Lecture 6 Acids and Bases v2

8/9/2019 Lecture 6 Acids and Bases v2

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Pick Up a

Clicker!

Schedule

8:10-9:00 Questions

9:00-9:45 Test on Equilibrium

9:45-12:00 Acids and Bases

Good Morning

by joseph o. holmes, manhattan, march 12


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Recall:

1) Strong acids == completely dissociate in water. Examples: HCl, HBr, HI

(haloacids excluding HF); HNO3 (nitric); HClO4 (perchloric); 1st

dissociation of sulfuric H2SO4a. We can write the dissociation as:

b. HCl + H2O H3O+ + Cl-(aq) or shorthand notation

c. HCl H+ + Cl-

Theres all sorts of stuff you need to see in this:

a. Anytime you have an ion and the solvent is not explicitly given,

assume water

b. H+ is called proton and is meant to symbolize the hydronium ion

(H3O+ )c. The Ka is assumed to be very large

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Recall:

2) Strong bases == completely dissociate in water. Examples: Group 1

hydroxides (NaOH, KOH, CsOH, RbOH) and Ca(OH)2a. We can write the dissociation as:

b. NaOH(s) + H2O Na+ (aq) + OH-(aq)

c. NaOH Na+ (aq) + OH-(aq)

You need to see in this:

a. The Kb is assumed to be very large

H2O

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Calculate:

3) Strong acids. If I have a solution that is labeled 1.0M HCl. What does

that mean exactly?

a) because it has not been specified otherwise, the solvent is water

b) The label tells me that ifthe HCl were completed associated, then I

would have 1.0 moles of HCl for every liter of solutionc) But since HCl is completely dissociated, then for every 1 mole of HCl I

have, I really have 1 mole of proton and one mole of chloride ion in

solution: [H+] = 1.0M; [Cl-]=1.0M.

d) If I bubble 18.2g of HCl through a container of 1.5 liters of water,

what is present in the solution and in what concentration?

present: H+ and Cl-, no HCl[H+] = 18.2/36.5/1.5 = 0.33 M

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Calculate:

4) Strong bases. If I have a container that is labeled 0.75M KOH, what does

that mean exactly?

a. I have 0.75 moles of KOH for every liter of solution

b. Since KOH dissociated completely, I actually have NO associated

KOH it is all dissociated.

c. [K+]=0.75M; [OH-]=0.75M

d. If I weigh 52.5 g of KOH and dissolve it in water to make a 2.5L

solution, what would I write on the bottle and what is actually

present in solution (and in what concentration).

Moles KOH = 52.5/56.1 = 0.94 molesBottle label: 0.94/2.5 = 0.37M KOH

Actually present: K+ and OH-; [K+]=0.37M; [OH-]=0.37M

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Models of Acids and Bases

Bronsted-Lowry Acid: donates a proton (proton donor)

Bronsted-Lowry Base: accepts a proton (proton acceptor)

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

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

These are examples of weak acids or bases theyincompletely dissociate in water. These are equilibrium

reactions.

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Visualize the dissociation

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Recall: Weak Acid/Base Equilibria

HC2H3O2(l) + H2OAcid base

H3O+ (aq) + C2H3O2

- (aq)conjugate conjugate

acid base

H

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Recall: Weak Acid/Base Equilibria

Equilibrium constants

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

- (aq)Acid base conjugate conjugate

acid base

][

]][[

][

]][[

232

3232

HA

HA

OHHC

OHOHCK

a

!!

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Weak Acid/Base Equilibria

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

conjugate conjugate

acid base

Base acid

H

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Weak Acid/Base Equilibria

Equilibrium constants

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

Base acid conjugate conjugate

acid base

][

]][[

][

]][[

3

4

B

OHBH

NH

OHNHK

b

!!

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Autoionization of Water

Water is both a Bronsted-Lowry acid and base (amphoteric).

H2O +H2O H3O+ (aq) + OH-(aq)

MOHbutOH

OHOHK 55][

][

]][[22

2

3}!

