chemistry review round 3 kinetics, equilibrium, and acids and bases
TRANSCRIPT
Chemistry ReviewRound 3
Kinetics, Equilibrium, and Acids and Bases
Kinetics The study of reaction rates. Spontaneous reactions are reactions
that will happen - but we can’t tell how fast.
Diamond will spontaneously turn to graphite – eventually.
Reaction mechanism- the steps by which a reaction takes place.
Reaction Rate
Rate = Conc. of A at t2 -Conc. of A at t1
t2- t1 Rate =[A]
t Change in concentration per unit time For this reaction N2 + 3H2 2NH3
As the reaction progresses the concentration H2 goes down
Concentration
Time
[H[H22]]
As the reaction progresses the concentration N2 goes down 1/3 as fast
Concentration
Time
[H[H22]]
[N[N22]]
As the reaction progresses the concentration NH3 goes up.
Concentration
Time
[H[H22]]
[N[N22]]
[NH[NH33]]
Calculating Rates Average rates are taken over long
intervals Instantaneous rates are determined by
finding the slope of a line tangent to the curve at any given point because the rate can change over time
Average slope method
Concentration
Time
[H[H22]]
tt
Instantaneous slope method.
Concentration
Time
[H[H22]]
tt
Defining RateWe can define rate in terms of the
disappearance of the reactant or in terms of the rate of appearance of the product.
In our example N2 + 3H2 2NH3
-[N2] = -3[H2] = 2[NH3] t t t
Rate Laws Reactions are reversible. As products accumulate they can begin
to turn back into reactants. Early on the rate will depend on only the
amount of reactants present. We want to measure the reactants as
soon as they are mixed. This is called the Initial rate method.
Two key points The concentration of the products do
not appear in the rate law because this is an initial rate.
The order must be determined experimentally,
can’t be obtained from the equation
Rate LawsRate Laws
You will find that the rate will only depend on the concentration of the reactants.
Rate = k[NO2]n
This is called a rate law expression. k is called the rate constant. n is the order of the reactant -usually a
positive integer.
2 NO2 2 NO + O2
Types of Rate Laws Differential Rate law - describes how
rate depends on concentration. Integrated Rate Law - Describes how
concentration depends on time. For each type of differential rate law
there is an integrated rate law and vice versa.
Rate laws can help us better understand reaction mechanisms.
Determining Rate Laws The first step is to determine the form of
the rate law (especially its order). Must be determined from experimental
data. For this reaction
2 N2O5 (aq) 4NO2 (aq) + O2(g)
The reverse reaction won’t play a role
The method of Initial Rates This method requires that a reaction be
run several times. The initial concentrations of the
reactants are varied. The reaction rate is measured bust after
the reactants are mixed. Eliminates the effect of the reverse
reaction.
An example For the reaction
BrO3- + 5 Br- + 6H+ 3Br2 + 3 H2O
The general form of the Rate Law is
Rate = k[BrO3-]n[Br-]m[H+]p
We use experimental data to determine the values of n,m,and p
Initial concentrations (M)
Rate (M/s)
BrOBrO33-- BrBr-- HH++
0.100.10 0.100.10 0.100.10 8.0 x 108.0 x 10--
44
0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--
33
0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--
33
0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--
33
Now we have to see how the rate changes with concentration
Integrated Rate Law Expresses the reaction concentration as
a function of time. Form of the equation depends on the
order of the rate law (differential). Changes Rate = [A]n
t We will only work with n=0, 1, and 2
First Order For the reaction 2N2O5 4NO2 + O2
We found the Rate = k[N2O5]1
If concentration doubles rate doubles. If we integrate this equation with respect
to time we get the Integrated Rate Law ln[N2O5] = - kt + ln[N2O5]0
ln is the natural log [N2O5]0 is the initial concentration.
General form Rate = [A] / t = k[A] ln[A] = - kt + ln[A]0
In the form y = mx + b y = ln[A] m = -k x = t b = ln[A]0
A graph of ln[A] vs time is a straight line.
First Order
By getting the straight line you can prove it is first order
Often expressed in a ratio
First Order
By getting the straight line you can prove it is first order
Often expressed in a ratio
First Order
lnA
A = kt0
Half Life The time required to reach half the
original concentration. If the reaction is first order [A] = [A]0/2 when t = t1/2
Half Life• The time required to reach half the
original concentration.
