NH4NO3(aq) solution that has a pH

= 4.80?

Chapter Fifteen

47


The Common Ion Effect

• Consider a solution of acetic acid.

• If we add acetate ion as a second solute (i.e., sodium acetate), the pH of the solution increases:

LeChâtelier’s principle: What happens to [H3O+] when the

equilibrium shifts to the left?

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The Common Ion Effect (cont’d)

The common ion effect is the suppression of the ionization of a weak acid or a weak base by the presence of a common ion from a strong electrolyte.

Acetic acid solution at equilibrium: a

few H3O+ ions and a few CH3COO– ions

When acetate ion is added, and equilibrium

reestablished: more acetate ions, fewer

H3O+ ions

Chapter Fifteen

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HC2H3O2 = H+ + C2H3O2-

We calculated before that a 1.0 X 10-2 M solution would have [H+] = 4.2 X 10-4 M and pH = 3.37How does Le Chatelier suggest the above equilibrium would shift if NaC2H3O2 was added?

How would the pH change?Say we added enough NaC2H3O2 to make the solution 1.0 X 10-2 M in

C2H3O2-. What would the pH be

Initial 1.0 X 10-2 0 * 1.0 X 10-2

Change -x +x +x

Equilibrium

1.0 X 10-2 – x X 1.0 X 10-2 + x

Ka = x (1.0 X 10-2 + x) Assume x is negligible compared to 1.0 X 10-2 M

(1.0 X 10-2 – x )

x = 1.8 X 10-5 = [H+] => pH = 4.74

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Example 15.16

Calculate the pH of an aqueous solution that is both 1.00 M CH3COOH and 1.00 M CH3COONa.

Chapter Fifteen

51


Buffer Solutions

• Many industrial and biochemical reactions—especially enzyme-catalyzed reactions—are sensitive to pH.

• To work with such reactions we often need a solution that maintains a nearly-constant pH.

• A buffer solution is a solution that changes pH only slightly when small amounts of a strong acid or a strong base are added.

• A buffer contains significant concentrations of both

– a weak acid and its conjugate base, or

– a weak base and its conjugate acid.

Chapter Fifteen

52


Buffer Solutions (cont’d)

• The acid component of the buffer neutralizes small added amounts of OH–, forming the weaker conjugate base which does not affect pH much:

HA + OH– H2O + A–

• The base component neutralizes small added amounts of H3O+, forming the weaker conjugate acid which does not affect pH much.

A– + H3O+ H2O + HA

• Pure water does not buffer at all …

Chapter Fifteen

53


Pure water increases in pH by about 5 pH units when the OH– is added, and decreases by about 5

pH units when the H3O+ is added.

In contrast, the same amounts of OH– and H3O+ added to a buffer solution barely change the pH.

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An Equation for Buffer Solutions

• In certain applications, there is a need to repeat the calculations of the pH of buffer solutions many times. This can be done with a single, simple equation, but there are some limitations.

• The Henderson–Hasselbalch equation:

• To use this equation, the ratio [conjugate base]/[weak acid] must have a value between 0.10–10 and both concentrations must exceed Ka by a factor of 100 or more.

[conjugate base]pH = pKa + log –––––––––––––– [weak acid]

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55


Comparing three weak acids

HF Ka = 6.8 X 10-4 pKa = 3.17

HC2H3O2 1.8 x 10-5 4.74

HCN 4.9 X 10-10 9.31

As you can see, the stronger the acid, the smaller pKa. You will have a similar trend with pKb.

Chapter Fifteen

56


Buffer Capacity andBuffer Range

• There is a limit to the ability of a buffer solution to neutralize added acid or base.

• This buffer capacity is reached before either buffer component has been consumed.

• In general, the more concentrated the buffer components in a solution, the more added acid or base the solution can neutralize.

• As a rule, a buffer is most effective if the concentrations of the buffer acid and its conjugate base are equal or nearly so.

• Therefore, a buffer is most effective when the desired pH of the buffer is very near pKa of the weak acid of the buffer.

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Human blood is a buffered solution

Source: Visuals Unlimited

Chapter Fifteen

58


Molecular model: HC2H3O2, C2H3O2-

Chapter Fifteen

59


Example 15.17

A buffer solution is 0.24 M NH3 and 0.20 M NH4Cl. (a) What is the pH of this buffer? (b) If 0.0050 mol NaOH is added to 0.500 L of this solution, what will be the pH?

Example 15.18

What concentration of acetate ion in 0.500 M CH3COOH(aq) produces a buffer solution with pH = 5.00?

