acids and bases

Acids and Bases

…and Airman 1st class HCl flew back over the front lines, confident

that he had neutralized the enemy’s strongest base.

NaOH into HCl

0

2

4

6

8

10

12

14

0 10 20 30

NaOH into HC2H3O2

0

2

4

6

8

10

12

14

0 10 20 30

NaOH into NH4Cl

0

2

4

6

8

10

12

14

0 10 20 30

0

2

4

6

8

10

12

14

0 10 20 30

Arrhenius acids and bases

• Substances that ionize in water to form H+ ions are acids.

• Substances that ionize in water to form OH- ions are bases.




• Remember strong and weak electrolytes?




• Remember strong and weak electrolytes? Strong acids and bases ionize completely Weak acids and bases ionize partially.

Examples?

• Acids • Bases

Properties of Acids and Bases• Acids • Bases

Properties of Acids and Bases• AcidsAre electrolyte solutions• Make ions in solution• Affect indicators• Have low pH• Taste sour• Neutralize basesCan cause serious burnsCorrode reactive metalsHave more H+ than OH-

(in solution)

• BasesAre electrolyte solutions• Make ions in solution• Affect indicators• Have high pH• Taste bitter• Neutralize acidsCan cause serious burnsCorrode aluminum onlyHave more OH- than H+ (in

solution)

Properties of Both• AcidsAre electrolyte solutions• Make ions in solution• Affect indicators• Have low pH• Taste sour• Neutralize basesCan cause serious burnsCorrode reactive metalsHave more H+ than OH-

(in solution)

• BasesAre electrolyte solutions• Make ions in solution• Affect indicators• Have high pH• Taste bitter• Neutralize acidsCan cause serious burnsCorrode aluminum onlyHave more OH- than H+ (in

solution)

BrØnsted-Lowry Definition

• Substances that donate a proton (H+ ion) in a reaction are acids.

• Substances that accept a proton (H+ ion) are bases.

Lewis Definition

• Substances that accept an electron pair in a reaction are acids.

• Substances that donate an electron pair are bases.

Conjugates

• After an acid has donated a proton, the rest of the species is the conjugate base.

HAA- + H+

• After a base has accepted a proton, the resulting species is the conjugate acid.

B- + H+ HB

What is the conjugate base of…

• HCl• CH3COOH• H2SO4

• HSO4-

• H2O• NH4

+

• NH3


ACID (loses H+ to form its) Conjugate base• HCl• CH3COOH• H2SO4

• HSO4-

• H2O• NH4

+

• NH3


ACID (loses H+ to form its) Conjugate base• HCl ( H+ and) Cl- • CH3COOH• H2SO4

• HSO4-

• H2O• NH4

+

• NH3


ACID (loses H+ to form its) Conjugate base• HCl ( H+ and) Cl- • CH3COOH( H+ and) CH3COO-

