Transcript Chapter 1

Homework


Chapter 0 - 0, 1, 2, 4
Chapter 1 – 15, 16, 19, 20, 29, 31, 34
Question:
What is the molarity of a 10% (w/v) solution
of glucose?
Parts per million (PPM)
PPM

Parts per million is a convenient way to express
dilute concentrations. Historically, 1 mg per liter or
per 1000 ml is referred to as 1 ppm. However, this
is not really the case, as parts per million should
be expressed as:
c ppm
mg solute
=
1,000,000m
g solution
Show that the above equation is equivalent to mg per liter.
PPM
c ppm
mg solute
=
1,000,000m
g solution
For dilute solutions, the density of the solution will be the same as water.
Density of solution = Density of water=
1.0 g/ml
c ppm
mg solute
1000mgsolution 1g solution
=


1,000,000m
g solution
1ml solution
1g solution
c ppm
mg solute
=
1000 ml solution
Question
Converting PPM to Molarity
The town of Canton prohibits the dumping of copper solutions
that have concentrations greater than 0.3969 ppm. When
cleaning the quant lab, Dr. Skeels found a bottle labeled
“copper standard - 7 mM”, is it permissible to dump this
solution down the drain?
Volunteers??
Preparation of Stock Solutions


Solids
Liquids
Solution preparation cont’d
Describe the Preparation of a 500.0 mL of a solution that
contains 8.00 mM Cu2+ using CuSO4.5H2O (MW 149.69).
8.00mmolCu 2
1mol
8.00103 m ol Cu 2


1L
1000 mmol
1L
8.00103 m ol Cu 2 1mol CuSO 4  5 H 2O 8.00103 m olCuSO 4  5H 2O


2
1L
1L
1mol Cu
8.00103 m ol CuSO 4  5H 2O
 0.5000 L  4.00103 molCuSO4  5H 2O
1L
Solution preparation cont’d
Describe the Preparation of a 500.0 mL of a solution that
contains 8.00 mM Cu2+ using CuSO4.5H2O (MW 149.69).
4.00103 molCuSO4  5H 2O 
249.69g CuSO 4  5H 2O
1 molCuSO 4  5H 2O
 0.999g CuSO4  5H 2O
Thus …
Add ______g CuSO4.5H2O
Into a volumetric flask

Add about _____ ml of water
Swirl to dissolve
And fill to the _____ ml mark
Question

Using the 8 mM Cu2+ solution, prepare
20 mL of a 0.25 mM Cu2+ solution.
Dilutions

To make dilutions of a solution, the
following equation should be employed:
M i  Vi = M f  Vf
Question
Using the 8 mM Cu2+ solution, prepare 20 mL of a
0.25 mM solution.
M i  Vi = M f  Vf
8mM  Vi = 0.25mM 20ml
0.25mM  20ml
= 0.625mL
Vi =
8mM
From a liquid – consider
concentrated HCl
A more difficult example

Prepare a 500.0 mL of 1 M HCl.
MW
Wt %
Density
Try it out …
Consider it in two steps:
(1) Determine concentration of Stock
(2) Make dilution
(1) Concentration of Stock

Must find grams of HCl per liter of solution
dHCl=1.19 g/ml
%HCl (w/w)=37%
MW=36.46 g/mol
Mass
HCl per
Liter
Molarity
1.19g solution 1000m l
37g HCl



m L solution
1L
100g Solution
4.40102 g HCl  1m ol HCl

36.46g HCl
L
 12 .1M
4.40102 g HCl

L
Dilution

Determined concentration of stock is
______ M HCl. We want a 500.0 mL
solution that is 1M.
Mi  Vi = Mf  Vf
12.1M  Vi = 1M 500.0mL
1M  500 .0mL
Vi =
= 41.322mL
12.1M
NOTE
Care must be exercised
when handling strong acids!!
(Always, Always add acid to water)

