Download Chem. 31 * 9/15 Lecture

Chem. 1B – 9/22 Lecture
Announcements I
• Next Stuff that is due
– Mastering (15b – on Saturday, 16a - buffers on
Tues.)
– Quiz 4 (Experiment 2 + Polyprotic Acids + Buffers)
– Exam 1 (Thursday)
– Experiment 7 due (Wed./Thurs.)
• Exam #1
– Next week on Thursday (9/29)
– Same format as last year
– Covers material covered through buffers (covering
today)
Announcements II
• Exam 1 – cont.
– Possible Help Session Tues., 4:00 to 5:00 time frame
in Sequoia 452
– PAL lead session Monday 4-9; Sequoia 443
– Everyone here but Lab Sect. 7 (in Sequoia 426)
– Need to bring Scantron Form SC982-E
• Today’s Lecture
– Buffers (Chapter 16)
– Exam 1 Review (part I – Chapter 14; rest on Tues.)
– Titrations (Chapter 16)
Chem 1B – Aqueous Chemistry
Buffers (Chapter 16)
•
•
•
•
We have discussed some mixtures briefly (e.g.
strong acid + weak acid)
One particular type of mixture: weak acid +
conjugate base (or weak base + conjugate
acid) makes a solution called a buffer
Buffers are desirable because they keep the
pH nearly constant even if an acid or base is
added
Buffers are very important in biology because
many enzymes (protein catalysts) will only
work over a narrow pH range
Chem 1B – Aqueous Chemistry
Buffers (Chapter 16)
•
Example: Determine pH of a mix of 0.010 M HC2H3O2
and 0.025 M Na+C2H3O2- solution (Ka of HC2H3O2 = 1.8
x 10-5)
Chem 1B – Aqueous Chemistry
Buffers (Chapter 16)
• Buffer Solutions:
– Question: Was the ICE Problem set up
needed?
– Answer: No. The assumption of x << [HA],
[A-] is valid for all “traditional” buffers
– Traditional Buffer
• Weak acid (3 < pKa < 11)
• Ratio of weak acid to conjugate base in range 0.1
to 10
• mM+ concentration range
Chem 1B – Aqueous Chemistry
Buffers (Chapter 16)
• Buffer Solutions:
– Since ICE not needed, can just use Ka
equation
– Ka = [H+][A-]/[HA] = [H+][A-]o/[HA]o
(always valid)
(valid for traditional buffer)
– But log version more common
– pH = pKa + log([A-]/[HA])
– Also known as Henderson-Hasselbalch
Equation
Chem 1B – Aqueous Chemistry
Buffers (Chapter 16)
• Addition of small amounts of acid to a
buffer:
– Example: let’s say we have a buffer made to
be 0.050 M NH3 + 0.100 M NH4Cl in 1.00 L
– Calculate the pH
– Now lets add 0.005 moles of HCl. What is the
new pH?
Chem 1B – Aqueous Chemistry
Buffers (Chapter 16)
• Change of pH and Buffer Capacity
– A buffer is designed to minimize the change in pH
from addition of base or acid
– What is the most effective pH?
• (show next slide)
– Can we add too much acid or base to a buffer?
– Does the absolute concentration of acid/base affect
pH? Why is a higher concentration better?
– Buffer Capacity is the ability of the buffer to absorb
acid or base without the pH changing significantly
Chem 1B – Aqueous Chemistry
Buffers (Chapter 16)
• Most Effective pH Range
– NH3 + NH4Cl in 1.00 L Example [NH3] +
[NH4+] = 0.150 M
– Addition of 0.005 moles HCl
[NH3] (M)
[NH4+] (M)
pH
DpH
0.005
0.145
7.78
-2.75
0.030
0.120
8.64
-0.09
0.050
0.100
8.94
-0.07
0.075
0.075
9.24
-0.06
0.100
0.050
9.55
-0.07
0.120
0.030
9.84
-0.08
0.145
0.005
10.71
-0.32
Best region
(equal
moles of
WA and CB)
Chem 1B – Aqueous Chemistry
Buffers (Chapter 16)
• A Second Way to Make Buffers (not on
Exam 1)
– To make a traditional buffer, we need both an acid
and its conjugate base, but this also can be “made”
through other combinations such as:
– A weak acid and a strong base or a weak base and a
strong acid
– This is not covered in the text for buffers, but is
needed for titrations (section 16.4)