• H2SO4 ( H+ and) HSO4-

• HSO4- ( H+ and) SO4

-2

• H2O ( H+ and) OH-

• NH4+ ( H+ and) NH3

• NH3 ( H+ and) NH2-

What is the conjugate acid of…

• NO3-

• C2O4-2

• HPO4-2

• HSO4-

• H2O

• F-


Base (gains H+ to form its) Conjugate acid

• NO3-

• C2O4-2

• HPO4-2

• HSO4-

• H2O

• F-



• NO3- (+H+ ) HNO3

• C2O4-2

• HPO4-2

• HSO4-

• H2O

• F-



• NO3- (+H+ ) HNO3

• C2O4-2 (+H+ ) HC2O4

-

• HPO4-2 (+H+ ) H2PO4

-

• HSO4- (+H+ ) H2SO4

• H2O (+H+ ) H3O+

• F- (+H+ ) HF

Nomenclature

• If the anion name then the acid name

• ends in…. is…

Fill in the blanks

• HCl is _____________acid

• HClO4 is _____________acid

• HClO3 is _____________acid

• HClO2 is _____________acid

• HClO is _____________acid

Fill in the blanks

• HCl is _____________acid

• HClO4 is _____________acid

• HClO3 is _____________acid

• HClO2 is _____________acid

• HClO is _____________acid

Hydrogen chlorideH

ydro

gen

chlo

rate

Hydrogen perchlorate

Hydrogen hypochlorite

Hydrogen chlorite

Nomenclature

• If the anion name then the acid name

• ends in…. is…

• --ide Hydro___ic acid

• (hypo--) --ite Hypo___ous acid

• --ite ___ous acid

• --ate ___ic acid

• (per--) –ate Per ___ic acid

Fill in the blanks

• HNO3 is _____________acid

• HIO4 is _____________acid

• H2CO3 is _____________acid

• H3PO3 is _____________acid

• HBrO is _____________acid

Fill in the blanks

• _____________is hydrocyanic acid

• _____________ is perbromic acid

• _____________ is phosphoric acid

• _____________ is sulfurous acid

• _____________ is hypoiodous acid

Show the conjugate acid/base pairs in the following reactions.

• C2O4-2 + H3O+ HC2O4

- + H2O

• CH3COOH + NH2- NH3 + CH3COO-


• C2O4-2 + H3O+ HC2O4

- + H2O


Acid

Base

ConjugateBase

Conjugate Acid


• C2O4-2 + H3O+ HC2O4

- + H2O


Acid

Acid

Base

Base

ConjugateBase

Conjugate Acid

ConjugateBase

Conjugate Acid

Water dissociates!

H2O H+ + OH-

This makes an equilibrium for water where:

Kw=[H+][OH-]=1 x 10-14

(at 25oC)

Water dissociates!

H2O H+ + OH-


Kw=[H+][OH-]=1 x 10-14

(at 25oC)

(endothermic or exothermic?)

(Does Kw increase or decrease at higher T?)

Water dissociates!

H2O H+ + OH-


Kw=[H+][OH-]=1 x 10-14

(at 25oC)

(endothermic or exothermic?)

(Does Kw increase or decrease at higherT?)

[H+] is inversely related to [OH-]

• When [H+] increases, [OH-] decreases in a water solution, and vice versa.

[H+] is inversely related to [OH-]

• When [H+] decreases, [OH-] increases in a water solution, and vice versa.

(Why?)

pH

• The basic (and acidic) definitions are:

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

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

Kw=[H+][OH-]=1 x 10 -14 (at 25oC)

pH + pOH = 14 (at 25oC)

pH practice

• If pH is 3.38….

1) What is the pOH?

pH practice

• If pH is 3.38….

1) What is the pOH? 14-pH= 10.62

pH practice

• If pH is 3.38….


2) What is [H+]?

pH practice

• If pH is 3.38….


2) What is [H+]? 10-3.38= 4.17 x 10-4M

pH practice

• If pH is 3.38….


2) What is [H+]? 10-3.38= 4.17 x 10-4M

3) What is [OH-]?

pH practice

• If pH is 3.38….


2) What is [H+]? 10-3.38= 4.17 x 10-4M

3) What is [OH-]? 10-10.62=2.40x10-11M and Kw/4.17x10-4M=2.40x10-11 M!

pH practice

• If [OH-]= 4.8 x 10-6 M…

pH practice

• If [OH-]= 4.8 x 10-6 M…

1) What is pOH?

pH practice

• If [OH-]= 4.8 x 10-6 M…

1) What is pOH? -log (4.8 x 10-6 )= 5.32

pH practice

• If [OH-]= 4.8 x 10-6 M…

1) What is pOH? -log (4.8 x 10-6 )= 5.32

2) What is pH?

pH practice

• If [OH-]= 4.8 x 10-6 M…

1) What is pOH? -log (4.8 x 10-6 )= 5.32

2) What is pH? 14-5.32= 8.68

pH practice

• If [OH-]= 4.8 x 10-6 M…

1) What is pOH? -log (4.8 x 10-6 )= 5.32

2) What is pH? 14-5.32= 8.68

3) What is [H+]?

pH practice

• If [OH-]= 4.8 x 10-6 M…

1) What is pOH? -log (4.8 x 10-6 )= 5.32

2) What is pH? 14-5.32= 8.68

3) What is [H+]? 10-8.68= 2.08 x 10-9 M and Kw/ 4.8 x 10-6 = 2.08 x 10-9 M !

Homework

pH problems

• Back of the forecast (3/30-4/10)

Please recall:

• Strong acids and bases dissociate completely in a water environment. Weak acids and bases do not.

