Download What is the pH of a 0.100 M

Review pH
pH is a measure of the concentration of H+ in the
solution. (Chapter 15.4)
pH = -log[H+]
What is the pH of a 0.100 M solution of HNO3?
Chemistry 103 Spring 2011
Equilibrium of weak acids and weak bases
(Chapter 15.5 and 15.7)
What is the pH of 0.100 M CH3COOH?
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO-(aq)
I:
C:
Eq:
Kc =
Kacid = [H3O+][A-] / [HA] = Ka
acid ionization constant: Ka, an equilibrium
constant for a weak acid.
Ka =
x=
pH = -log[H3O+] =
2
Chemistry 103 Spring 2011
% ionization =
amount ionized x 100%
amount initial (total)
% ion. for 0.100 M CH3COOH = [H3O+]equilibrium
[CH3COOH]init
= 1.34 x 10-3 M / 0.100 M x 100% = 1.34%
(% ionization is also called % dissociation)
Practice: What is the pH and % ionization of
0.200 M hydrocyanic acid (HCN)?
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Chemistry 103 Spring 2011
CH3COOH vs. HA
-COOH = functional group = carboxylic acid
(p. 544, Section 15.2)
Lewis structure
HA emphasizes the acidic proton, but can also
be confusing.
CH3COOH(aq) + H2O(l)
H3O+(aq) + CH3COO-(aq)
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
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Chemistry 103 Spring 2011
Equilibrium is similar for weak bases …
What is [OH-] of 0.150 M CH3NH2?
I:
CH3NH2 + H2O(l)
0.150 M
CH3NH3+ + OH-
C:
Eq:
Kc = [CH3NH3+][OH-] / [CH3NH2] = Kbase = Kb
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Chemistry 103 Spring 2011
-NH2 and variations = functional group = amine
(p. 545, Section 15.2)
Lewis structures of -NH2 and variations
To emphasize the base reaction:
CH3NH3+ + OH-
CH3NH2 + H2O(l)
B + H2O(l)
BH+ + OH-
base ionization constant: Kb, an equilibrium
constant for a weak base.
Kb = [BH+][OH-] / [B]
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Chemistry 103 Spring 2011
More pH (Chapter 15.3 and 15.4)
Why is the pH of water at 25 °C equal to 7?
Because water itself ionizes in small amounts.
H2O(l) + H2O(l)
H3O+(aq) + OH-(aq)
K = [H3O+][OH-] = Kwater = Kw
H2O(l) + H2O(l)
H3O+(aq) + OH-(aq)
I:
C:
Eq:
At 25 °C, [H3O+] is measured to be 1.0 x 10-7 M.
Kw =
7
Chemistry 103 Spring 2011
Kw = [H3O+][OH-] = 1.0 x 10-14
Le Chatelier:
What happens if more H+ is added to water?
What happens if more OH- is added to water?
The equilibrium of water, H3O+ and OH- means
if we know either the concentration of H3O+ or
the concentration of OH- in an aqueous solution,
then we know the concentration of the other.
Example: What is the pH of 0.150 M CH3NH2?
8
Chemistry 103 Spring 2011
An alternative method …
Kw = [H3O+][OH-] = 1.0 x 10-14
-logKw = -log[H3O+] + -log[OH-] = -log(1.0 x 10-14)
pKw = pH + pOH = 14
Example: What is the pH of 0.150 M CH3NH2?
9
Chemistry 103 Spring 2011
Practice: Calculate the pH of 0.010 M NaOCl.
10
Chemistry 103 Spring 2011
Review of some unit conversions
Example: 0.63% by weight SnF2
1. weight % = mass fraction x 100%
0.63% = mass fraction x 100%
mass fraction = 0.63% / 100% = 0.0063
(grams of SnF2 / grams of solution = 0.0063)
2. molarity = moles solute / L of solution
• Assume water is the solvent
• Assume 100 grams of solution to simplify math
• Assume the solution is dilute enough that we
can treat the 100 grams of solution as 100 grams
of water => 100 grams = 100 mL = 0.100 L.
mass fraction = 0.0063 grams SnF2
1 gram of solution
mass fraction = 0.63 grams of SnF2
100 grams of solution
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Chemistry 103 Spring 2011
molarity = 0.63 g SnF2 x (1 mole / 156.707 g)
0.100 L of solution
molarity = 0.040 M
3. 1 ppm = 1 g solute / 106 g solution
ppm is similar to mass fraction
=> ratio of grams solute / grams solution
Conversion factor:
1 ppm x 106 g solution = 1
1 g solute
Convert mass fraction to ppm:
0.0063 g solute x 1 ppm x 106 g solution =
1 g solution
1 g solute
ppm = 0.0063 x 106 ppm
ppm = 6.3 x 103 ppm
Note: same as
ppm = mass fraction x 106 ppm
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