chm116a lecture 13-student slides-1
TRANSCRIPT
CHM 116 Spring 2012
Lecture 13
Section 18.9, 18.4
Next Lecture
Read Sec. 19.1, 19.2
In Silberberg
Lab: LeChatelier’s Principle (Ch 5)
Perform test procedures on 3 systems:
Fe3+(aq) + SCN-(aq) ⇌ FeSCN2+(aq)
CH3CO2H(aq) + H2O(l) ⇌ CH3CO2-(aq) + H3O+(aq)
HBrm + H2O ⇌ Brm- + H3O+
determine the direction of the equilibrium shiftidentify substances that increased and decreased
F
B
F F
H
N
H H
+
F
B
F F
H
N
H H
acid base adduct
A Lewis acid is an electron-pair acceptor.
A Lewis base is an electron-pair donor.
M2+H2O(l)
M(H2O)42+(aq)
adduct
Molecules as Lewis Acids
3 electron pairs, 1
available orbital
1 unshared electron
pair
Protons act as Lewis acids in that they accept an electron pair in all reactions: B: + H + B H +
New covalent
bond
Reactions of Acids and BasesChemical Reactivity
Lewis Bases(Nucleophile: electron-rich
species that attacks an electron deficient center)
1. Lone pair of electrons
e.g. H3N:, H2O:
2. Anion
e.g. OH-, Cl-
3. electrons
e.g. H2C=CH2
Lewis Acids(Electrophile: electron-deficient species attracted to a source of
electron density)
1. Partial (+)
2. Carbocations
3. H+
4. Empty “p” orbital
e.g. H3B, H3Al
. .
Acid/Base Equilibrium Calculations
Calculations are the same as before:
Find K Find missing [concentration] Find x and [concentration]eq
1. Assume x is small (weak acids!)2. Assume reaction goes to completion
then calculate x and [ ]eq
3. Use some method of reaching an exact solution (e.g., quadratic equation)
We will see some new assumptions, however.
A study of phenylacetic acid (C6H5CH2COOH, simplified here as HPAc) shows that the pH of 0.12 M HPAc is 2.62. What is the Ka of phenylacetic acid?
Find the Ka from the pH and [HA]i
HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)
pH = 2.62 [H3O+] = 10-pH = 10-2.62 = 2.4 x 10-3 M
[PAc] = [H3O+] = 2.4 x 10-3 M
Ka = 4.8 x 10-5
Concentration(M) HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)
Initial 0.12 - ~1x10-7 0
Change --x +x [H3O+] +x
What simplifying assumptions have we made?
Equilibrium -0.12 - x x[H3O+] +(<1x10-7)
1. We can neglect H3O+ from water
Find the Ka from the pH and [HA]i
Concentration(M) HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)
So Ka =(2.4x10-3) (2.4x10-3)
0.12= 4.8 x 10-5
Checking for % error. = 4x10-3 %
x 100[HPAc]dissn;
2.4x10-3M0.12M
[H3O+]from water; 1x10-7M
2.4x10-3Mx 100
= 2.0 %
Initial 0.12 - ~1x10-7 0
Change --x +x [H3O+] +x
Equilibrium -0.12 - x x[H3O+] +(<1x10-7)
Find the Ka from the pH and [HA]i
What simplifying assumptions have we made?
Determining Concentrations from Ka and initial [HA]
Propanoic acid (CH3CH2COOH, which we simplify as HPr) is an organic acid. What is the equilibrium pH of 0.10 M HPr? (Ka = 1.3 x 10-5)?
HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq) Ka = 1.3 x 10-5
initial concentration
Doccam
Equilibria Involving A Weak Base
What is the pH of 0.010 M NH3?
NH3 + H2O NH4+ + OH-
Kb = 1.8 x 10-5
Step 1. Define equilibrium concs. in ICE table
[NH3] [NH4+] + [OH-]
initial
change
equil.
You have 0.010 M NH3. Calculate the pH.
NH3 + H2O NH4+ + OH- Kb = 1.8 x 10-5
Step 2. Solve the equilibrium expression.
Assume x is small, so
Equilibria Involving A Weak Base
You have 0.010 M NH3. Calculate the pH.
NH3 + H2O NH4+ + OH- Kb = 1.8 x 10-5
Step 3. Calculate pH
Equilibria Involving A Weak Base
Stronger acid weaker conjugate base
Stronger base weaker conjugate acid
HF + H2O F- + H3O+
F- + H2O HF + OH-
Add equations
2 H2O H3O+ + OH-
Ka x Kb = [H3O+][OH-] = Kw
Remember: When we add equations, we can multiply K’s!
For a conjugate acid/base pair:
Consider HF acid:
And its conjugate base:
Determining the pH of a Solution of a Conjugate Base of a Weak Acid
What is the pH of 0.25M sodium acetate (CH3COONa, or NaAc for this problem).
Ka of acetic acid (HAc) is 1.8x10-5.
Plan: 1. Sodium salts are soluble in water so [Ac-] = 0.25M.
2. Write the association equation for acetic acid; use the Ka to find the Kb.
Doccam
Calculate the pH of a 0.10 M solution of Na2CO3.
Ka = 4.7 x 10-11 for dissociation of HCO3-
CO32- + H2O HCO3
- + OH-
base acid acid base
Determining the pH of a Solution of a Conjugate Base of a Weak Acid