weak acids & bases

WEAK ACIDS & BASES

SCH 4U

WEAK ACIDS Ionize only partially in water,

exist primarily in molecule form

Dynamic equilibrium established between unreacted molecules and ions formed from rxn with water

Like all equilibria, can be shifted by removal/ addition of reactants or products

WEAK BASES Have a weak attraction for protons

Recall: the conjugate base of a strong acid is a weak base

HA + H2O A– + H3O+

Usually non-hydroxide bases(Recall Arrhenius vs. Bronsted-Lowry

definitions)

PERCENT IONIZATIONFor weak acids:

p = [H3O+] x 100%

[HA]

For weak bases:p = [OH-] x 100%

[B]

Ex. The pH of a 0.10 mol/L methanoic acid solution is 2.38. What is the percent ionization of methanoic acid?

PERCENT IONIZATION EXAMPLEEx. The pH of a 0.10 mol/L methanoic acid

solution is 2.38. What is the percent ionization of methanoic acid?

HCO2H(aq) H+(aq) + HCO2

-(aq)

[H+(aq)] = 10-pH

= 10-2.38

= 4.2 x 10-3 mol/L

PERCENT IONIZATION EXAMPLE

p = conc. of acid ionized x 100% conc. of acid solute

p = 4.2 x 10-3 mol/L x 100% 0.10 mol/L

p = 4.2 %

Therefore methanoic acid ionizes 4.2% in a 0.10 mol/L solution

i.e. HCO2H(aq) H+(aq) + HCO2

–(aq)

4.2 %

IONIZATION CONSTANTS (Ka) FOR WEAK ACIDS

Equilibrium constant found as before, called “acid ionization constant,” Ka

E.g. for acetic acid:HC2H3O2(aq) H+

(aq) + C2H3O2-(aq)

Ka = [H+][C2H3O2-]

[HC2H3O2]

CALCULATING Ka FROM PERCENT IONIZATION Calculate the acid ionization constant, Ka,

of acetic acid if a 0.1000 M solution at equilibrium at SATP has a percent ionization of 1.3%

HC2H3O2(aq) H+(aq) + C2H3O2

-(aq)

Ka = [H+][C2H3O2-]

[HC2H3O2]

A percent ionization of 1.3% means initial [HC2H3O2] is diminished by 1.3% by time system reaches equilibrium

Note in this example molar ratio is 1:1:1

Use an ICE table:

Solve for x:x = 0.1 mol/L X 0.013x = 0.0013 mol/L

HC2H3O2(aq)

H+(aq)

+ C2H3O2

-

(aq)

Initial 0.1000 0 0

Change - x + x + x

Equilibrium 0.1000 - x 0 + x 0 + x

HC2H3O2(aq)

H+(aq)

+ C2H3O2

-

(aq)

Initial 0.1000 0 0

Change - 0.0013 + 0.0013 + 0.0013

Equilibrium 0.0987 0.0013 0.0013

Now we can calculate Ka:

Ka = [H+][C2H3O2-]

[HC2H3O2]

= (0.0013) (0.0013) (0.0987)

Ka = 1.7 x 10-5

HC2H3O2(aq)

H+(aq)

+ C2H3O2

-

(aq)

Initial 0.1000 0 0

Change - 0.0013 + 0.0013 + 0.0013

Equilibrium 0.0987 0.0013 0.0013

PERCENT IONIZATION & CONCENTRATION Ka values provide a means of comparing

relative strengths of acids

Can also compare % ionization values, but only when acids are equal in initial conc.

Data for acetic acid

PERCENT IONIZATION & CONCENTRATION More dilute the solution, greater the

degree of ionization

Can explain using Le Chatelier’s Principle:

HA(aq) A–(aq) + H+

(aq)

IONIZATION CONSTANTS (Kb) FOR WEAK BASES

Equilibrium constant called “base ionization constant,” Kb

E.g. for ammonia:

NH3(aq) + H2O(l) OH-(aq) + NH4

+(aq)

Kb = [OH-][NH4+]

[NH3]

Note: many weak bases contain one or more N atom, others are conjugate bases of strong acids

RELATIONSHIP BETWEEN Ka AND Kb

Consider an acetic acid solution at equilibrium:

HC2H3O2(aq) + H2O(l) H3O+(aq) +

C2H3O2-(aq)

As a base, the acetate ion also reacts with water, establishing an equilibrium:

C2H3O2-(aq) + H2O(l) OH-

(aq) + HC2H3O2(aq)

Ka = [H3O+][C2H3O2

-] [HC2H3O2]

Kb = [OH-][HC2H3O2] [C2H3O2

-]

RELATIONSHIP BETWEEN Ka AND Kb

Ka x Kb =

Kw = Ka x Kb

[H3O+] [C2H3O2-] x [OH-]

[HC2H3O2] [HC2H3O2] [C2H3O2

-]

= [H3O+] [OH-]

Ka = Kw

Kb

Kb = Kw

Ka

STRENGTH GENERALIZATIONS The conjugate base of a strong acid is a

______________ base

The conjugate base of a weak acid is a ______________ base

The conjugate base of a very weak acid is a ______________ base

* Go over page 562 together

very weak

weak

strong

STRENGTH GENERALIZATIONS The conjugate acid of a strong base is a

______________ acid

The conjugate acid of a weak base is a ______________ acid

The conjugate acid of a very weak base is a ______________ acid

very weak

weak

strong

CALCULATING CONC & pH OF A WEAK ACID GIVEN Ka

Calculate the hydrogen ion conc. and pH of a 0.10 mol/L acetic acid solution. Ka for acetic acid is 1.8 x 10-5.

*First need to compare Ka’s of all equilibria that may contribute H+ to the system...

Ka = 1.8 x 10-5 Kw = 1.0 x 10-14

HC2H3O2(aq) H+(aq) + C2H3O2

-

(aq)

I 0.10 0 0C - x +x +xE 0.10 – x x x

Ka = [H+][C2H3O2-]

[HC2H3O2]

1.8 x 10-5 = x2

0.10 –x

Could solve quadratic or could make it simpler...

FINISH... on p.563!

FINDING Kb, GIVEN CONC & pHFor diethylamine (CH3CH2)2NH, the pH of a 2.6 x 10-2 M solution is 11.56. What is Kb for diethylamine?

B(aq) + H2O(l) OH-(aq) + HB+

(aq)

Can use pH to det. [OH]...

pH = 11.56 therefore pOH = 14.00 – 11.56 = 2.44

 pOH = -log[OH-][OH-] = 10-2.44

[OH-] = 3.6 x 10-3 M

B(aq) + H2O(l) OH-(aq) + HB+

(aq)

I 2.6 x 10-2 0 0C -x +x +xE 2.6 x 10-2 – x x x

x = 3.6 x 10-3  

Kb = [OH-][HB+]

[B]

Kb = (3.6 x 10-3)2

2.6 x 10-2 – 3.6 x 10-3

= 5.8 x 10-4 See summary of problem-solving steps p.574

POLYPROTIC ACIDS E.g. Sulfuric acid H2SO4, boric acid H3BO3

Different Ka values (Ka1, Ka2, etc.)

In general,

Ka1 > Ka2 > Ka3 ...

See p. 574, 575

* Because Ka1 is usually >> Ka2, Ka3, etc., typically use just Ka1 to determine pH

HOMEWORK

p. 579 # 3-6, 11-13