– Example: What is the pH when we add 10.0 mL of
0.50 M KOH with 34.1 mL of 0.19 M HC2H3O2?
Chem 1B – Aqueous Chemistry
Buffers (Chapter 16)
• One Final Question:
– How many mL of 1.00 M HCl must be added
to 200.0 mL of 0.065 M NH3 to make a pH
10.00 buffer?
Exam 1 Review
• General Advice
– Multiple choice questions:
• Read full question and all answers before selecting (some
answers may “sound right” until you see the actual right
answer)
• If it is taking time to calculate or if you are uncertain how to
proceed, skip that problem and get back to later (circle
question so you know you still need to do it)
– General Question (12 pts):
• Will be multiple parts and involve calculations (e.g. like
determination of equilibrium concentrations on quiz)
• Need to show work (don’t just punch numbers into your
calculator and write answer)
• I will give partial credit for parts done correctly
Exam 1 Review
• Topics – Chapter 14
– Understand how equilibrium relates to kinetics
– Understand how to get equilibrium equations from
equilibrium reactions (chemical equations)
– Understand what the K value tells you about
equilibrium conditions
– Know the difference and be able to convert between
KP and KC (for given reaction with R and T given)
– Know how equilibrium reaction “manipulation” affects
K values (e.g. reversing reaction leads to Knew =
1/Kold)
Exam 1 Review
• Topics – Chapter 14 – cont.
– Be able to determine K from all equilibrium
concentrations or from initial and equilibrium
conditions
– Be able to determine equilibrium concentrations from
“at equilibrium” conditions and K values or from initial
conditions and K values
– Be able to determine whether reaction will proceed to
products or reactants from K and initial
concentrations
Exam 1 Review
• Topics – Chapter 14 – cont.
– Know how to apply Le Châtelier’s principle to systems
initially at equilibrium with the following changes:
• reactant/product addition/subtration
• increase/decrease in volume (including dilutions)
• changes in T (with DH known)
Chem 1B – Aqueous Chemistry
Titrations (Chapter 16)
• Review from 4.8
– Titrations are a way to accurately tell when
we have reached a stoichiometric point in a
reaction
– Generic reaction: aA + bB → products
– The equivalence point is defined where
(moles A)/(moles B) = a/b
– mL of titrant (either A or B) are carefully
measured (usually to determine the
concentration of B or A)
Chem 1B – Aqueous Chemistry
Titrations (Chapter 16)
• Review from 4.8
– Example question: An unknown H2SO4
solution is pipeted (25.00 mL) into a flask. It
is titrated with KOH until reaching an endpoint
(where the equivalence point is observed). It
requires 39.1 mL of 0.150 M KOH. What is
the concentration of H2SO4 in the unknown
solution?
Chem 1B – Aqueous Chemistry
Titrations (Chapter 16)
• Titrations – pH behavior
– Review material covered how we can use
titrations, but not the exact behavior during
the titration
– In Chapter 16, we cover, through calculations,
the pH of the full titration curve
– The most valuable titrations are accurate
(observed end point gives the equivalent
point) and precise (reproduceable)
– This occurs when change in pH in a titration is
rapid
Chem 1B – Aqueous Chemistry
Titrations (Chapter 16)
• Strong Acid – Strong
Base Titration
– How does pH change as
NaOH is added?
– 3 regions to titrations
(different calculations in
each region):
• before equivalence point
• at equivalence point
• after equivalence point
– Show pH at 5 mL, 12.5 mL,
and 15 mL
0.100 M NaOH
0.050 M HCl, 25 mL
Chem 1B – Aqueous Chemistry
Titrations (Chapter 16)
Titration Plot
0.100 M NaOH
Titration Plot
14.00
12.00
pH
10.00
8.00
0.050 M HCl, 25 mL
6.00
4.00
2.00
0.00
0
5
10
15
V(HCl)
20
25