• If the reaction is first order
• [A] = [A]0/2 when t = t1/2
ln
A
A = kt0
01 2
2
ln(2) = kt1/2
Half Life t1/2 = 0.693/k The time to reach half the original
concentration does not depend on the starting concentration.
An easy way to find k
Second Order Rate = -[A] / t = k[A]2 integrated rate law 1/[A] = kt + 1/[A]0 y= 1/[A] m = k x= t b = 1/[A]0 A straight line if 1/[A] vs t is graphed Knowing k and [A]0 you can calculate [A]
at any time t
Second Order Half Life [A] = [A]0 /2 at t = t1/2
1
20
2[ ]A = kt +
1
[A]10
22[ [A]
- 1
A] = kt
0 01
tk[A]1 =
1
02
1
[A] = kt
01 2
Zero Order Rate Law Rate = k[A]0 = k Rate does not change with concentration. Integrated [A] = -kt + [A]0
When [A] = [A]0 /2 t = t1/2
t1/2 = [A]0 /2k
Most often when reaction happens on a surface because the surface area stays constant.
Also applies to enzyme chemistry.
Zero Order Rate Law
Time
Concentration
Time
Concentration
A]/t
t
k =
A]
More Complicated Reactions BrOBrO33
-- + 5 Br + 5 Br-- + 6H + 6H++ 3Br 3Br22 + 3 H + 3 H22OO
For this reaction we found the rate law For this reaction we found the rate law to beto be
Rate = k[BrORate = k[BrO33--][Br][Br--][H][H++]]22
To investigate this reaction rate we To investigate this reaction rate we need to control the conditions need to control the conditions
Summary of Rate Laws
Reaction Mechanisms The series of steps that actually occur
in a chemical reaction. Kinetics can tell us something about the
mechanism A balanced equation does not tell us
how the reactants become products.
2NO2 + F2 2NO2F Rate = k[NO2][F2] The proposed mechanism is NO2 + F2 NO2F + F (slow) F + NO2 NO2F (fast) F is called an intermediate It is formed
then consumed in the reaction
Reaction Mechanisms
Each of the two reactions is called an elementary step .
The rate for a reaction can be written from its molecularity .
Molecularity is the number of pieces that must come together.
Reaction Mechanisms
Unimolecular step involves one molecule - Rate is rirst order.
Bimolecular step - requires two molecules - Rate is second order
Termolecular step- requires three molecules - Rate is third order
Termolecular steps are almost never heard of because the chances of three molecules coming into contact at the same time are miniscule.
A products Rate = k[A] A+A products Rate= k[A]2
2A products Rate= k[A]2
A+B products Rate= k[A][B] A+A+B Products Rate= k[A]2[B] 2A+B Products Rate= k[A]2[B] A+B+C Products Rate= k[A][B]
[C]
Collision theory Molecules must collide to react. Concentration affects rates because
collisions are more likely. Must collide hard enough. Temperature and rate are related. Only a small number of collisions
produce reactions.
Potential Energy
Reaction Coordinate
Reactants
Products
Potential Energy
Reaction Coordinate
Reactants
Products
Activation Energy Ea
Potential Energy
Reaction Coordinate
Reactants
Products
Activated complex
Potential Energy
Reaction Coordinate
Reactants
ProductsE}
Terms Activation energy - the minimum energy
needed to make a reaction happen. Activated Complex or Transition State -
The arrangement of atoms at the top of the energy barrier.
Arrhenius Said the at reaction rate should
increase with temperature. At high temperature more molecules
have the energy required to get over the barrier.
The number of collisions with the necessary energy increases exponentially.
Arrhenius Number of collisions with the required
energy = ze-Ea/RT
z = total collisions e is Euler’s number (opposite of ln) Ea = activation energy
R = ideal gas constant T is temperature in Kelvin
Arrhenius Equation
k = zpe-Ea/RT = Ae-Ea/RT
A is called the frequency factor = zp
ln k = -(Ea/R)(1/T) + ln A
Another line !!!!
ln k vs t is a straight line
Catalysts Speed up a reaction without being used
up in the reaction. Enzymes are biological catalysts. Homogenous Catalysts are in the same
phase as the reactants. Heterogeneous Catalysts are in a
different phase as the reactants.