• Strong acids= nitric, hydrochloric, sulfuric, hydrobromic, hydroiodic, perchloric

• Strong bases-Group 1 & 2 hydroxides—(group 2’s might not dissolve well)

Please recall:

1. What is the concentration (M) of NaOH if .35 mole NaOH is dissolved in .120 L solution?

2. What is the molarity of HCl if 12 g HCl is dissolved in .85 L of solution?

3. What is [OH-] if .35 g Ba(OH)2 is dissolved in .250 L solution?

4. What mass of H2SO4 is in 55 ml of .38 M H2SO4?

Please recall:

1. What is the concentration (M) of NaOH if .35 mole NaOH is dissolved in .120 L solution?

2. What is the molarity of HCl if 12 g HCl is dissolved in .85 L of solution?

3. What is [OH-] if .35 g Ba(OH)2 is dissolved in .250 L solution?

4. What mass of H2SO4 is in 55 ml of .38 M H2SO4?

Did you notice?

Analyze these solutions

Contents pH [H+] [OH-] pOH Acidic or Basic

1 .023 mol HCl /L

2 1.5g NaOH /L

3 ?mol LiOH/50ml 8.5

4 ?mol KOH/25ml 2.5

5 ?gHClO4/150ml .02

6 ?molBa(OH)2/L .007

Right!

Contents pH [H+] (M)

[OH-] (M)

pOH Acidic or Basic

1 .023 mol HCl /L 1.64 .023 4.3 x 10-13

12.36 Acidic

2 1.5g NaOH /L 12.57 2.7 x10-13

.0375 1.43 Basic

3 1.6 x10-7 mol LiOH/ 50ml

8.5 3.2 x10-9

3.2 x10-

6

5.5 basic

4 7.9 x10 -5 mol KOH/25ml

11.5 3.2 x10-13

3.2 x10-

3

2.5 basic

5 .30 gHClO4

/150ml1.70 .02 5.0 x10-

13

12.30 acidic

6 .0035mol Ba(OH)2/L

11.85 1.4 x10-12

.007 2.15 basic

Review question:

125 ml of a KOH solution is mixed so that the pH is 12.23

1) What is the pOH, [OH-] and [H+]?

2) What is the [KOH] ?

3) How many moles KOH was used?

4) What mass of KOH was used?

(FMKOH= 56.1 g/mol)

Review question:

125 ml of a KOH solution is mixed so that the pH is 12.23

1)pOH=1.77;[OH-]=.0170M;[H+]=5.88x10-13M

2) [KOH]=[OH-]= .0170M (it’s a strong base!)

3)moles=MxV=.0170Mx.125L=.00213mol

4) massKOH =molesKOHx FMKOH

= .00213mol x 56.1 g/mol

=.119 g

Strength of acids and bases.

• HCl

• H2CO3

• CH4


• HCl -- strong acid

• H2CO3 -- weak acid

• CH4 --so weak it’s pathetic

Strength is determined by

amount of dissociation


• HCl -- strong acid, it dissociates completely

• H2CO3 -- weak acid, it does not dissociate completely.

• CH4 --so weak it’s pathetic, it does not dissociate to any measurable extent.

• What about their conjugates?


• Cl-

• HCO3-

• CH3-


• Cl- -- pathetic base

• HCO3- -- weak base

• CH3- --so strong a base, it associates

completely with water, leaving hydroxide, a strong base.

Strength is determined by

amount of association


• Cl- -- pathetic base, it does not associate with water to any measurable extent.

• HCO3- -- weak base, it does not associate

completely with water.

• CH3- --so strong a base, it associates

completely with water, leaving hydroxide, a strong base.

• What about their conjugates?


The conjugate of a strong acid is a pathetic base

The conjugate of a weak acid is a weak base

The conjugate of a pathetic acid is a strong base


The conjugate of a strong acid is a pathetic base

The conjugate of a weak acid is a weak base

the stronger the acid, the weaker the base and vice versa

The conjugate of a pathetic acid is a strong base

When comparing weak acids and bases…

• For a weak acid,HA (aq)H+ (aq) +A- (aq)

Ka=[H+][A-]/[HA] • For a weak base,

B- (aq) +H2O (l) HB (aq) +OH- (aq)

Kb=[HB][OH-]/[B-]• The position of the equilibrium is the

strength of the acid or base.

