lecture 6 acids and bases v2
TRANSCRIPT
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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
What is the value of x?
-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
What is the value of 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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