Chemistry Review Round 3 Kinetics, Equilibrium, and Acids and Bases.

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

[H[H22]]

As the reaction progresses the concentration N2 goes down 1/3 as fast

Concentration

[H[H22]]

[N[N22]]

As the reaction progresses the concentration NH3 goes up.

Concentration

[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

[H[H22]]

Instantaneous slope method.

Concentration

[H[H22]]

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--

0.200.20 0.100.10 0.100.10 1.6 x 101.6 x 10--

0.200.20 0.200.20 0.100.10 3.2 x 103.2 x 10--

0.100.10 0.100.10 0.200.20 3.2 x 103.2 x 10--

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

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

A = kt0

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

2[ ]A = kt +

22[ [A]

A] = kt

tk[A]1 =

[A] = kt

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

Concentration

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]

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

Hydrogen bonds to surface of metal.

Break H-H bonds

Heterogenous Catalysts

Pt surface

C HH C

Pt surface

C HH C

The double bond breaks and bonds to the catalyst.

Pt surface

C HH C

The hydrogen atoms bond with the carbon

Pt surface

C HH C

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 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

Forward Reaction

Reverse reaction

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

[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.

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-]

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

0 1 3 5 7 9 11 13 14

Basic 10010-110-310-510-710-910-1110-1310-14

013579111314

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

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- ]

Strength of Bases

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

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.

Chemistry Review Round 3 Kinetics, Equilibrium, and Acids and Bases.

reaction rate rate

rate changes

changes rate

rate lawsthe

rate of appearance

rate constant

rate law differential

rate law expression

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