Write the reaction and equilibrium constant expression for:

• Ammonia associating with water

• Ammonium dissociating in water



NH3(aq) + H2O (l) NH4+ (aq) + OH- (aq)


NH4+ (aq) H+ (aq) + NH3 (aq)



NH3(aq) + H2O (l) NH4+ (aq) + OH- (aq)

Kb = [NH4+][OH-]/[NH3]


NH4+ (aq) H+ (aq) + NH3 (aq)

Ka = [H+ ][NH3]/[NH4+]

And…

Kb Ka = ([NH4+][OH-][H+ ][NH3])/([NH4

+][NH3])

And…

Kb Ka = ([NH4+][OH-][H+ ][NH3])/([NH4

+][NH3])

= [OH-][H+ ]=Kw

This is true for any conjugate pair in a water solution

For example:

Ka=1.8x10-5 for acetic acid

Ka=6.5x10-5 for benzoic acid

For example:



• Benzoic acid is a stronger acid.

For example:




• .10M solutions of each would have a lower pH for benzoic acid.

For example:




• .10M solutions of each would have a lower pH for benzoic acid.

• An acetic acid solution could have a lower pH, but only at a higher concentration.

?pH

• What is the pH of a .25 M acetic acid solution?

?pH


It’s a weak acid! [H+] is nowhere near .25M

• Use the ICE method.

CH3COOH CH3COO- + H+

I) .25 M 0M 0M

?pH





I) .25 M 0M 0M

C) -x +x +x

?pH





I) .25 M 0M 0M

C) -x +x +x

E) .25-xM xM xM

• Ka= [CH3COO-][H+] / [CH3COOH]


=(xM)(xM)/(.25-xM)=1.8 x 10-5


=(xM)(xM)/(.25-xM)=1.8 x 10-5

The two sides are equal where x=[H+]=.0021M, pH=2.68

For a weak acid solution…

• For a weak acid solution …



• For a WEAK acid solution…




• x is very small. The .25 M doesn’t change very much. Try it.





• Ka=(x)(x)/(.25M)=1.8 x 10-5





• Ka=(x)(x)/(.25M)=1.8 x 10-5

The two sides are equal where x=[H+]=√[(Ka)(.25)]=.0021M, pH=2.67

?pH


• What is the pH of .20 M carbonic acid?

• What is the pH of a 1.2 M NH3 solution?

• What is the pH of .80 M acetic acid which is also .35 M sodium acetate?

• What is the pH of a .30 M NH3 solution that is also 1.0 M ammonium chloride?

Titration

Titration

• An acid is neutralized by a base—at least one should be strong.

Titration


• The comparison of two volumes, with one known concentration, will give you the other concentration.

Titration


• The comparison of two volumes, with one known concentration, will give you the other concentration.

• That’s a titration.

H+ + OH-H2O

• Just remember–

One H+ neutralizes one OH-

• It’s true for strong and weak acids and bases.

H+ + OH-H2O

• Just remember–

One H+ neutralizes one OH-

• It’s true for strong and weak acids and bases.

M1V1=M2V2

Moles of H+

Moles of OH-

For example:

• If 23.56 ml of a .115M NaOH solution neutralizes 10.00 ml of HCl solution, what is the concentration of the original acid?

For example:


• Step 1: Find moles of NaOH

For example:



moles=MxV=.115Mx.02356L=

For example:



moles=MxV=.115Mx.02356L=.00271 mol

For example:




• Step 2: Find moles of HCl

For example:





.00271 mol NaOHx1HCl/1NaOH=

For example:





.00271 mol NaOHx1HCl/1NaOH=.00271mol

For example:






• Step 3: Find molarity of HCl

For example:







M=Moles/L=.00271mol/.01000 L

For example:







M=Moles/L=.00271mol/.01000L=.271 M HCl

For example:

• If 14.20 ml of a .850M HNO3 solution neutralizes 25.00 ml of a KOH solution, what is the concentration of the original base?

For example:

• If 14.20 ml of a .850M HNO3 solution neutralizes 25.00 ml of a KOH solution, what is the concentration of the original base?

(.483M KOH)

For example:

• If 9.87 ml of a 1.32M H2SO4 solution neutralizes 20.00 ml of a LiOH solution, what is the concentration of the original base?

For example:


Did you notice?

For example:


Did you notice?

The ratio for the conversion from acid to base (Step 2) is :

2 LiOH

1 H2SO4

What volume…

What volume of a .365 M NaOH solution is needed to neutralize 15.00 ml of a .211 M nitric acid solution?