How Catalysts Work Catalysts allow reactions to proceed by
a different mechanism - a new pathway. New pathway has a lower activation
energy. More molecules will have this activation
energy. Do not change E
Pt surface
HH
HH
HH
HH
Hydrogen bonds to surface of metal.
Break H-H bonds
Heterogenous Catalysts
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
The double bond breaks and bonds to the catalyst.
Pt surface
HH
HH
Heterogenous Catalysts
C HH C
HH
The hydrogen atoms bond with the carbon
Pt surface
H
Heterogenous Catalysts
C HH C
HH
H HH
Homogenous Catalysts Chlorofluorocarbons catalyze the
decomposition of ozone. Enzymes regulating the body
processes. (Protein catalysts)
Catalysts and rate Catalysts will speed up a reaction but
only to a certain point. Past a certain point adding more
reactants won’t change the rate. Zero Order
Catalysts and rate.
Concentration of reactants
Rate
Rate increases until the active sites of catalyst are filled.
Then rate is independent of concentration
Reactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the
forward reaction is possible As C and D build up, the reverse
reaction speeds up while the forward reaction slows down.
Eventually the rates are equal
Rea
ctio
n R
ate
Time
Forward Reaction
Reverse reaction
Equilibrium
B
E
Equilibrium
What is equal at Equilibrium? Rates are equal Concentrations are not. Rates are determined by concentrations
and activation energy. The concentrations do not change at
equilibrium. or if the reaction is verrrry slooooow.
Law of Mass Action For any reaction jA + kB lC + mD
K = [C]l[D]m PRODUCTSpower
[A]j[B]k REACTANTSpower
K is called the equilibrium constant.
is how we indicate a reversible reaction
Playing with K If we write the reaction in reverse. lC + mD jA + kB Then the new equilibrium constant is
K’ = [A]j[B]k = 1/K [C]l[D]m
Playing with K If we multiply the equation by a constant njA + nkB nlC + nmD Then the equilibrium constant is
K’ =
[A]nj[B]nk = ([A] j[B]k)n
= Kn
[C]nl[D]nm ([C]l[D]m)n
The units for K Are determined by the various powers
and units of concentrations. They depend on the reaction.
Equilibrium Constant
One for each Temperature
Calculate K N2 + 3H2 3NH3
Initial At Equilibrium [N2]0 =1.000 M [N2] = 0.921M
[H2]0 =1.000 M [H2] = 0.763M
[NH3]0 =0 M [NH3] = 0.157M
Calculate K N2 + 3H2 3NH3
Initial At Equilibrium [N2]0 = 0 M [N2] = 0.399 M
[H2]0 = 0 M [H2] = 1.197 M
[NH3]0 = 1.000 M [NH3] = 0.157M
K is the same no matter what the amount of starting materials
Equilibrium and Pressure Some reactions are gaseous PV = nRT P = (n/V)RT P = MRT M is a concentration in moles/Liter M = P/RT
Equilibrium and Pressure 2SO2(g) + O2(g) 2SO3(g)
Kp = (PSO3)2
(PSO2)2 (PO2)
K = [SO3]2
[SO2]2 [O2]
Equilibrium and Pressure K = (PSO3/RT)2
(PSO2/RT)2(PO2/RT)
K = (PSO3)2 (1/RT)2
(PSO2)2(PO2) (1/RT)3
K = Kp (1/RT)2 = Kp RT
(1/RT)3
General Equation jA + kB lC + mD
Kp = K (RT)(l+m)-(j+k) = K (RT)n
n=(l+m)-(j+k)=Change in moles of gas
Homogeneous Equilibria So far every example dealt with
reactants and products where all were in the same phase.
We can use K in terms of either concentration or pressure.
Units depend on reaction.
Heterogeneous Equilibria If the reaction involves pure solids or
pure liquids the concentration of the solid or the liquid doesn’t change.
As long s they are not used up they are not used up we can leave them out of the equilibrium expression.
For example
For Example H2(g) + I2(s) 2HI(g)
K = [HI]2 [H2][I2]
But the concentration of I2 does not
change.
