Strength of Acids and Bases

Strength of Strength of Acids and BasesAcids and Bases

Do they ionize 100%?Do they ionize 100%?

Strong Acids :Strong Acids : Give up HGive up H++ easily easily

Dissociate completely (100%) in waterDissociate completely (100%) in water

HCl, HBr, HI, HNOHCl, HBr, HI, HNO33, H, H22SOSO44, HClO, HClO44, HClO, HClO33

Weak acids: (all others)Weak acids: (all others)Hold onto HHold onto H++

Few molecules dissociateFew molecules dissociate

Ex: HCEx: HC2HH33OO2 2

Strong/Weak Acid Strong/Weak Acid AnimationAnimationhttp://educypedia.karadimov.info/library/acid13.swf

HA

Let’s examine the behavior of an acid, HA, in aqueous solution.

What happens to the HA molecules in solution?

HA

H+

A-

Strong Acid

100% dissociation of HA

Would the solution be conductive?

Oh yeah…

HA

H+

A-

Weak Acid

Partial dissociation of HA

Would the solution be conductive?

Not really…

HA

H+

A-

Weak Acid

HA H+ + A-

At any one time, only a fraction of

the molecules

are dissociated.

Strong BasesStrong Bases: : Dissociate completely (100%) in waterDissociate completely (100%) in water

- Group I metal hydroxides (NaOH, LiOH, etc.)- Group I metal hydroxides (NaOH, LiOH, etc.)

- Some Group II metal hydroxides- Some Group II metal hydroxides

Ca(OH)Ca(OH)22, Ba(OH), Ba(OH)22, ,

Sr(OH)Sr(OH)22

Weak BasesWeak Bases

Only a few ions dissociateOnly a few ions dissociate

Ex: NHEx: NH 3 3 (ammonia) (ammonia)

Strength and ReactivityStrength and Reactivity

Acids/bases of the same initial molar concentration Acids/bases of the same initial molar concentration can react differently and conduct electricity can react differently and conduct electricity differently if one is weak and the other strong.differently if one is weak and the other strong.

Ex: Ex: 2M HCl 2M HCl Strong Acid, Strong Acid,

very conductive very conductive very reactivevery reactive

2M 2M HCHC22HH33OO2 2 Weak Acid Weak Acid

Weak ConductionWeak Conduction

Salad Dressing!!!Salad Dressing!!!

Conjugate Acid/Base PairsConjugate Acid/Base Pairs

Strong acid will have a weak conjugate baseStrong acid will have a weak conjugate baseStrong base will have a weak conjugate acidStrong base will have a weak conjugate acid

HydrolysisHydrolysis

Opposite reaction to neutralizationOpposite reaction to neutralization

Salt + Water Salt + Water Acid + BaseAcid + Base

Parent Acid/BaseParent Acid/Base

If you know the salt involved you should If you know the salt involved you should be able to determine which acid and base be able to determine which acid and base it would form if water is added.it would form if water is added.

Salt + Water Salt + Water Acid + BaseAcid + Base

Ex: Ex:

NaCl with water (HOH) would form HCl and NaOHNaCl with water (HOH) would form HCl and NaOH

You Try ItYou Try It

Name the “parent” acid and base that Name the “parent” acid and base that would be produced from these salts.would be produced from these salts.

Ex:Ex: Potassium chloridePotassium chloride

Magnesium carbonateMagnesium carbonate

pH and HydrolysispH and Hydrolysis

Salts can yield neutral, acidic or basic Salts can yield neutral, acidic or basic solutions depending on what type of acid solutions depending on what type of acid or base they produce.or base they produce.

SA/SB = NeutralSA/SB = Neutral

SA/WB = AcidicSA/WB = Acidic

WA/SB = BasicWA/SB = Basic

WA/WB = UndeterminedWA/WB = Undetermined

The Acid and Base The Acid and Base Dissociation Constant, Dissociation Constant,

Ka & Kb Ka & Kb

Setting up Ka and Kb ExpressionsSetting up Ka and Kb Expressions

weak acid:weak acid: CHCH33COOH + HCOOH + H22O ↔ HO ↔ H33OO++ + CH + CH33COOCOO--

Acid ionization constantAcid ionization constant: K: Kaa = [H = [H33OO++][CH][CH33COOCOO--]]

[CH[CH33COOH]COOH]

weak base:weak base: NHNH33 + H + H22O ↔ NHO ↔ NH44++ + OH + OH--

Base ionization constantBase ionization constant: K: Kbb = [NH = [NH44++][OH][OH--]]

[NH[NH33]]

Acid and base ionization constants are the measure of theAcid and base ionization constants are the measure of thestrengths of acids and bases.strengths of acids and bases.The larger the Ka/Kb value the stronger the acid or baseThe larger the Ka/Kb value the stronger the acid or base

KaKa Weak acidsWeak acids::

only ionize to a small extent only ionize to a small extent come to a state of chemical equilibrium.come to a state of chemical equilibrium.

