Acid and BaseEquilibrium

Some Properties of Acids

• Produce H3O+ ions in water (the hydronium ion is a hydrogen

ion attached to a water molecule)

• Taste sour

• Corrode metals

• Electrolytes

• React with bases to form a salt and water

• pH is less than 7

• Turns blue litmus paper to red

Some Properties of Bases

Produce OHProduce OH-- ions in water ions in water

Taste bitter, chalkyTaste bitter, chalky

Are electrolytesAre electrolytes

Feel soapy, slipperyFeel soapy, slippery

React with acids to form salts and waterReact with acids to form salts and water

pH greater than 7pH greater than 7

Turns red litmus paper to blue “Basic Blue”Turns red litmus paper to blue “Basic Blue”

Arrhenius acid is a substance that produces (H3O+) in water

Arrhenius base is a substance that produces OH- in water

Acid/Base definitionsDefinition 1: Arrhenius

Acid/Base Definitions

Definition #2: Brønsted – LowryDefinition #2: Brønsted – Lowry

Acids – proton donorAcids – proton donor

Bases – proton acceptorBases – proton acceptor

A “proton” is really just a hydrogen A “proton” is really just a hydrogen atom that has lost it’s electron!atom that has lost it’s electron!

A Brønsted-Lowry acid is a proton donorA Brønsted-Lowry base is a proton acceptor

acidconjugate

basebase conjugate

acid

ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES

The Brønsted definition means NHThe Brønsted definition means NH33 is a BASE is a BASE

in water — and water is itself an ACIDin water — and water is itself an ACID

BaseAcidAcidBaseNH4

+ + OH-NH3 + H2OBaseAcidAcidBase

NH4+ + OH-NH3 + H2O

Conjugate PairsConjugate Pairs

Learning Check!

Label the acid, base, conjugate acid, and Label the acid, base, conjugate acid, and conjugate base in each reaction:conjugate base in each reaction:

HCl + OHHCl + OH--   Cl   Cl-- + H + H22OO HCl + OHHCl + OH--   Cl   Cl-- + H + H22OO

HH22O + HO + H22SOSO44   HSO   HSO44-- + H + H33OO

++ HH22O + HO + H22SOSO44   HSO   HSO44-- + H + H33OO

++

AcidAcidAcidAcid

AcidAcidAcidAcid

BaseBaseBaseBase

BaseBaseBaseBase

Conj.Conj.BaseBaseConj.Conj.BaseBase

Conj.Conj.BaseBaseConj.Conj.BaseBase

Conj.Conj.AcidAcidConj.Conj.AcidAcid

Conj.Conj.AcidAcidConj.Conj.AcidAcid

Acids & Base DefinitionsAcids & Base Definitions

Lewis acid - a substance Lewis acid - a substance that accepts an electron that accepts an electron pairpair

Lewis base - a substance Lewis base - a substance that donates an electron that donates an electron pairpair

Definition #3 – Lewis Definition #3 – Lewis

Lewis Acids & BasesLewis Acids & BasesFormation of hydronium ion is also an Formation of hydronium ion is also an

excellent example.excellent example.

•Electron pair of the new O-H bond Electron pair of the new O-H bond originates on the Lewis base.originates on the Lewis base.

HH

H

BASE

••••••

O—HO—H

H+

ACID

Lewis Acid/Base ReactionLewis Acid/Base Reaction

More About WaterMore About Water

KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC

In a neutral solution [HIn a neutral solution [H33OO++] = [OH] = [OH--]]

and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M

OH-

H3O+

OH-

H3O+

AutoionizationAutoionization

More About WaterMore About WaterHH22O can function as both an ACID and a BASE.O can function as both an ACID and a BASE.

In pure water there can be AUTOIONIZATIONIn pure water there can be AUTOIONIZATION

Equilibrium constant for water = KEquilibrium constant for water = Kww

KKww = 1.00 x 10 = 1.00 x 10-14-14 = [H = [H33OO++] [OH] [OH--] at 25 ] at 25 ooCC

Take -logs of both sidesTake -logs of both sides

14 = pH + pOH14 = pH + pOH

The pH scale is a way of The pH scale is a way of expressing the strength of expressing the strength of acids and bases. Instead of acids and bases. Instead of using very small numbers, we using very small numbers, we just use the NEGATIVE power just use the NEGATIVE power of 10 on the Molarity of the of 10 on the Molarity of the HH33OO++ (or OH (or OH--) ion.) ion.