K[I2]= [HI]2 = K’ [H2]
Solving Equilibrium Problems
Type 1
The Reaction Quotient Tells you the directing the reaction will
go to reach equilibrium Calculated the same as the equilibrium
constant, but for a system not at equilibrium
Q = [Products]coefficient
[Reactants] coefficient
Compare value to equilibrium constant
What Q tells us If K>Q
Not enough productsShift to right
If K<QToo many productsShift to left
If Q=K system is at equilibrium
Example for the reaction 2NOCl(g) 2NO(g) + Cl2(g) K = 1.55 x 10-5 M at 35ºC In an experiment 0.10 mol NOCl,
0.0010 mol NO(g) and 0.00010 mol Cl2 are mixed in 2.0 L flask.
Which direction will the reaction proceed to reach equilibrium?
Solving Equilibrium Problems Given the starting concentrations and
one equilibrium concentration. Use stoichiometry to figure out other
concentrations and K. Learn to create a table of initial and final
conditions.
Consider the following reaction at 600ºC 2SO2(g) + O2(g) 2SO3(g) In a certain experiment 2.00 mol of SO2,
1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol were found to be present.
Calculate the equilibrium concentrations of O2 and SO2, K and KP
Consider the same reaction at 600ºC In a different experiment 0.500 mol SO2
and 0.350 mol SO3 were placed in a
1.000 L container. When the system reaches equilibrium 0.045 mol of O2 are
present. Calculate the final concentrations of SO2
and SO3 and K
What if you’re not given equilibrium concentration?
The size of K will determine what approach to take.
First let’s look at the case of a small value of K ( <0.01).
Allows us to make simplifying assumptions.
Example H2(g) + I2(g) 2HI(g) K = 7.1 x 10-3 at 25ºC Calculate the equilibrium concentrations
if a 5.00 L container initially contains 15.9 g of H2 294 g I2 .
[H2]0 = (15.7g/2.02)/5.00 L = 1.56 M [I2]0 = (294g/253.8)/5.00L = 0.232 M [HI]0 = 0
Q= 0<K so more product will be formed. Assumption since K is small reaction is
reactant favored at equilibrium. Set up table of initial, final and change in
concentrations.
Using to stoichiometry we can find Change in H2 = -X
Change in HI = -X Change in HI = +2X
H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M changefinal
Now we can determine the final concentrations by adding.
H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M
change -X -X +2Xfinal
Now plug these values into the equilibrium expression
K = (2X)2 = 7.1 x 10-3 (1.56-X)(0.232-X)
H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M
change -X -X +2X
final 1.56-X 0.232-X 2X
Why we chose X K = (2X)2 = 7.1 x 10-3
(1.56-X)(0.232-X) Since X is going to be small, we can
ignore it in relation to 0.464 and 1.328 So we can rewrite the equation 7.1 x 10-3 = (2X)2
(1.56)(0.232) Makes the algebra easy
Checking the assumption The rule of thumb is that if the value of
X is less than 5% of all the other concentrations, our assumption was valid.
If not we would have had to use the quadratic equation
More on this later. Our assumption was valid.
Practice For the reaction Cl2 + O2 2ClO(g)
K = 156 In an experiment 0.100 mol ClO, 1.00
mol O2 and 0.0100 mol Cl2 are mixed in
a 4.00 L flask. If the reaction is not at equilibrium, which
way will it shift? Calculate the equilibrium concentrations.
Another problem with small K
K< .01
For example For the reaction
2NOCl 2NO +Cl2 K= 1.6 x 10-5 If 1.20 mol NOCl, 0.45 mol of NO and 0.87
mol Cl2 are mixed in a 1 L container What are the equilibrium concentrations
Q = [NO]2[Cl2] = (0.45)2(0.87) = 0.15 M
[NOCl]2 (1.20)2
K = (0.45+2X)2(0.87+X) = 1.6 x 10-5 (1.20-2X)2
Figure out X
2NOCl 2NO Cl2
Initial 1.20 0.45 0.87
Change -2X +2X +X
Final 1.20-2X 0.45+2X 0.87+X
Make assumptions K = (0.45)2(0.87) = 1.6 x 10-5
(1.20-2X)2
X= 8.2 x 10-3
2NOCl 2NO Cl2
Initial 1.20 0.45 0.87
Change -2X +2X +X
Final 1.20-2X 0.45+2X 0.87+X
Check assumptions 2(.0082)/1.20 = 1.2 % OKAY!!!