Determine how much it ionizes by calculating the equilibrium Determine how much it ionizes by calculating the equilibrium

constant (constant (KaKa) )

Larger Larger Ka:Ka:stronger acid stronger acid more ions found in solution more ions found in solution more easily donate a protonmore easily donate a proton

HCOOH HCOOH (aq)(aq) ++ H H22O O (l)(l) < -- > H < -- > H33OO++ (aq)(aq) + HCOO + HCOO-- (aq)(aq)

Ka Ka = = [H[H33OO++][HCOO][HCOO--]]

[HCOOH][HCOOH]

Notice how the Ka ignores the water since Notice how the Ka ignores the water since we are dealing with dilute solutions of we are dealing with dilute solutions of acids, water is considered a constant and acids, water is considered a constant and doesn’t have a concentration valuedoesn’t have a concentration value

Practice:Practice:

1.1. A solution of a weak acid, “HA”, is made A solution of a weak acid, “HA”, is made up to be 0.15 M. Its pH was found to be up to be 0.15 M. Its pH was found to be 2.96. Calculate the value of 2.96. Calculate the value of KaKa..

Steps to follow:Steps to follow:1.1. Write balanced equationWrite balanced equation

2.2. Calculate [HCalculate [H++] using 10] using 10-pH-pH

3.3. Set up chart for equilibrium (ICESet up chart for equilibrium (ICE))

4.4. Solve using Ka expressionSolve using Ka expression

AnswerAnswer

1.1. HA HA (aq)(aq) + H + H22O O (l)(l) < -- >H < -- >H33OO++ (aq)(aq) + A + A--(aq)(aq)

2.2. [H[H33O+]= 10 O+]= 10 -2.96-2.96 = 0.0011 M = 0.0011 M

HA H2O < -- > H3O+ A-

0.15

-.0011

0.139 0.0011 0.0011

I

C

E

4.4. Ka = [Ka = [HH33OO++][ A][ A--]]

[HA][HA]

= = [0.0011][0.0011][0.0011][0.0011]

[0.139][0.139]

= 8.7 x 10 = 8.7 x 10 -6-6

Percent IonizationPercent IonizationThe fraction of acid molecules that dissociateThe fraction of acid molecules that dissociate

compared with the initial concentration of the acid.compared with the initial concentration of the acid.

Percent Ionization = Percent Ionization = [H[H33OO++]] x 100% x 100%

[HA [HA ii]]

For the previous question:For the previous question:

Percent Ionization = Percent Ionization = [0.0011][0.0011] x 100% =0.73 % x 100% =0.73 %

[0.15][0.15]

Practice:Practice:KaKa, for a hypothetical weak acid, HA, at 25°C , for a hypothetical weak acid, HA, at 25°C

is 2.2 x 10is 2.2 x 10-4-4..

a)a) Calculate [HCalculate [H33OO++] of a 0.20 M solution of HA.] of a 0.20 M solution of HA.

b) Calculate the percent ionization of HA.b) Calculate the percent ionization of HA.

AnswerAnswer

a) HA(aq) + H2O(l) < -- >H3O+(aq) + A-(aq)a) HA(aq) + H2O(l) < -- >H3O+(aq) + A-(aq)

Ka Ka = = [H[H33OO++] [A-] ] [A-] = 2.2 x 10= 2.2 x 10-4-4

[HA][HA]

2.2 x 102.2 x 10-4-4 = = ((xx)()(xx)) (0.20M - x)(0.20M - x)

xx22 = (2.2 x 10 = (2.2 x 10-4-4) (0.20 M)) (0.20 M)x x = 0.0066 M= 0.0066 M

The [HThe [H33OO++] is 0.0066 M.] is 0.0066 M.

Because it is a 1:1 ratio they are both the same concentration (x)

Since Ka is rather small this number can be disregarded

b) % ionization = b) % ionization = 0.0066 M0.0066 M x 100% = 3.3% x 100% = 3.3%

0.20 M0.20 M

The pH of 0.100 M H2C2O4 is 1.28.

Calculate the Ka for the weak acid.