Under 7 = acidUnder 7 = acid7 = neutral7 = neutral

Over 7 = base Over 7 = base

Calculating the pHpH = - log [H3O+]

((Remember that the [ ] mean Molarity or concentration)

Example: If [H3O+] = 1 X 10-10

pH = - log 1 X 10-10

pH = - (- 10)

pH = 10

Example: If [H3O+] = 1.8 X 10-5, what is the pH?

pH = - log 1.8 X 10-5

pH = - (- 4.74)

pH = 4.74

pH calculations – Solving for HpH calculations – Solving for H33OO++pH calculations – Solving for HpH calculations – Solving for H33OO++

If the pH of Coke is 3.12, what is the [If the pH of Coke is 3.12, what is the [H3O+] ?] ?

Because pH = Because pH = -- log [ log [H3O+] then] then

-- pH = log [ pH = log [H3O+]]

Take antilog (10Take antilog (10xx) of both) of both sides and get sides and get

1010--pH pH == [[H3O+]]

[[H3O+] = 10] = 10--3.123.12 = 7.58 x 10 = 7.58 x 10-4-4 M M

*** to find antilog on your calculator, look for “Shift” *** to find antilog on your calculator, look for “Shift” or “2or “2nd nd function” and then the log buttonfunction” and then the log button

Strong and Weak Acids/BasesStrong and Weak Acids/Bases

• Generally acids and bases are divided into Generally acids and bases are divided into STRONG or WEAK ones.STRONG or WEAK ones.

• Strong acids – 6 (HNOHNO33, HCl, HBr, HI, HClO, HCl, HBr, HI, HClO44, , HH22SOSO44) all others are weak) all others are weak

• Strong Bases – group 1 and 2 hydroxides (all others are weak) except Be(OH)2

Strong acids and basesStrong acids and bases

Find the pH of these:Find the pH of these:

1)1) A 0.15 M solution of A 0.15 M solution of Hydrochloric acidHydrochloric acid

2) 2) A 3.00 X 10A 3.00 X 10-7-7 M M solution of Nitric acidsolution of Nitric acid

3) What is the pH of 3) What is the pH of 0.0034M H0.0034M H22SOSO44??

pH = - log [pH = - log [H3O+]]

pH = - log 0.15pH = - log 0.15

pH = - (- 0.82)pH = - (- 0.82)

pH = 0.82pH = 0.82

pH = - log 3 X 10pH = - log 3 X 10-7-7

pH = - (- 6.52)pH = - (- 6.52)

pH = 6.52pH = 6.52

• Strong acids and bases - are 100 % ionized.Strong acids and bases - are 100 % ionized.• No equilibrium is set upNo equilibrium is set up• [Acid] = [H30+] ( 1:1 ratio)[Acid] = [H30+] ( 1:1 ratio)• ACID ACID H A + HH A + H22O O H3O+ + A-• BASE B + HH22O O BH BH++ + OH + OH--

Finding pH Finding pH

pH of strong basespH of strong bases

What is the pH of the What is the pH of the 0.0010 M NaOH solution? 0.0010 M NaOH solution?

[OH-] = 0.0010 (or 1.0 X 10[OH-] = 0.0010 (or 1.0 X 10-3-3 M) M)

pOH = - log 0.0010pOH = - log 0.0010

pOH = 3pOH = 3

pH = 14 – 3 = 11pH = 14 – 3 = 11

OR KOR Kww = [H = [H33OO++] [OH] [OH--]]

[H[H3OO++] = 1.0 x 10] = 1.0 x 10-11-11 M M

pH = - log (1.0 x 10pH = - log (1.0 x 10-11-11) = 11.00) = 11.00

What is the pH of a 2 x 10-3 M HNO3 solution?

HNO3 is a strong acid – 100% dissociation.

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)

pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7

Start

End

0.002 M

0.002 M 0.002 M0.0 M

0.0 M 0.0 M

What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?

Ba(OH)2 is a strong base – 100% dissociation.

