2014

Chemistry Acid Base Equilibrium

Adv

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ent Presenter Copy

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Ce

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Pr

Pa

Nd U

Pm Np

Sm Pu

Eu

Am

Gd

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Tb Bk

Dy Cf

Ho

Es

Er

Fm

Tm Md

Yb

No

Lu Lr

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2 10 18 36 54 86

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Mg

Ca Sr

Ba

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B Al

Ga In Tl

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Ge

Sn

Pb

N P As

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O S Se Te Po

F Cl

Br I At

Sc Y La Ac

Ti Zr Hf

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Cr

Mo W Sg

Mn Tc Re

Bh

Fe Ru

Os

Hs

Co

Rh Ir Mt

Ni

Pd Pt §

Cu

Ag

Au §

Zn Cd

Hg §

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ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS

Throughout the test the following symbols have the definitions specified unless otherwise noted.

L, mL = liter(s), milliliter(s) mm Hg = millimeters of mercury g = gram(s) J, kJ = joule(s), kilojoule(s) nm = nanometer(s) V = volt(s) atm = atmosphere(s) mol = mole(s)

ATOMIC STRUCTURE

E = hν

c = λν

E = energy ν = frequency

λ = wavelength

Planck’s constant, h = 6.626 × 10−34 J s

Speed of light, c = 2.998 × 108 m s−1

Avogadro’s number = 6.022 × 1023 mol−1

Electron charge, e = −1.602 × 10−19 coulomb

EQUILIBRIUM

Kc = [C] [D]

[A] [B]

c d

a b, where a A + b B c C + d D

Kp = C

A B

( ) ( )

( ) ( )

c dD

a b

P P

P P

Ka = [H ][A ][HA]

+ -

Kb = [OH ][HB ][B]

- +

Kw = [H+][OH−] = 1.0 × 10−14 at 25°C

= Ka × Kb

pH = − log[H+] , pOH = − log[OH−]

14 = pH + pOH

pH = pKa + log [A ][HA]

-

pKa = − logKa , pKb = − logKb

Equilibrium Constants

Kc (molar concentrations)

Kp (gas pressures)

Ka (weak acid)

Kb (weak base)

Kw (water)

KINETICS

ln[A] t − ln[A]0 = − kt

[ ] [ ]0A A1 1

t

- = kt

t½ = 0.693k

k = rate constant

t = time t½ = half-life

ADVANCED PLACEMENT CHEMISTRY EQUATIONS AND CONSTANTS

Throughout the test the following symbols have the definitions specified unless otherwise noted.

L, mL = liter(s), milliliter(s) mm Hg = millimeters of mercuryg = gram(s) J, kJ = joule(s), kilojoule(s) nm = nanometer(s) V = volt(s)atm = atmosphere(s) mol = mole(s)

ATOMIC STRUCTURE

E = hνc = λν

E = energy ν = frequency

λ = wavelength

Planck’s constant, h = 6.626 × 10−34 J s

Speed of light, c = 2.998 × 108 m s−1

Avogadro’s number = 6.022 × 1023 mol−1

Electron charge, e = −1.602 × 10−19 coulomb

EQUILIBRIUM

Kc = [C] [D]

[A] [B]

c d

a b, where a A + b B c C + d D

Kp = C

A B

( ) ( )

( ) ( )

c dD

a b

P P

P P

Ka = [H ][A ][HA]

+ -

Kb = [OH ][HB ][B]

- +

Kw = [H+][OH−] = 1.0 × 10−14 at 25°C

= Ka × Kb

pH = − log[H+] , pOH = − log[OH−]

14 = pH + pOH

pH = pKa + log [A ][HA]

-

pKa = − logKa , pKb = − logKb

Equilibrium Constants

Kc (molar concentrations)

Kp (gas pressures)

Ka (weak acid)

Kb (weak base)

Kw (water)

KINETICS

ln[A] t − ln[A]0 = − kt

[ ] [ ]0A A1 1

t

- = kt

t½ = 0.693k

k = rate constant

t = time t½ = half-life

GASES, LIQUIDS, AND SOLUTIONS

PV = nRT

PA = Ptotal × XA, where XA =moles A

total moles

Ptotal = PA + PB + PC + . . .