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Acid–Base Indicators

• An acid–base indicator is a weak acid or base.

• The acid form (HA) of the indicator has one color, the conjugate base (A–) has a different color. One of the “colors” may be colorless.

• In an acidic solution, [H3O+] is high. Because H3O+ is a common ion, it suppresses the ionization of the indicator acid, and we see the color of HA.

• In a basic solution, [OH–] is high, and it reacts with HA, forming the color of A–.

• Acid–base indicators are often used for applications in which a precise pH reading isn’t necessary.

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61


Different indicators have different values of Ka, so they exhibit color changes at different values of pH …

Chapter Fifteen

62


Figure 8.10: The pH curve for the titration of 50.0

Chapter Fifteen

63


Table 8.4: A Summary of Various Points in the Titration of a

Triprotic Acid

Chapter Fifteen

64


Figure 8.11: A summary of the important equilibria at various points

in the titration of a triprotic acid

Chapter Fifteen

65


Example 15.19 A Conceptual Example Explain the series of color changes of thymol blue indicator

produced by the actions pictured in Figure 15.14:

(a) A few drops of thymol blue are added to HCl(aq).

(b) Some sodium acetate is added to solution (a).

(c) A small quantity of sodium hydroxide is added to solution (b).

(d) An additional, larger quantity of sodium hydroxide is added to solution (c).

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66


Neutralization Reactions

• At the equivalence point in an acid–base titration, the acid and base have been brought together in precise stoichiometric proportions.

• The endpoint is the point in the titration at which the indicator changes color.

• Ideally, the indicator is selected so that the endpoint and the equivalence point are very close together.

• The endpoint and the equivalence point for a neutralization titration can be best matched by plotting a titration curve, a graph of pH versus volume of titrant.

Chapter Fifteen

67


Titration Curve, Strong Acid with Strong Base

Bromphenol blue, bromthymol blue, and

phenolphthalein all change color at very

nearly 20.0 mL

At about what volume would we see a color

change if we used methyl violet as the indicator?

Chapter Fifteen

68


Example 15.20Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M HCl with 0.500 M NaOH:

H3O+ + Cl– + Na+ + OH– Na+ + Cl– + 2 H2O

(a) before the addition of any NaOH

(b) after the addition of 10.00 mL of 0.500 M NaOH

(c) after the addition of 20.00 mL of 0.500 M NaOH

(d) after the addition of 20.20 mL of 0.500 M NaOH

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69


Titration Curve, Weak Acid with Strong Base

The equivalence-point pH is NOT 7.00 here.

Why not??

Bromphenol blue was ok for the strong acid/strong

base titration, but it changes color far too early

to be useful here.

Chapter Fifteen

70


Figure 8.4: The pH curves for the titrations of 50.0

Chapter Fifteen

71


Figure 8.6: The indicator phenolphthalein is pink

in basic solution and colorless in acidic solution.

Chapter Fifteen

72


Example 15.21Calculate the pH at the following points in the titration of 20.00 mL of 0.500 M CH3COOH with 0.500 M NaOH:

CH3COOH + Na+ + OH– Na+ + CH3COO– + H2O

(a) before the addition of any NaOH

(b) after the addition of 8.00 mL of 0.500 M NaOH

(c) after the addition of 10.00 mL of 0.500 M NaOH

(d) after the addition of 20.00 mL of 0.100 M NaOH

(e) after the addition of 21.00 mL of 0.100 M NaOH

Chapter Fifteen

73


Example 15.22 A Conceptual Example

This titration curve shown in Figure 15.18 involves 1.0 M solutions of an acid and a base. Identify the type of titration it represents.

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74


Lewis Acids and Bases

• In organic chemistry, Lewis acids are often called electrophiles (“electron-loving”) and Lewis bases are often called nucleophiles (“nucleus-loving”).

• There are reactions in nonaqueous solvents, in the gaseous state, and even in the solid state that can be considered acid–base reactions which Brønsted–Lowry theory is not adequate to explain.

• A Lewis acid is a species that is an electron-pair acceptor and a Lewis base is a species that is an electron-pair donor.

Sulfur accepts an electron pair from the oxygen of CaOCaO(s) + SO2(g) CaSO3(s)

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75


Cumulative Example

A 0.0500 M aqueous solution of cyanoacetic acid, CNCH2COOH, has a freezing point of –0.11 °C. Calculate the freezing point of a 0.250 M aqueous solution of cyanoacetic acid.

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