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

Start

End

0.018 M

0.018 M 0.036 M0.0 M

0.0 M 0.0 M

pH = 14.00 – pOH = 14.00 - (-log(0.036) )= 12.5615.4

Weak acidsWeak acids are much less than 100% ionized in water.are much less than 100% ionized in water. Equilibrium is set up. Common weak acids are acetic acid Equilibrium is set up. Common weak acids are acetic acid

and weak base is ammoniaand weak base is ammonia

Weak Acids/BasesWeak Acids/Bases

Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases

Consider acetic acid, HCConsider acetic acid, HC22HH33OO22 (HOAc) (HOAc)

HCHC22HH33OO2(aq)2(aq) + H + H22OO(l)(l) H H33OO++ (aq)(aq) + C + C22HH33OO22 ––(aq)(aq)

AcidAcid Conj. base Conj. base

Ka [H3O+][OAc- ]

[HOAc] 1.8 x 10-5Ka

[H3O+][OAc- ][HOAc]

1.8 x 10-5

((K is designated KK is designated Kaa for ACID) for ACID)

K gives the ratio of ions (split up) to molecules (don’t split up)K gives the ratio of ions (split up) to molecules (don’t split up)

Ionization Constants for Acids/BasesIonization Constants for Acids/Bases

AcidsAcids ConjugateConjugateBasesBases

Increase strength

Increase strength

Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids

Equilibrium Constants Equilibrium Constants for Weak Acidsfor Weak Acids

Weak acid has KWeak acid has Kaa < 1 < 1

Leads to small [HLeads to small [H33OO++] and a pH of 2 - 7] and a pH of 2 - 7

Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases

Equilibrium Constants Equilibrium Constants for Weak Basesfor Weak Bases

Kb – base dissociation constant and Kb – base dissociation constant and

is temp dependentis temp dependent

Weak base has KWeak base has Kbb < 1 < 1

Leads to small [OHLeads to small [OH--] and a pH of 12 - 7 ] and a pH of 12 - 7

Relation Relation

of Kof Kaa, K, Kbb, ,

[H[H33OO++] ]

and pHand pH

Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidYou have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the

equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,

and the pH.and the pH.

Step 1. Define equilibrium concs. in ICE Step 1. Define equilibrium concs. in ICE

table.table.

[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]

initialinitial

changechange

equilibequilib

1.001.00 00 001.001.00 00 00

-x-x +x+x +x+x-x-x +x+x +x+x

1.00-x1.00-x xx xx1.00-x1.00-x xx xx

Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid

Step 2. Write KStep 2. Write Kaa expression expression

You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.

Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

This is a quadratic. Solve using quadratic This is a quadratic. Solve using quadratic formula.formula.

or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)or you can make an approximation if x is very or you can make an approximation if x is very small! (Rule of thumb: 10small! (Rule of thumb: 10-5-5 or smaller is ok) or smaller is ok)

Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid

Step 3. Solve KStep 3. Solve Kaa expression expression

You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.

Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

First assume x is very small because First assume x is very small because KKaa is so small. is so small.

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

Now we can more easily solve this Now we can more easily solve this approximate expression.approximate expression.

Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid

Step 3. Solve KStep 3. Solve Kaa approximate approximate expression expression

You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

x = [x = [HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M

pH = - log [pH = - log [HH33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) = 2.37) = 2.37

Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidCalculate the pH of a 0.0010 M solution of formic Calculate the pH of a 0.0010 M solution of formic

acid, HCOacid, HCO22H.H.

HCOHCO22H + HH + H22O O HCO HCO22-- + H + H33OO++

KKaa = 1.8 x 10 = 1.8 x 10-4-4

Approximate solutionApproximate solution

[H[H33OO++] = 4.2 x 10] = 4.2 x 10-4-4 M, pH = 3.37 M, pH = 3.37

Exact SolutionExact Solution

[H[H33OO++] = [HCO] = [HCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M

[HCO[HCO22H] = 0.0010 - 3.4 x 10H] = 0.0010 - 3.4 x 10-4-4 = 0.0007 M = 0.0007 M

pH = 3.47 pH = 3.47

Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseYou have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 1. Define equilibrium concs. in ICE tableStep 1. Define equilibrium concs. in ICE table