n = mM

K = °C + 273

D = mV

KE per molecule = 12

mv2

Molarity, M = moles of solute per liter of solution

A = abc

1 1

pressurevolumetemperaturenumber of molesmassmolar massdensitykinetic energyvelocityabsorbancemolarabsorptivitypath lengthconcentration

Gas constant, 8.314 J mol K

0.08206

PVTnm

DKE

Aabc

R

Ã

- -

=============

==

M

1 1

1 1

L atm mol K

62.36 L torr mol K 1 atm 760 mm Hg

760 torr

STP 0.00 C and 1.000 atm

- -

- -====

THERMOCHEMISTRY/ ELECTROCHEMISTRY

products reactants

products reactants

products reactants

ln

ff

ff

q mc T

S S S

H H H

G G G

G H T S

RT K

n F E

qI

t

D

D

D D D

D D D

D D D

=

= -Â Â

= -Â Â

= -Â Â

= -

= -

= -

�

heatmassspecific heat capacitytemperature

standard entropy

standard enthalpy

standard free energynumber of moles

standard reduction potentialcurrent (amperes)charge (coulombs)t

qmcT

S

H

Gn

EIqt

============ ime (seconds)

Faraday’s constant , 96,485 coulombs per moleof electrons1 joule

1volt1coulomb

F =

=

ACID – BASE EQUILIBRIUM Ka, Kb, and Salt Hydrolysis

What I Absolutely Have to Know to Survive the AP Exam The following might indicate the question deals with acid – base equilibrium:

pH, pOH, [H3O+], [OH−], strong and weak, salt hydrolysis, solubility product, Ka, Kb, acid or base dissociation constant, percent ionized, Bronsted-Lowry, Arrhenius, hydronium ion, etc…

Acids and Bases: By Definition! Arrhenius

• Acids: Hydrogen ion, H+, donors• Bases: Hydroxide ion, OH− donors

HCl → H+ + Cl− NaOH → Na+ + OH−

Bronsted-Lowry • Acids: proton donors (lose H+)• Bases: proton acceptors (gain H+)

HNO3 + H2O → H3O+ + NO3−

NH3 + H2O ⇌ NH4+ + OH−

Hydrogen Ion, Hydronium Ion, Which is it!!!? Hydronium − H+ riding piggy-back on a water molecule; water is polar and the + charge of the naked proton is greatly attracted to “Mickey's chin” (i.e. the oxygen atom)

• H3O+ “Anthony”• H+ “Tony” • Often used interchangeably in problems; if H3O+ is used be sure water is in the equation!

Bronsted−Lowry and Conjugate Acids and Bases: What a Pair! Acid and conjugate base pairs differ by the presence of one H+ ion.

HC2H3O2 + H2O → H3O+ + C2H3O2−

• HC2H3O2 is the acid; thus C2H3O2− is its conjugate base (what remains after the H+ has been donated to the H2O

molecule)• H2O behaves as a base in this reaction. The hydronium ion is its conjugate acid (what is formed after the H2O

accepts the H+ ion)NH3 + H2O → NH4

+ + OH− • NH3 is the base; thus NH4

+ is it’s conjugate acid (what is formed after the NH3 accepts the H+ ion)• H2O behaves as an acid in this reaction. The hydroxide ion is its conjugate base (what remains after the H+ has

been donated to NH3)

Understanding conjugate acid/base pairs is very important in understanding acid-base chemistry; this concepts allows for the understanding of many complex situations (buffers, titrations, etc…)

Important Notes… Amphiprotic/amphoteric--molecules or ions that can behave as EITHER acids or bases; water, some anions of weak acids, etc… fit this bill. Monoprotic − acids donating one H+ Diprotic − acids donating two H+ Polyprotic − acids donating 3+ H+

Regardless, always remember: ACIDS ONLY DONATE ONE PROTON AT A TIME!!!

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Acid – Base Equilibrium

Ionization: That’s What it’s All About! Relative Strengths A strong acid or base ionizes completely in aqueous solution (100% ionized [or very darn close to it!])

• The equilibrium position lies far, far to the right (favors products)…• Since a strong acid/base dissociates into the ions, the concentration of the H3O+/OH− ion is equal to the

original concentration of the acid/base respectively.• They are strong electrolytes.

