Acids and Bases
Properties of Acids and Bases
Pg 236
ACIDS
BASES
Sour taste (vinegar)
Bitter taste (baking soda)
• Why do Acids and bases change
React with some metal to form H2 gas Feels slippery (soap)
(Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
Turns blue litmus red
Turns red litmus blue
Arrhenius Definition
• Acid: produces H+ (or H3O+) when dissolved in
water
HCl(aq)  H+(aq) + Cl-(aq)
H+(aq) + H2O(l)  H3O+(aq)
NOTE: H3O+ = hydronium ion
• Base: produces OH- when dissolved in water.
NaOH(s)  Na+(aq) + OH-(aq)
Bronsted-Lowry Definition
Acids: proton (H+) donors
HF(aq)  H+(aq) + F-(aq)
H+(aq) + H2O(l)  H3O+(aq)
Bases: proton acceptors
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
H
H
H
N
H
+
H
H
N+
H
H
H2O: acts as an acid and a base = AMPHOTERIC
Strength of Acids
Strong Acids:  ionize (splits up into ions) almost 100% in
water
 mostly ions in solution
 amount of HCl present is negligeable
HCl(aq)  H+(aq) + Cl-(aq)
Weak acids:  ionize poorly in water
 not many of these ions present in solution
 mostly acetic acid (HC2H3O2)
HC2H3O2(aq)
C2H3O2-(aq) + H+(aq)
NOTE: strong acids are strong electrolytes and will conduct
electricity better than weak acids.
Strength of Bases
Strong Bases: ionize almost 100% in water
NaOH(s)  Na+(aq) + OH-(aq)
Weak Bases: ionize poorly in water
NH3(l) + H2O(l)
NH4+(aq) + OH-(aq)
NOTE: strong bases are strong electrolytes
Conjugate Acids and
Conjugate Bases
• these differ by only one proton
• Examples
Lose a proton
HCl
ClAcid
Conjugate base of
HCl
Gain a proton
SO42Base
HSO4Conjugate acid of
SO42-
Reactions with Water
H
H
C
H
O
+
C
H
acetic acid
H
O
O H
H
H
C
H
water
O
C
H
H
+
O
-
O
acetate ion
Conjugate acid-base pair: CH3CO2H/CH3CO2Conjugate acid-base pair: H2O/H3O+
H
hydronium
ion
Monoprotic, Diprotic and Triprotic
Monoprotic

donates one acidic proton

eg: HCl + H2O  H3O+ + Cl
only one H+ to donate
Diprotic

donates two acidic protons

eg: H2SO4 + H2O  H3O+ + HSO4
HSO4- + H2O  H3O+ + SO42-
Triprotic

donates three acidic protons

eg: H3PO4 + H2O  H3O+ + H2PO4
three H+ to donate
Homework
• Pg 251 #1, 2
• Pg 253 #4, 5, 6
pH < 7
pH = 7
pH > 7
acidic
neutral
basic
pH
 a measure of acid strength
 By definition all acids contain at least one acidic
proton = H+
 HA is a symbol used to represent any general acid
HA
H+(aq) + A-(aq)
H+ + H 2 O  H3 O +
  [H+] = [H3O+]
 If a lot of H3O+ is produced the solution is very
acidic.
 pH is directly related to [H3O+].
Self-Ionization of Water
H2O(l) + H2O(l)  H3O+(aq) + OH-(aq)
This reaction does not occur to any great extent.
[H3O+] = 1 x 10-7 mol/L
[OH-] = 1 x 10-7 mol/L
Because both concentrations are equal water is
said to be neutral.
Therefore, if
[H3O+] = [OH-]
neutral
[H3O+] > [OH-]
acidic
[H3O+] < [OH-]
basic
NOTE: [H3O+][OH-] = 1.0 x 10-14
pH = -log[H3O+]
• Expressing hydronium concentrations in
scientific notation isn’t very convenient.
The pH scale was developed to make the
expression of H3O+ concentration more
convenient.
• [H3O+] is the concentration in mol/L
Example 1: pH of Water
The concentration of H3O+ is 1.0 x 10-7. Calculate
the pH.
pH = -log[H3O+]
= -log(1.0 x 10-7)
= -(-7)
=7
The pH of water is 7. Therefore pH 7 is neutral.
Example 2
Determine the pH of a 1M solution of HCl.
HCl (aq)  H+ + Cl1M
x
Therefore [H3O+] = 1
Therefore, pH = -log[H3O+]
= -log(1)
=0
Therefore a 1M solution of HCl has pH 0.
Example 3
What is the pH of a 0.01M solution of HCl?
HCl (aq)  H+ + Cl-
[H3O+] = 0.01 M
Therefore, pH = -log[H3O+]
= -log(0.01)
=2
Example 4
What is the pH of a 1M NaOH solution?
pOH = -log[OH-]
= -log(1)
=0
pH + pOH = 14
pH = 14 – pOH
pH = 14
Therefore a 1M solution of NaOH has pH 14. The pH
of a very basic solution.
Example 5
Determine the pH of a 0.01M NaOH solution.
pOH = -log[OH-]
= -log(0.01)
=2
pH + pOH = 14
pH = 14 – pOH
pH = 12
Therefore a 1M solution of NaOH has pH 12. The pH
of a basic solution.
Example 6
The pH reading of a solution is 10.33. What is its
hydrogen ion concentration?
10-pH = [H+]
10-10.33 = [H+]
4.7  10-11 mol/L = [H+]
Base ten logarithm represents
an exponent log10(100) =2
102
Example 7
Calculate the pH of a 0.00242 M H2SO4 solution .
H2SO4  2H+ + SO420.00242 M 0.00484M
pH = - log[H+]
= -log(0.00484)
= -(-2.315)
= 2.32
Homework
• Pg 239 #1, 2
• Pg 242 #5, 7, 9, 10
Acid-Base Indicators
• Can determine if the solution is acidic, basic or
natural using various indicators
• Litmus paper, bromothyomol blue,
phenolphthalein are some examples.
• Depending on the indicator they change
colour at varying pH levels.
• Need to use various indicators to solve the pH
level
Chart
• Back of book