[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial

changechange

equilibequilib

0.0100.010 00 000.0100.010 00 00

-x-x +x+x +x+x-x-x +x+x +x+x

0.010 - x0.010 - x x x xx0.010 - x0.010 - x x x xx

Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseYou have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 2. Solve the equilibrium expressionStep 2. Solve the equilibrium expression

Kb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x2

0.010 - xKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x2

0.010 - x

Assume x is small, soAssume x is small, so x = [OHx = [OH--] = [NH] = [NH44

++] = 4.2 x 10] = 4.2 x 10-4-4 M M

and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 ≈ 0.010 M ≈ 0.010 M

The approximation is validThe approximation is valid !!

Equilibria Involving A Weak Equilibria Involving A Weak BaseBase

You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 3. Calculate pHStep 3. Calculate pH[OH[OH--] = 4.2 x 10] = 4.2 x 10-4-4 M Mso pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37Because pH + pOH = 14,Because pH + pOH = 14,

pH = 10.63pH = 10.63

Percent Ionization for Weak Acids Most weak acids ionize < 50% Percent ionization (p)

General Weak Acid: HA(aq) + H2O(l) H3O+ (aq) + A- (aq)

• p varies depending on concentration: increase [HA ] decreases p

• This is caused by Le Chatelier’s Principle

• Remember, for strong acids we assume complete ionization (100%)

100xacidofionconcentrat

onHydroniumiofionconcentrat=p

100)(3 x][HA

]O[H=p

(aq)

aq+

ExamplesThe pH of a 0.10mol/L methanoic acid (HCOOH)

solution is 2.38. Calculate the percent ionization of methanoic acid Ans: 4.2%

Calculate the acid ionization constant (Ka )of acetic acid if a 0.1000mol/L solution at equilibrium at SATP has a percent ionization of 1.3% (Hint: ICE table Ans: 1.7x10-5

Relationship Between Ka and Kb for Conjugate Base Pairs

• Recall: Conjugate Pairs – an acid and base that differ by one hydrogen

• Lets consider the hypothetical weak acid, HA, and its conjugate base, A -

HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

HA

]

aq

3

][

(aq)(aq)][AO[H=K

)(

+

a

• Now consider the hypothetical weak base, A- in water

A- (aq) + H2O (l) HA (aq) + OH- (aq)

• Now let’s put that together

HA (aq) + H2O (l) H3O+ (aq) + A- (aq) Ka

A- (aq) + H2O (l) HA (aq) + OH- (aq) Kb

Add both equations

2H2O (l) H3O+ (aq) + OH- (aq) Kw

Relationship Between Ka and Kb for Conjugate Base Pairs

)(

]

]aq[A

(aq)(aq)][HA[OH=Kb

Relationship Between Ka and Kb for Conjugate Base Pairs

HA aq

aqaq3

][

]][AO[H=K

)(

)()(+

a

OHHA

aq

aqaq

][A

]][[=K

)(

)()(

b

ba KxK OHHA

HA aq

aqaq

aq

aqaq3

][A

]][[x

][

]][AO[H=

)(

)()(

)(

)()(+

OH aqaq3 ]][O[H= )()(+ Kw

Recall: Autoionization of water H2O(l) + H2O(l) H3O+(aq) + OH-(aq)

Kw=1.00x10-14 **must remember this value**

Relationship Between Ionization Constants for Conjugate Base Pairs

For acids and bases whose chemical formulas differ by only one hydrogen (conjugate pairs) the following apply:

Kw = Ka x Kb Kb =Kw/Ka Ka = Kw/Kb

• Therefore if only the Ka value is available in the table, we can determine the conjugate pairs Kb by using the

above equations Note: these equations show the larger the Ka the

smaller the KbStronger acid weaker conjugate baseWeaker acid stronger conjugate base

Learning check!

1. What is the value of the base ionization constant (Kb) for the acetate

ion, C2H3O2- (aq)

Ans: 5.6x10-10

2. Calculate the percent ionization of propanoic acid, HC3H5O2(aq), if a 0.050

mol/L solution has a pH of 2.78Ans. 3.3%

Rapid changes in pH can kill fish and other

organisms in lakes and streams.

Soil pH is affected and can kill plants and create

sinkholes