H2C2O4 ⇄ H+ + HC2O4-

I 0.100 CE

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 CE

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 CE

pH = 1.28

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 CE

pH = 1.28

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 CE

pH = 1.28

[H+] = 10-1.28

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 CE

pH = 1.28

[H+] = 10-1.28

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 CE

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 CE

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 CE 0.05248 0.05248

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 CE 0.05248 0.05248

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 C - 0.05248E 0.04752 0.05248 0.05248

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 C - 0.05248E 0.04752 0.05248 0.05248

pH = 1.28

[H+] = 10-1.28

[H+] = 0.05248 M

1. The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 C - 0.05248E 0.04752 0.05248 0.05248

Ka = [H+][HC2O4-] =(0.05248)2

[H2C2O4] 0.04752

Ka = 5.8 x 10-2

1. The pH of 0.100 M H2C2O4 is 1.28.


H2C2O4 ⇄ H+ + HC2O4-

I 0.100 C - 0.05248E 0.04752 0.05248 0.05248

Ka = [H+][HC2O4-] =(0.05248)2

[H2C2O4] 0.04752

Ka = 5.8 x 10-2

KbKb

When using weak bases, the same rules When using weak bases, the same rules apply as with weak acids, except you are apply as with weak acids, except you are solving for pOH and using [OHsolving for pOH and using [OH--]]

If the pH of 0.40 M NH3 @ 25 oC is 11.427,

calculate the Kb.

NH3 + H2O ⇄ NH4+ + OH-

I 0.40 CE


calculate the Kb.

NH3 + H2O ⇄ NH4+ + OH-

I 0.40 CE


calculate the Kb.

NH3 + H2O ⇄ NH4+ + OH-

I 0.40 CE

pH = 11.427pOH = 2.573


calculate the Kb.

NH3 + H2O ⇄ NH4+ + OH-

I 0.40 CE

pH = 11.427pOH = 2.573


calculate the Kb.

NH3 + H2O ⇄ NH4+ + OH-

I 0.40 CE

pH = 11.427pOH = 2.573[OH-] =10-2.573

[OH-] =0.002673 M


calculate the Kb.

NH3 + H2O ⇄ NH4+ + OH-

I 0.40 CE

pH = 11.427pOH = 2.573[OH-] =10-2.573

[OH-] =0.002673 M


calculate the Kb. NH3 + H2O ⇄ NH4

+ + OH-

I 0.40 C - 0.002673E 0.3973 0.002673 0.002673

[OH-] = 0.002673 M



+ + OH-

I 0.40 C - 0.002673E 0.3973 0.002673 0.002673

[OH-] = 0.002673 M



+ + OH-

I 0.40 C - 0.002673E 0.3973 0.002673 0.002673

[OH-] = 0.002673 M

Kb = [NH4+][OH-] = (0.002673)2

[NH3] 0.3973

Kb = 1.8 x 10-5



+ + OH-

I 0.40 C - 0.002673E 0.3973 0.002673 0.002673

[OH-] = 0.002673 M

Kb = [NH4+][OH-] = (0.002673)2

[NH3] 0.3973

Kb = 1.8 x 10-5

The pH of 0.20 M NaCN is 11.456, calculate the Kb for CN-.



CN- + H2O ⇄ HCN+ OH-

pH = 11.456pOH = 2.55[OH-] = 10-2.55

[OH-] = .002858M


CN- + H2O ⇄ HCN+ OH-

pH = 11.456pOH = 2.55[OH-] = 10-2.55

[OH-] = .002858M


CN- + H2O ⇄ HCN + OH-

I 0.20C - 0.002858E 0.1971 0.002858 0.002858



I 0.20C - 0.002858E 0.1971 0.002858 0.002858



I 0.20C - 0.002858E 0.1971 0.002858 0.002858

Kb = [HCN][OH-] = (0.002858)2= 4.1 x 10-5

[CN-] 0.1971



I 0.20C - 0.002858E 0.1971 0.002858 0.002858

Kb = [HCN][OH-] = (0.002858)2= 4.1 x 10-5

[CN-] 0.1971

Calculate the pH of 0.020 M H3BO3 (Ka = 3.8 x 10-10)

(weak acid dissociates one H+ at a time)

H3BO3 ⇄ H+ + H2BO3-

I 0.020 MC - xE 0.020 - x x x

Calculate the pH of 0.020 M H3BO3 (Ka = 3.8 x 10-10)

(weak acid dissociates one H+ at a time)

H3BO3 ⇄ H+ + H2BO3-

I 0.020 MC - xE 0.020 - x x x

disregard small Ka

x2 = 3.8 x 10-10

0.020

x = [H+] = 2.76 x 10-6 M

pH = -Log[2.76 x 10-6]

pH = 5.42

2 sig figs due to molarity and Ka

Strength of Acids and Bases

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