2NOCl 2NO Cl2
Initial 1.20 0.45 0.87
Change -2X +2X +X
Final 1.20-2X 0.45+2X 0.87+X
Practice Problem For the reaction
2ClO(g) Cl2 (g) + O2 (g) K = 6.4 x 10-3 In an experiment 0.100 mol ClO(g), 1.00
mol O2 and 1.00 x 10-2 mol Cl2 are mixed in a 4.00 L container.
What are the equilibrium concentrations.
Problems Involving Pressure Solved exactly the same, with same
rules for choosing X depending on KP
For the reaction N2O4(g) 2NO2(g)
KP = .131 atm. What are the equilibrium
pressures if a flask initially contains 1.000 atm N2O4?
Le Chatelier’s Principle If a stress is applied to a system at
equilibrium, the position of the equilibrium will shift to reduce the stress.
3 Types of stress
Change amounts of reactants and/or products
Adding product makes Q>K Removing reactant makes Q>K Adding reactant makes Q<K Removing product makes Q<K Determine the effect on Q, will tell you
the direction of shift
Change Pressure By changing volume System will move in the direction that
has the least moles of gas. Because partial pressures (and
concentrations) change a new equilibrium must be reached.
System tries to minimize the moles of gas.
Change in Pressure By adding an inert gas Partial pressures of reactants and
product are not changed No effect on equilibrium position
Change in Temperature Affects the rates of both the forward and
reverse reactions. Doesn’t just change the equilibrium
position, changes the equilibrium constant.
The direction of the shift depends on whether it is exo- or endothermic
Exothermic H<0 Releases heat Think of heat as a product Raising temperature push toward
reactants. Shifts to left.
Endothermic H>0 Produces heat Think of heat as a reactant Raising temperature push toward
products. Shifts to right.
D
D
B
B
A
Arrhenius Definition Acids produce hydrogen ions in
aqueous solution. Bases produce hydroxide ions when
dissolved in water. Limits to aqueous solutions. Only one kind of base. NH3 ammonia could not be an
Arrhenius base.
Bronsted-Lowry Definitions And acid is an proton (H+) donor and a
base is a proton acceptor. Acids and bases always come in pairs. HCl is an acid.. When it dissolves in water it gives its
proton to water.
HCl(g) + H2O(l) H3O+ + Cl-
Water is a base makes hydronium ion
Pairs General equation HA(aq) + H2O(l) H3O+(aq) + A-(aq) Acid + Base Conjugate acid +
Conjugate base This is an equilibrium. Competition for H+ between H2O and A-
The stronger base controls direction. If H2O is a stronger base it takes the H+ Equilibrium moves to right.
Acid dissociation constant Ka The equilibrium constant for the general
equation.
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Ka = [H3O+][A-]
[HA]
H3O+ is often written H+ ignoring the
water in equation (it is implied).
Acid dissociation constant Ka HA(aq) H+(aq) + A-(aq)
Ka = [H+][A-]
[HA] We can write the expression for any acid. Strong acids dissociate completely. Equilibrium far to right. Conjugate base must be weak.
Back to Pairs Strong acids
Ka is large
[H+] is equal to [HA] A- is a weaker base
than water
Weak acids
Ka is small
[H+] <<< [HA] A- is a stronger
base than water
Types of Acids
Polyprotic Acids- more than 1 acidic hydrogen (diprotic, triprotic).
Oxyacids - Proton is attached to the oxygen of an ion.
Organic acids contain the Carboxyl group -COOH with the H attached to O
Generally very weak.
Amphoteric Behave as both an acid and a base. Water autoionizes 2H2O(l) H3O+(aq) + OH-(aq) KW= [H3O+][OH-]=[H+][OH-] At 25ºC KW = 1.0 x10-14 In EVERY aqueous solution. Neutral solution [H+] = [OH-]= 1.0 x10-7 Acidic solution [H+] > [OH-] Basic solution [H+] < [OH-]
pH pH= -log[H+] Used because [H+] is usually very small As pH decreases, [H+] increases exponentially Sig figs only the digits after the decimal place
of a pH are significant [H+] = 1.0 x 10-8 pH= 8.00 2 sig figs pOH= -log[OH-] pKa = -log K
Relationships KW = [H+][OH-] -log KW = -log([H+][OH-]) -log KW = -log[H+]+ -log[OH-] pKW = pH + pOH
KW = 1.0 x10-14
14.00 = pH + pOH [H+],[OH-],pH and pOH Given
any one of these we can find the other three.