• Do Not confuse concentration (M or mol/L) with strength!• Strong Acids

• Hydrohalic acids: HCl, HBr, HI; Nitric: HNO3; Sulfuric: H2SO4; Perchloric: HClO4

• Oxyacids! More oxygen atoms present, the stronger the acid WITHIN that group. The H+ that is “donated” isbonded to an oxygen atom. The oxygen atoms are highly electronegative and are pulling the bonded pair ofelectrons AWAY from the site where the H+ is bonded, “polarizing it”, which makes it easier [i.e. requires lessenergy] to remove − thus the stronger the acid!

H O Br < H O Br O < H O Br

O

O• Strong Bases

• Group IA and IIA (1 and 2) metal hydroxides; be cautious as the poor solubility of Be(OH)2 and Mg(OH)2 limits the effectiveness of these 2 strong bases.

• IT’S A 2 for 1 SALE with the Group 2 (IIA) ions), i.e. 0.10M Ca(OH)2 is 0.20M OH −

Where the fun begins… A weak acid or base does not completely ionize (usually < 10%)

• They are weak electrolytes.• The equilibrium position lies far to the left (favors reactants)…• [H+] is much less than the acid concentration − thus to calculate this amount and the resulting pH you must

return to the world of EQUILIBRIUM Chemistry!• The vast majority of acid/bases are weak. Remember, ionization not concentration!!!!• Acids and Bases ionize one proton (or H+) at a time!• For weak acid reactions: HA + H2O → H3O+ + A−

K a =[H3O

+ ][A-][HA]

where K a is typically much less than 1

• For weak base reactions: B + H2O → HB+ + OH−

Kb =[HB+ ][OH-]

[B] where K b is typically much less than 1

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Acid – Base Equilibrium

Ionization: That’s What it’s All About! con’t. Working it Out! Calculating the pH of weak acids simply requires some quick problem solving…

• MONUMENTAL CONCEPT…• On the AP Exam the weak acid will be significantly weak so that the initial [HA]o is mathematically the

same as the equilibrium concentration − i.e. there is no need to subtracting the amount of weak acid ionizedas it is mathematically insignificant.

• If given the concentration of the weak acid, HA and the ionization constant for the acid, Ka, plug both into the equilibrium expression and solve. Both H3O+ and A− are equal (remember it's 1:1 always) so call them x and solve.

K a =[H3O

+ ][A- ][HA]

• Thus all you need to know is…

K a =[x][x][Mo ]

where x = [H3O+ ] or K b =

[x][x][Mo ]

where x = [OH- ]

• Never forget 141 10a bK K −× = ×

• Very important when you are given the ionization constant for the acid but need the conjugate base and vice-versa.• Percent Ionization

• Often will be asked to determine how ionized the weak acid or base is…

% =[x]

[Mo ]×100 where x = [H3O

+ ]or[OH− ]

Salts and pH: It’s all about Hydrolysis A salt is the PRODUCT of an acid base reaction; SOME salts affect the pH of the solution. So… an easy way to tell, you ask…?

• Ask yourself, which acid and which base are needed to make that salt?….were they strong or weak?1. A salt such as NaNO3 does not affect the pH of the solution; if it could hydrolyze water the following

would happen: Na+ + 2 H2O → NaOH + H3O+ NO3

− + H2O → HNO3 + OH− − Neither of the salt’s ions can hydrolyze water because it would result in the formation of a strong

acid (HNO3) or a strong base (NaOH), which would ionize 100% back into the ions. 2. K2S would make the solution more basic—SB (KOH) & WA (H2S)

− NOTE: Only the ion that forms a weak acid or base through water hydrolysis can actaully hydrolyze water.

In this case: S2− + H2O → H2S + OH−

The increase in OH− ions means the anion of this salt makes the solution more basic… 3. NH4Cl would make the solution more acidic—since WB (NH4OH – or NH3) & SA (HCl)

In this case: NH4

+ + H2O → NH3 + H3O+ The increase in H3O+ ion means the solution is more acidic…

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Acid – Base Equilibrium

Salts and pH: It’s all about Hydrolysis con’t Working it Out! Calculating the pH of a salt solution

• A salt in solution is either behaving as a weak acid or a weak base.• Remember, this means they do not ionize much. So treat them just like any other weak acid or base.• The MAJOR DIFFERENCE; no Ka or Kb value will be provided. The salt is the conjugate acid or base to

some weak acid or base that is provided - that K will be given you need to convert it. • Example, the salt NaC2H3O2; the Ka for HC2H3O2 is typically provided. The Kb needs to be

calculated.• K a × K b = 1×10−14 Very important!