C25at

10]][[

14

3

Q!! OHOHK

w

For pure water [H3O+] = [OH-] = 10-7 M

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It is very important to have a mental picture of what is in the

solution. Try some clicker questions:

A solution 0.2M nitric acid contains:

H2O 1. Major 2. Minor 3. None

HNO3 1. Major 2. Minor 3. None

H+ 1. Major 2. Minor 3. None

NO3- 1. Major 2. Minor 3. None

OH- 1. Major 2. Minor 3. None

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Now estimate concentrations

The approximate concentrations of all species in a solution of 0.2M

nitric acid is:

H2O A. 55M B. 1M C. 0.2M D.0.2x10-7M E. 5x 10-14M

HNO3 A. 55M B. 0.2M C. 0.2x10-7M D. 5x10-14 E. not present

H+ A. 55M B. 0.2M C. 0.2x10-7M D. 5x10-14 E. not present

NO3- A. 55M B. 0.2M C. 0.2x10-7M D. 5x10-14 E. not present

OH- A. 55M B. 0.2M C. 0.2x10-7M D. 5x10-14 E. not present

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Now lets do some calculations

Calculating

EquilibriumPositions

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Consider the reaction

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

Suppose we start with ammonia dissolved in water before dissociation of the

ammonia. Imagine we have a 1M solution of ammonia initial. What concentrations of

ammonium and hydroxide do we have at equilibrium? Kb = 1.8x10

-5

Make a table:

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

Init 1.0 M 0 10

-7

Change -x x +x

At eq 1.0-x x 10-7 + x

What is the value of x?

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Consider the reaction

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

Suppose we start with ammonia dissolved in water before dissociation of the

ammonia. Imagine we have a 1M solution of ammonia initially (meaning before

dissociation). What concentrations of ammonium and hydroxide do we have at

equilibrium? Kb = 1.8x10-5

Make a table:

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

InitChange

At eq


-x x +x

1.0-x x 10-7 + x

1.0M 0M 10-7

M

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Calculatingx

5

7-

3

4108.1

0.1

)x(10

][

]][[

!

!! x

x

x

NH

OHNHK

b

0108.1)10108.1(5752

!

xxxx

2

)108.1(4)10108.1()10108.1( 527575 s!

xxxx

x=4.2x10-3 M

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CalculatingxNH3(g) + H2O NH4+ + OH- (aq)

Init 1.0 M 0 10-7

Change -x x +x

At eq 1.0-x x 10-7 + x


x=4.2x10-3 M

So at equilibrium:

[NH3] = 1.0-x = 1.0M

[NH4+] = x= 4.2x10-3M

[OH-] = 10-7+x = 4.2x10-3M

There are some interesting

approximations here!!

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Funny Addition

Notice that when adding two numbers of very different magnitudes, the smaller

number does not contribute much if anything to the sum. In other words, how do you

think giving Bill Gates $100 will change his net worth? On the other hand, how would

you feel about getting $100? Thus we can make the following set of approximately:

1.0 x ~ 1.0 if x is smaller than about 0.1

x + 10-7 ~ x if x is bigger than about 10-6

Abstractly: x + ysometimes equalsx, sometimes equals y, and sometimes is the

sum after all!

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Return to the concentration tableNH3(g) + H2O NH4

+ + OH- (aq)

Init 1.0 M 0 10-7

Change -x x +x

At eq 1.0-x x 10-7 + x

IfK is small (like less than 10-2) then the reaction will notshift

muchto the right. We can then make the following approximations:

1.0-x ~ 1.0; x + 10-7~ x

NH3(g) + H

2O NH

4

+ + OH- (aq)

Init 1.0 M 0 10-7

Change -x x +x

At eq 1.0 x x

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Compare these two calcs

0108.1)10108.1( 5752 !

xxxx

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

Init 1.0 M 0 10-7

Change -x x +x

At eq 1.0 x x

)102.4(108.1

108.10.1

)(

][

]][[

352

5

3

4

!!

!!!

xxxx

xxx

NH

OHNHK

b

Vs.

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Another calcThe Ka of nitrous acid is 4.0x10-4. If I dissolve 0.5 moles of the acid in 500 mL of

water, then what species and how much of each species are present in solution?

Write the equation (look at table A5.1 if you need to)

HNO2(aq) + H2O > H3O+ + NO2

- notes

Do the eq chart

Init 0.5/0.5 0 0 neglect the conc of H+ from water

Change -x xx

At eq 1-x x x now with approximations

Approx 1 x x

X2/1 = 4.0x10-4; x=2x10-2 [HNO2] = 1-.02 = 0.98M; [H3O+]=0.02M; [NO2

-]=0.02M

Is the approximation valid?

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Check the validity5% rule if there is less than a 5% error, then the approximation is valid

Mathematical statement:

1 - x ~ 1

Error is estimated as x/1 = 0.02/1 = 2%. Valid approximation.

Second approximation

10-7 + x = x

Error is estimated as 10-7/x since 10-7/0.02 = 5x10-4 less than 1% error.