BasicAcidic Neutral
100 10-1 10-3 10-5 10-7 10-9 10-11 10-13 10-14
[H+]
0 1 3 5 7 9 11 13 14
pH
Basic 10010-110-310-510-710-910-1110-1310-14
[OH-]
013579111314
pOH
Calculating pH of Solutions Always write down the major ions in
solution. Remember these are equilibria. Remember the chemistry. Don’t try to memorize there is no one
way to do this.
Strong Acids HBr, HI, HCl, HNO3, H2SO4, HClO4 ALWAYS WRITE THE MAJOR SPECIES Completely dissociated [H+] = [HA] [OH-] is going to be small because of
equilibrium 10-14 = [H+][OH-] If [HA]< 10-7 water contributes H+
Weak Acids Ka will be small. ALWAYS WRITE THE MAJOR
SPECIES. It will be an equilibrium problem from
the start.
Example Calculate the pH of 2.0 M acetic acid
HC2H3O2 with a Ka 1.8 x10-5
Calculate pOH, [OH-], [H+]
A mixture of Weak Acids The process is the same. Determine the major species. The stronger will predominate. Bigger Ka if concentrations are
comparable Calculate the pH of a mixture 1.20 M HF
(Ka = 7.2 x 10-4) and 3.4 M HOC6H5 (Ka
= 1.6 x 10-10)
Percent dissociation = amount dissociated x 100
initial concentration For a weak acid percent dissociation
increases as acid becomes more dilute. Calculate the % dissociation of 1.00 M
and .00100 M Acetic acid (Ka = 1.8 x 10-5)
As [HA]0 decreases [H+] decreases but %
dissociation increases. Le Chatelier
The other way What is the Ka of a weak acid that is 8.1
% dissociated as 0.100 M solution?
Bases The OH-
is a strong base. Hydroxides of the alkali metals are strong
bases because they dissociate completely when dissolved.
The hydroxides of alkaline earths Ca(OH)2
etc. are strong dibasic bases, but they don’t dissolve well in water.
Used as antacids because [OH- ] can’t build
up.
Bases without OH-
Bases are proton acceptors.
NH3 + H2O NH4+ + OH-
It is the lone pair on nitrogen that accepts the proton.
Many weak bases contain N B(aq) + H2O(l) BH+(aq) + OH- (aq)
Kb = [BH+][OH- ]
[B]
Strength of Bases
N:
Hydroxides are strong. Others are weak. Smaller Kb weaker base.
Calculate the pH of a solution of 4.0 M pyridine (Kb = 1.7 x 10-9)
Polyprotic acids Always dissociate stepwise. The first H+ comes of much easier than
the second. Ka for the first step is much bigger than
Ka for the second. Denoted Ka1, Ka2, Ka3
Polyprotic acid H2CO3 H+ + HCO3
-
Ka1= 4.3 x 10-7
HCO3- H+ + CO3
-2
Ka2= 4.3 x 10-10
Base in first step is acid in second. In calculations we can normally ignore
the second dissociation.
Calculate the Concentration Of all the ions in a solution of 1.00 M
Arsenic acid H3AsO4
KaKa11 = 5.0 x 10 = 5.0 x 10-3-3
KaKa22 = 8.0 x 10 = 8.0 x 10-8-8
KaKa33 = 6.0 x 10 = 6.0 x 10-10-10
Sulfuric acid is special In first step it is a strong acid.
Ka2 = 1.2 x 10-2
Calculate the concentrations in a 2.0 M solution of H2SO4
Calculate the concentrations in a 2.0 x 10-3 M solution of H2SO4
Salts as acids an bases Salts are ionic compounds. Salts of the cation of strong bases and
the anion of strong acids are neutral. for example NaCl, KNO3
There is no equilibrium for strong acids and bases.
We ignore the reverse reaction.