• Then solve as shown below to find the pH…

K b =[x][x][Mo ]

where x = [OH- ]

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Acid – Base Equilibrium

Acid – Base Equilibrium Cheat Sheet

Relationships

Equilibrium Expression

K a =[H3O

+ ][A-][HA]

Kb =[HB+ ][OH-]

[B]+

3o

[x][x] where x = [H O ][M ]aK = -

o

[x][x] where x = [OH ][M ]bK =

+3

o

[x]% 100 where x = [H O ]or[OH ][M ]

−= ×

141 10a bK K −× = × pH = −log[H+]

pH + pOH = 14 pOH = −log[OH−]

Kw = [H3O+][OH−] = 1.0×10−14 @ 25°C Acidic pH < 7 [H3O+] > [OH−]

Neutral pH = 7 [H3O+] = [OH−] Basic pH > 7 [H3O+] < [OH−]

Conjugate acids and bases Ksp = [M+]x[A−]y MA(s) ⇌ xM+(aq)+ yA−(aq)

Connections

Equilibrium Buffers and Titrations

Precipitation and Qualitative Analysis Bonding and Lewis Structures − justify oxyacid strengths

Potential Pitfalls

Ka or Kb with salt pH

Weak acids and bases − be sure you know what you are using, acid or base – Ka or Kb; solving a Kb problem gives [OH−] thus you are finding the pOH!

Weak is about IONIZATION not CONCENTRATION

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Acid – Base Equilibrium

NH3(aq) + H2O() ⇌ NH4+(aq) + OH−(aq) Kb = 1.80 × 10-5

1. Ammonia reacts with water as indicated in the reaction above.

(a) Write the equilibrium constant expression for the reaction represented above.

+4

3

[NH ][OH ][NH ]bK

−

= 1 point is earned for the correct expression

(b) Calculate the pH of a 0.150 M solution of NH3

Kb =x2

M where x = [OH− ]

1.80x10−5 = x2

0.1501.64x10−3 = x− log x = pOH = 2.784pH = 11.216

1 point is earned for the correct set up and for calculating the concentration of hydroxide ions

1 point is earned for the correct pH

(c) Determine the percent ionization of the weak base NH3.

o3

[x]% 100[M ]

1.64 10% 100 1.09%[0.150]

−

= ×

×= × =

1 point is earned for the correct percent ionization

(d) Calculate the hydronium ion, H3O+, concentration in the above solution. Be sure to include units with your answer.

+ 143

14 14+ 12

3 3

+ pH3

+ 11.216 123

[H O ][OH ] = 1.00 10

1.0 10 1.00 10[H O ]= 6.10 10[OH ] 1.64 10

OR[H O ] = 10

[H O ] = 10 6.08 10

M

M

− −

− −−

− −

−

− −

×

× ×= = ××

= ×

1 point is earned for the concentration of hydronium ions with units of mol L−1 or M

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NMSI SUPER PROBLEM

Acid – Base Equilibrium

When a specified amount of ammonium nitrate (NH4NO3) is dissolved in water, the ammonium ions hydrolyze the water according to the partial reaction shown below. The resulting solution has a pH of 4.827.

(e) Complete the reaction above by drawing the complete Lewis structures for both products of the hydrolysis reaction.

+

+N

HH

HO

HH

H

1 point is earned for each correct Lewis structure.

2 points possible

(f) Determine the

(i) molarity (M) of the ammonium ions in this solution

[H3O+ ] = 10−pH = 10−4.827 = 1.49×10−5

Ka =Kw

Kb

= 1.00×10−14

1.80×10−5 = 5.56×10−10

5.56×10−10 = [1.49×10−5][1.49×10−5][M ]

M = 0.400

1 point is earned for calculating the Ka for ammonium ions 1 point is earned for calculating the concentration of ions in the solution.

(ii) number of moles ammonium ions in 250 mL of the above solution.

M = 0.400

0.400M = x mol0.250 L

x = 0.100 mol

1 point is earned for calculating the molar concentration and the number of moles of ammonium ions

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