What volume…

What volume of a .365 M NaOH solution is needed to neutralize 15.00 ml of a .211 M nitric acid solution?

What happens if you add more NaOH?

Overtitration

• Once you’ve neutralized a weak acid or base, ignore it.

• The pH is based on the excess of strong acid or base added afterwards.

• Subtract what was used in the neutralization, divide the excess by the total volume

Practice problem:

10.00 ml of .416 M HCl is titrated with .104M NaOH. What is the pH after:

• 0 ml?• 10 ml?• 20 ml?• 30 ml?• 40 ml?• 50 ml?

Practice problem:

10.00 ml of .416 M HC2H3O2 is titrated with .104 M NaOH. What is the pH after:

• 0 ml?• 10 ml?• 20 ml?• 30 ml?• 40 ml?• 50 ml?

Buffer solutions

• Resist changes in pH

• Composed of significant amounts of a weak acid and its conjugate base

• Can be a partly neutralized weak acid, or mixed with the sodium salt as the base

• In the buffer range, the amount shifting is insignificant—ignore it. At the edges, use ICE

Salt hydrolysis

• Q: What happens to the pH of a solution when you add a salt?

Salt hydrolysis


• A: It depends. Is it an acidic salt or a basic salt?

Salt hydrolysis


• A: It depends. Is it an acidic salt or a basic salt?

In the words of Glenda the Good:

• “I don’t know. Are you a good witch or a bad witch?”

Neutral salts

• If the cation is a group 1 metal, and the anion is the conjugate base of a strong acid, no effect.

Neutral salts

• If the cation is a group 1 metal, and the anion is the conjugate base of a strong acid, no effect.

• Examples: KCl, NaNO3, LiClO4

If not…

• The cation will show some tendency to associate itself with hydroxide—making more H+ in solution

And/Or• The anion will show some tendency to

associate itself with H+, leaving more OH- in solution

• -

If not…

• The cation will show some tendency to associate itself with hydroxide—making more H+ in solution

• Examples: FeCl3, Al(NO3)3

And/Or• The anion will show some tendency to

associate itself with H+, leaving more OH- in solution

• Examples: NaF, Li2CO3

Acidic or basic solutions?

Aqueous:NaBr

KNO3

ZnCl2Al(NO3)3

CuCl2Li3PO4

NaC2H3O2

Acidic or basic solutions?

Aqueous:NaBr

KNO3

ZnCl2Al(NO3)3

CuCl2Li3PO4

NaC2H3O2

Neutral Acidic

Basic

Solubility Equilibria

• When a minimally soluble salt dissolves in water,



• Cu(OH)2 (s) Cu+2 (aq) + 2 OH - (aq)

• Ksp=[Cu+2] [OH-]2

No denominator--it’s a product of the ions! The solid is not part of the equilibrium.

sp for “solubility product”



• In general:

• MxBy (s) X M+? (aq) + Y B -? (aq)

• Ksp=[M+?]X[B -?]Y

No denominator--it’s a product of the ions! The solid is not part of the equilibrium.

sp for “solubility product”

Write the reaction for dissolving a minimally soluble ionic compound

For:

1. AgCl

2. BaCO3

3. Fe(OH)2

4. Zinc (II) phosphate

Write the equilibrium expression

For:

1. AgCl

2. BaCO3

3. Fe(OH)2


Given a solubility—calculate Ksp

For:

1. AgCl (solubility=1.34 x 10-5 mol/L)

2. BaCO3 (solubility=1.40 x 10-2 g/L)

3. Fe(OH)2 (solubility=5.23 x 10-5 g/100 ml)


Given Ksp—calculate solubility

• For:

1. AgI Ksp=8.3 x 10-17

2. PbCO3 Ksp=3.3 x 10-14

3. Al(OH)3 Ksp=1.8 x 10-33


(let’s not; and say we did)

Calculate solubility in a solution with one of the ions already.

• Trying to dissolve lead chloride in a solution that already has chloride ions in it works even worse than dissolving in water

• Same thing for dissolving magnesium hydroxide in a solution that is already basic.

• This is the common ion effect.

Calculate solubility in a solution with one of the ions already.

• Solve for the unknown ion.

• The common ion will not change much

• Try the previous two problems if

a) [Cl-]= .85 M and

b) pH= 13.63

Are you ready for a test?

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