Error is: small number/large number x 100.

Now lets work problems.

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Practice Problems

1. What are the species and the concentrations of all species present if a 0.2 M

solution of benzoic acid is prepared? (Ka 6.4x10-5)

2. How about this: The same as above except with chlorous acid instead of

benzoic? (Ka 1.2x10-2)

3. Now try the basic calculation what are the species and the concentration ofall species present in a 0.4M solution of triethylamine ((C2H5)3N, Kb=4.0x10

-4

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Calculations involving salts

What are salts? Ionic compounds.

What are soluble salts? Ionic compounds that dissolve in

water.

Examples of salts: NaClKCl Na2CO3 KHSO4Na3PO4

You need to be able to 1) recognize an ionic compound, 2) identify the ions it

dissociates into.

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Recognizing Salts

1. Cations are metals

2. Cation is ammonium NH4+

3. Anions are halides or come from the polyatomic anions.

NaNO3Sodium

Nitrate

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Dissolving in Water

NaNO3 (s) + H2O (l) Na+(aq) + NO3

-(aq)

All sodium and potassium salts are water

soluble and dissociate completely

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Identify major species

Na+(aq) ; NO3-(aq)

Na+ is a spectator ion and does not react

with water

Anions of strong acids are so weak that theydont react with water

Do any species react with water?

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Try sodium nitrite

NaNO2 (s) + H2O (l) Na+(aq) + NO2

-(aq)

All sodium and potassium salts are water

soluble and dissociate completely

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Identify major species

Na+(aq) ; NO2-(aq)

Na+ is a spectator ion and does not react

with water

But nitrite is an anion of a weak acid andtherefore is itself a weak base!!

Do any species react with water?

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Anions reacting with water

NO2-(aq) + H2O (l) HNO2

-(aq)+ OH-(aq)

But, alas, what is the K for this equilibria?

It should be a Kb, but it is not in the Kb

table.

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Nitrite is the conjugate base of nitrous acid

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Ka and Kb are related

Ka x Kb = Kw if the Ka and Kb go with the acid/base

conjugate pair

Consider HCN: Ka = 6.2x10-10

Whats the conjugate base?

CN-

Whats Kb and to what reaction does that Kb correspondto?

CN- + H2O HCN + OH- Kb=10-14/Ka = 0.16x10-4

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Sample calculation

What species are present and in what concentrations for a solution that is

0.5M KCN?

1. Look at KCN acid, base, or salt?

2. SALT3. If Salt, dissociate it: K+, CN-

4. Any spectator ions? Yes K+

5. What is concentration of K+

6. [K+] = 0.5M

7. How about anion conjugate base of weak acid, therefore weak base

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Sample calculation cont

What species are present and in what concentrations for a solution that is

0.5M KCN?

8. How about CN-: reacts with water according to:

9. CN- + H2O HCN + OH-10. This is an equilibrium reaction so make take

CN- + H2O HCN + OH-

Init 0.5 0 0

Change -x +x +xEq 0.5-x +x +x

Approx 0.5 +x +x

Kb=1.6x10-5 x2/0.5 = kb x=sqrt(8x10-6) ~ 2.7x10-3

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One more thing -- pH

The proton concentration (and hydroxide concentration) are some

important we will follow them with a new concept pH. In general these

concentrations are small but if the hydrogen ion concentration in your

blood increases from 3x10-7 M to 3x10-6M, you are probably dead. One

order ofmagnitude makes a huge difference.

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Writing all these exponents for low

concentrations of H+ are a pain:

the pH scale

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

1x10-2 2

1x10-4 4

1x10-6 6

1x10-8 8

1x10-10 10 38


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Writing all these exponents for low

concentrations of H+ are a pain:

the pH scale

pH = - log10 [H+]

Similarly, pOH = - log10 [OH]

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Now lets work problems of the

following type:

What are the concentrations of all species present in a solution that is XM of Y?

What is the pH and pOH of the solution?

I am not so interesting in the cases where the approximations dont work so forthe moment, lets consider XM to be rather large. So the question will be

What are the concentrations of all species present in a solution that is 0.5M of Y?

What is the pH and pOH of the solution?

The only variable is what is Y

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What are the concentrations of all species present in a solution

that is 0.5M of Y? What is the pH and pOH of the solution?

All possible Ys:

1) Strong acid2) Strong base

3) Weak acid

4) Weak base

5) Acidic salt

6) Acidic base

We are going to practice all cases.

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Lecture 6 Acids and Bases v2

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