Basic Salts If the anion of a salt is the conjugate base of
a weak acid - basic solution. In an aqueous solution of NaF The major species are Na+, F-, and H2O
F- + H2O HF + OH-
Kb =[HF][OH-] [F- ]
but Ka = [H+][F-] [HF]
Basic Salts Ka x Kb = [HF][OH-] x [H+][F-]
[F- ] [HF]
Basic Salts Ka x Kb = [HF][OH-] x [H+][F-]
[F- ] [HF]
Ka x Kb =[OH-] [H+]
Ka x Kb = KW
Ka tells us Kb
The anion of a weak acid is a weak base.
Calculate the pH of a solution of 1.00 M NaCN. Ka of HCN is 6.2 x 10-10
The CN- ion competes with OH- for the H+
Acidic salts A salt with the cation of a weak base and the
anion of a strong acid will be basic. The same development as bases leads to
Ka x Kb = KW
Calculate the pH of a solution of 0.40 M NH4Cl
(the Kb of NH3 1.8 x 10-5).
Other acidic salts are those of highly charged metal ions.
Anion of weak acid, cation of weak base
Ka > Kb acidic
Ka < Kb basic
Ka = Kb Neutral
Structure and Acid base Properties
Any molecule with an H in it is a potential acid.
The stronger the X-H bond the less acidic (compare bond dissociation energies).
The more polar the X-H bond the stronger the acid (use electronegativities).
The more polar H-O-X bond -stronger acid.
Strength of oxyacids The more oxygen hooked to the central
atom, the more acidic the hydrogen. HClO4 > HClO3 > HClO2 > HClO
Remember that the H is attached to an oxygen atom.
The oxygens are electronegative Pull electrons away from hydrogen
Strength of oxyacidsStrength of oxyacids
Electron Density
Cl O H
Strength of oxyacidsStrength of oxyacids
Electron Density
Cl O HO
Strength of oxyacidsStrength of oxyacids
Cl O H
O
O
Electron Density
Strength of oxyacids
Cl O H
O
O
O
Electron Density
Acid-Base Properties of Oxides
Non-metal oxides dissolved in water can make acids.
SO3 (g) + H2O(l) H2SO4(aq)
Ionic oxides dissolve in water to produce bases.
CaO(s) + H2O(l) Ca(OH)2(aq)
A
B
E
B
Chapter 15
Applying equilibrium
The Common Ion Effect When the salt with the anion of a weak
acid is added to that acid, It reverses the dissociation of the acid. Lowers the percent dissociation of the acid. The same principle applies to salts with the
cation of a weak base.. The calculations are the same as last
chapter.
Buffered solutions A solution that resists a change in pH. Either a weak acid and its salt or a weak
base and its salt. We can make a buffer of any pH by
varying the concentrations of these solutions.
Same calculations as before. Calculate the pH of a solution that is .50
M HAc and .25 M NaAc (Ka = 1.8 x 10-5)
Adding a strong acid or base Do the stoichiometry first. (BAAM) A strong base will grab protons from the
weak acid reducing [HA]0
A strong acid will add its proton to the anion of the salt reducing [A-]0
Then do the equilibrium problem. What is the pH of 1.0 L of the previous
solution when 0.010 mol of solid NaOH is added?
General equation Ka = [H+] [A-]
[HA] so [H+] = Ka [HA]
[A-] The [H+] depends on the ratio [HA]/[A-] taking the negative log of both sides pH = -log(Ka [HA]/[A-]) pH = -log(Ka)-log([HA]/[A-]) pH = pKa + log([A-]/[HA])
This is called the Henderson-Hasselbach equation
pH = pKa + log([A-]/[HA]) pH = pKa + log(base/acid) Calculate the pH of the following mixtures 0.75 M lactic acid (HC3H5O3) and 0.25 M
sodium lactate (Ka = 1.4 x 10-4) 0.25 M NH3 and 0.40 M NH4Cl
(Kb = 1.8 x 10-5)
Buffer capacity The pH of a buffered solution is determined
by the ratio [A-]/[HA]. As long as this doesn’t change much the
pH won’t change much. The more concentrated these two are the
more H+ and OH- the solution will be able to absorb.
Larger concentrations bigger buffer capacity.
