Transcript the value of active learning: understanding acidity, northeast 2014

The Value of Active Learning:
Understanding Acidity
Group 2
Sarah Malmquist
Joanna Kiryluk
Stony Brook University
Stony Brook University
Chandrika Narayan
Khaliliah Reddie
University of Massachusetts,
Lowell
University of Massachusetts,
Lowell
Valerie Purdie-Vaughns
Brent Stockwell
Columbia University
Columbia University
Learning goals
• Appreciate active learning vs traditional teaching
strategies
• Understand what pH is and be able to determine pH
of sample
Learning objectives
You should be able to…
Distinguish between active learning and traditional lectures
Define pH
Determine the pH of a solution using litmus paper
Calculate the pH of a solution given the proton concentration
Setting the scene
Problem to solve: readiness for
active learning
Who are you?
• Students in Bio 101 (or any
introductory science class)
• First day of class
What will you do?
• Participate in two activities
Part 1
Lecture: acidity
Acidity is a critical feature of
biological systems
• Acidity is the concentration of H+
– measured using pH
• pH is important living systems
– fish in water
– human blood
– tumor growth
– digestion
pH measures the concentration
of protons in a solution
The pH typically varies from 0 to 14
[H+] = concentration of H+
in Molarity (M = moles/Liter)
pH =
- log [H+]
How does the Log function work?
Log 10 = Log (101) = 1
Log 100 = Log (102) = 2
Log 1000 = Log (103) = 3
Log 0.1 = Log (10-1) = -1
The log (base 10) of a number is the exponent the number 10
needs to be raised to in order to generate that number
e.g. 0.001 M H+ = 10-3 M H+ = pH of 3
pH 3 = 10-3M
Acids have removable protons that dissolve in solution
• Lower pH = higher [H+] = acidic
• Lemon juice = pH 2
If [H+] = 10-7 M, pH = 7
= neutral water
H2O
H+ + OH-
[H+] in water = 10-7
Solutions that differ by 1 pH unit have a ten
fold difference in [H+]
Lemon Juice
pH = 2
0.01 M H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
Vinegar
pH = 3
0.001 M H+
H+
H+
H+
H+
H+
H+
H+
H+
H+
Example of a pH calculation
Add 1 mL 1M HCl (1 mol/liter) to 1 L water,
pH drops from 7 to 3
•
•
•
•
•
Why?
1 mL of 1M HCl has 10-3 moles of H+
10-7 mol (0.0000001 mol) in 1 L water
(negligible by comparison)
[H+] when added to 1 L = 10-3 moles/liter
log[10-3] = -3
pH = -log[H+] = 3
Summary
• pH is a numerical scale for determining
the acidity of a solution
• pH = - log[H+]
• It is important to know how to calculate
changes in the pH of a solution
Clicker Question
You have 5 mL of a solution of pH 4 and 5 mL of
another solution of pH 7.
Which of the following is the closest estimate for
the resulting pH if you mix the two solutions
together?
A. 3.5
B. 4.5
C. 5.5
D. 6.5
E. 7.5
Part 2
Active Learning: acidity
A hands-on pH experiment
GOAL: Understand how to measure and calculate pH
1. How to use a litmus paper to measure pH of a solution.
• Litmus paper: a strip of indicator
paper that is dipped in a sample
liquid.
• The paper turns a different color
depending upon the pH of the
solution.
• The strip is compared to a color
standard to determine pH
A hands-on pH experiment
2. Measurements
Use tube : “A” (5 mL of soda), “B” (5 mL of water), “C” (40 mL of water)
Measurement 1:
• Take a strip of litmus paper and dip the end into a tube A with soda.
• Measure the pH with litmus paper, record the pH.
• Convert your measured pH to the [H+] in the sample and record it
Measurement 2:
• Pour water from tube B to soda in tube A.
• Now you have diluted the original soda 2-fold.
• Use a strip of litmus to measure the pH of the diluted soda
• Record the pH, convert it to the [H+] in the sample, and record it
Measurement 3:
• Pour water from tube C to tube A to further dilute the soda.
• Now you have diluted the original soda 10-fold.
• With a new strip of litmus paper, measure the pH.
• Convert your measured pH to the [H+] in the sample and record it
A hands-on pH experiment
pH = –log [H+]
Soda
volume
(mL)
Water
volume
(mL)
Volume
Fold
(soda+wat dilution of
er)
soda
(mL)
1.
5
0
5
2.
5
5
10
3.
5
45
50
pH
[H+]
(M)
1
• Please label each pH strip after you use it
• What was the pH change after a two-fold dilution?
• After a 10-fold dilution?
Clicker Question
You have 5 mL of a solution of pH 4 and 5 mL of
another solution of pH 7.
Which of the following is the closest estimate for
the resulting pH if you mix the two solutions
together? (Talk with your whole table.)
A. 3.5
B. 4.5
C. 5.5
D. 6.5
E. 7.5
Clicker Question
You have 5 mL of a solution of pH 4 and 5 mL of
another solution of pH 7.
Which of the following is the closest estimate for
the resulting pH if you mix the two solutions
together?
A. 3.5
B. 4.5
C. 5.5
D. 6.5
E. 7.5
SOLUTION
• [H+] = 10-4 M = 0.00010 M
• -log[10-4] = 4 = pH
• The number of protons in water is 10-7 M
(0.0000001 M), which is negligible by
comparison
• Two-fold dilution of the pH 4 solution
• [H+] = 5 x 10-5 = 0.00005
• -log(5x10-5) = 4.3 = pH of mixture
Brainstorm
• How did the hands-on
experiment help prepare you for
the clicker question?
• Complements lecture
• Adds excitement
Decreased failure rate
Freeman et al., 2014, Proceedings in the National Academy of Sciences
“I need the best possible grade in this
course”
Active learning
in class
Higher grades
Dream come true!
Learning goals realized…
You should now have…
an appreciation for active learning compared to
traditional teaching
An understanding of what pH is and the ability to
determine the pH of a solution
Thank you.
Results of clicker question after lecture
Results of clicker question after
hands-on activity
Supplemental handout
Example of a pH calculation
If you add 1 mL of 1M HCl (1
mol/liter) to 1 liter of water, the pH of
the water drops from 7 to 3
Solution for people who “think in equations” (include
physicists, mathematicians, etc) and want to be able to
calculate pH in the most general case
V = volume, éë H + ùû = concentration
Vtotal ´ éë H + ùû = VHCl ´ éë H + ùû + Vwater ´ éë H + ùû
final
HCl
water
mol
+ù
-3
é
Vtotal ´ ë H û = 10 L ´1
+
final
L
Concentration:
10 -3 mol
mol
1L ´10
L
-7
10-7 is negligible, compared with 10 -3 mol
éë H + ùû » VHCl ´ éë H + ùû = 10 -3 ´1M
final
HCl
Vtotal
pH:
pH final
æ +
ö
V
= - log10 çéë H ùû ´ HCl ÷ = - log10 (10 -3 ) = 3
HCl
Vtotal ø
è
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