PA RT 2

UNDERSTANDING ACID-BASE EQUILIBRIUM

CONJUGATE ACID-BASE PAIRS

• Conjugate acid-base pair:• Two molecules or ions that differ because of the transfer

of a proton

• Conjugate base of an acid:• Particle that remains when a proton is removed from the

acid

• Conjugate acid of a base• Is the particle formed when the base receives the proton

from the acid.

TRY THESE…

• Identify the conjugate acid-base pairs in the following reactions:

HBr(aq) + H2O(l) H3O+(aq) + Br-

(aq)

HS-(aq) + H2O(l) H2S(aq) + OH-

(aq)

HClO4 (aq) + H2O(l) H3O+(aq) + ClO4

-(aq)

O2-(aq) + H2O(l) 2OH-

(aq)

H2S(aq) + NH3(aq) NH4+

(aq) + HS-(aq)

LEWIS ACIDS & BASES

• Lewis defined them as:Acid – an electron pair acceptorBase – an electron pair donor

THE ACID-BASE EQUILIBRIUM

• Strong Acids • Examples of Strong Acids:• Binary Acids: HCl, HBr, HI• Oxyacids (contain polyatomics): HNO3, H2SO4,

HClO3, H3PO4

• Strong acids ionize completely (100%) in water (equilibrium favours products, lies to the right)

• This means 100% of the acid will turn into ions. • therefore a quantitative reaction and a poor

equilibrium condition. SO, if you have 0.2M HCl, it will turn into 0.2 M H3O+

THE ACID-BASE EQUILIBRIUM

•Strong Bases • Examples of Strong Bases:• Oxides & Hydroxides of alkali metals (group I) and alkaline

earth metals (group 2) below beryllium (eg NaOH, MgO, Na2O)

• Strong bases dissociate completely in water (equilibrium favours products, lies to the right)

• This means 100% of the base will turn into ions. • therefore a quantitative reaction and a poor equilibrium

condition. SO, if you have 0.2M NaOH, it will turn into 0.2 M OH-

CALCULATIONS THAT INVOLVE STRONG ACIDS & BASES

• Strong acids/bases (and strong electrolytes) completely dissociate into ions in water • [H3O]+

(aq) is equal to the [strong acid] • [OH]-

(aq) is equal to the [strong base]

• The stronger the acid, the weaker its conjugate base and in a similar manner, the weaker the acid the stronger its conjugate base.

• Weak acids & bases dissociate partially in water• You cannot determine the concentrations of ions of weak

acids/bases/electrolytes the same way you could with strong acids/bases

• This means we will have to use the concept of EQUILIBRIUM!!!!

EXAMPLE

• Find the concentration of hydronium ions in 4.5 mol/L HCl• Since HCl is a strong acid, it will completely

dissociate into ions.HCl + H2O H3O+ + Cl-

4.5 M 4.5 MSO, the [H3O+] is 4.5 M

EXAMPLE

• Is this solution acidic or basic: 31.9 mL of 2.75 M HCl added to 125 mL of 0.05 M Mg(OH)2?

1. Find the moles of H3O+ (since HCl is strong) 2. HCl + H2O → H3O+ + Cl- (1:1 ratio) 3. n = cv = (2.75 M)(0.0319 L) = 0.0877 mol 4. Find the moles of OH- (Mg(OH)2 is strong) 5. Mg(OH)2 + H2O → Mg2+ + 2OH- (1:2 ratio)6. n = cv = (0.05 M)(0.125L) = 0.00625 mol x 2 = 0.0125 mol

of hydroxide ion 7. Combine the two! 0.0877 mol HCl - 0.0125 mol OH- leaves

us with 0.0752 mol H3O+ 8. c = n/v = 0.0752 / 0.1569 L = 0.479 M [H3O+]

AMPHOTERIC (AMPHIPROTIC)

•These are substances that appear to act as Brønsted-Lowry acids or bases in different chemical reactions.•An equilibrium condition.•A substance that can donate or accept protons.•Like water or the bicarbonate ion in baking soda.

AMPHOTERIC (AMPHIPROTIC)

H2O(l) + HCO3- (aq) H2CO3 (aq) + OH-

(aq)Acid Base Conjugate Conjugate

acid base

H2O(l) + HCO3- (aq) CO3

-(aq) + H3O+

(aq)Base Acid Conjugate Conjugate

base acid

AMPHOTERIC (AMPHIPROTIC)

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

(aq)

Water acts both as an acid and a base in its “autoionization”.

In the above reaction there are two conjugate acid-base pairs.

ION-PRODUCT CONSTANT OF WATER (KW)

TRY THESE…

A solution of lithium hydroxide, LiOH(aq), is made by placing 2.00 mol of the base into 1.50L of solution. What is the concentration of hydronium ions in this solution at 25 degrees Celcius?

ESTABLISHING pH and pOH

• The pH of a solution is the negative logarithm of the hydronium ion concentration (in mol/L).

pH = -log [H3O+][H3O+] = 10-pH

pOH = -log [OH-][OH-] = 10-pH

THE RELATIONSHIP BETWEEN pH, pOH & pKw

pKw = pH + pOH = 14

TRY THIS