Buffer Capacity Calculate the change in pH that occurs
when 0.010 mol of HCl(g) is added to 1.0L of each of the following:
5.00 M HAc and 5.00 M NaAc 0.050 M HAc and 0.050 M NaAc Ka= 1.8x10-5
Buffer capacity The best buffers have a ratio
[A-]/[HA] = 1 This is most resistant to change True when [A-] = [HA] Make pH = pKa (since log1=0)
Titrations Millimole (mmol) = 1/1000 mol Molarity = mmol/mL = mol/L Makes calculations easier because we
will rarely add Liters of solution. Adding a solution of known
concentration until the substance being tested is consumed.
This is called the equivalence point. Graph of pH vs. mL is a titration curve.
Strong acid with Strong Base Do the stoichiometry. There is no equilibrium . They both dissociate completely. The titration of 50.0 mL of 0.200 M HNO3
with 0.100 M NaOH Analyze the pH
Weak acid with Strong base There is an equilibrium. Do stoichiometry. Then do equilibrium. Titrate 50.0 mL of 0.10 M HF (Ka
= 7.2 x 10-4) with 0.10 M NaOH
Titration Curves
pH
mL of Base added
7
Strong acid with strong Base Equivalence at pH 7
pH
mL of Base added
>7
Weak acid with strong Base Equivalence at pH >7
pH
mL of Base added
7
Strong base with strong acid Equivalence at pH 7
pH
mL of Base added
<7
Weak base with strong acid Equivalence at pH <7
Summary Strong acid and base just stoichiometry. Determine Ka, use for 0 mL base Weak acid before equivalence point
–Stoichiometry first–Then Henderson-Hasselbach
Weak acid at equivalence point Kb Weak base after equivalence - leftover
strong base.
Summary Determine Ka, use for 0 mL acid. Weak base before equivalence point.
–Stoichiometry first
–Then Henderson-Hasselbach Weak base at equivalence point Ka. Weak base after equivalence - leftover
strong acid.
Solubility Equilibria
Will it all dissolve, and if not, how much?
All dissolving is an equilibrium. If there is not much solid it will all
dissolve. As more solid is added the solution will
become saturated. Solid dissolved The solid will precipitate as fast as it
dissolves . Equilibrium
General equation M+ stands for the cation (usually metal). Nm- stands for the anion (a nonmetal).
But the concentration of a solid doesn’t change.
Ksp = [M+]a[Nm-]b
Called the solubility product for each compound.
Watch out Solubility is not the same as solubility
product. Solubility product is an equilibrium
constant. it doesn’t change except with temperature. Solubility is an equilibrium position for how
much can dissolve. A common ion can change this.
Calculating Ksp
The solubility of iron(II) oxalate FeC2O4
is 65.9 mg/L The solubility of Li2CO3 is 5.48 g/L
Calculating Solubility The solubility is determined by
equilibrium. Its an equilibrium problem. Calculate the solubility of SrSO4, with a
Ksp of 3.2 x 10-7 in M and g/L. Calculate the solubility of Ag2CrO4, with
a Ksp of 9.0 x 10-12 in M and g/L.
Relative solubilities Ksp will only allow us to compare the
solubility of solids the at fall apart into the same number of ions.
The bigger the Ksp of those the more soluble.
If they fall apart into different number of pieces you have to do the math.
Common Ion Effect If we try to dissolve the solid in a solution with
either the cation or anion already present less will dissolve.
Calculate the solubility of SrSO4, with a Ksp of
3.2 x 10-7 in M and g/L in a solution of 0.010 M Na2SO4.
Calculate the solubility of SrSO4, with a Ksp of
3.2 x 10-7 in M and g/L in a solution of 0.010 M SrNO3.
Precipitation Ion Product, Q =[M+]a[Nm-]b If Q>Ksp a precipitate forms. If Q<Ksp No precipitate. If Q = Ksp equilibrium. A solution of 750.0 mL of 4.00 x 10-3M
Ce(NO3)3 is added to 300.0 mL of 2.00 x 10-
2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-
10M)precipitate and if so, what is the concentration of the ions?
Selective Precipitations Used to separate mixtures of metal ions
in solutions. Add anions that will only precipitate
certain metals at a time. Used to purify mixtures.
Often use H2S because in acidic solution
Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will
precipitate.
Selective Precipitation In Basic adding OH-solution S-2 will
increase so more soluble sulfides will precipitate.
Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3,
Al(OH)3
B
D
A
D
E