Add about 300 ml of water first
Then add acid
Dilute to mark
Homework
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
Chapter 0 - 0, 1, 2, 4
Chapter 1 – 15, 16, 19, 20, 29, 31, 34
Chapter 3
Experimental Error
And propagation of
uncertainty
Suppose
You determine the density of some mineral by
measuring its mass

4.635 + 0.002 g
And then measured its volume

1.13 + 0.05 ml
g
m ass( g )
 4.1018
d
ml
volum e(m l)
What is its uncertainty?
Significant Figures (cont’d)

The last measured digit always has
some uncertainty.
3-1 Significant Figures

What is meant by significant figures?
Significant figures: minimum number of digits
required to express a value in scientific
notation without loss of accuracy.
Examples
How many sig. figs in:

a.
b.
c.
d.
e.
3.0130 meters
6.8 days
0.00104 pounds
350 miles
9 students
“Rules”
All non-zero digits are significant
Zeros:
1.
2.
a.
b.
c.
3.
Leading Zeros are not significant
Captive Zeros are significant
Trailing Zeros are significant
Exact numbers have no uncertainty
(e.g. counting numbers)
Reading a “scale”
What is the “value”?
When reading the scale of any apparatus, try to estimate to
the nearest tenth of a division.
3-2
Significant Figures in Arithmetic

We often need to estimate the uncertainty of
a result that has been computed from two or
more experimental data, each of which has a
known sample uncertainty.
Significant figures can provide a marginally
good way to express uncertainty!
3-2
Significant Figures in Arithmetic

Summations:

When performing addition and subtraction report
the answer to the same number of decimal places
as the term with the fewest decimal places
+10.001
+ 5.32
+ 6.130
21.451
?
___ decimal places
Try this one
+
1.632 x 105
4.107 x 103
0.984 x 106
0.1632
x 106
0.004107 x 106
6
0.984
x
10
+
1.151307
1.151307 xx 10
1066
3-2
Significant Figures in Arithmetic

Multiplication/Division:

When performing multiplication or division report
the answer to the same number of sig figs as the
least precise term in the operation
16.315 x 0.031 = 0
? .505765
___ sig figs
___ sig figs
____ sig figs
3-2
Logarithms and Antilogarithms
From math class:
log(100) = 2
Or log(102) = 2
But what about significant figures?
3-2
Logarithms and Antilogarithms
Let’s consider the following:
An operation requires that you take the log of
0.0000339. What is the log of this number?
log (3.39 x 10-5) = -4.469800302
Between -5 and -4
____ sig figs
3-2
Logarithms and Antilogarithms
Try the following:
23442 x 104
Antilog 4.37 = 2.3442

___ sigs
“Rules”
Logarithms and antilogs
1. In a logarithm, keep as many digits to the
right of the decimal point as there are
sig figs in the original number.
2. In an anti-log, keep as many digits are
there are digits to the right of the
decimal point in the original number.

3-4. Types of error

Error – difference between your answer and the
‘true’ one. Generally, all errors are of one of three
types.

Systematic (aka determinate) – problem with the
method, all errors are of the same magnitude and
direction (affect accuracy) Can be detected and
eliminated or lessened


Random – (aka indeterminate) causes data to be
scattered more or less symmetrically around a mean
value. (affect precision) Estimated
Gross. – occur only occasionally, and are often large.
Treated statistically
Absolute and Relative Uncertainty

Absolute uncertainty expresses the margin of
uncertainty associated with a measurement.
Consider a calibrated buret which has an
uncertainty + 0.02 ml. Then, we say that the
absolute uncertainty is + 0.02 ml
Absolute and Relative Uncertainty

Relative uncertainty compares the size of the
absolute uncertainty with its associated
measurement.
Consider a calibrated buret which has an
uncertainty is + 0.02 ml. Find the relative
uncertainty is 12.35 + 0.02, we say that the
relative uncertainty is
0.02
absolute uncertainty

 0.002
RelativeUncertainty 
magnitudeof measurement 12.35
3-5. Estimating Random Error
(absolute uncertainty)

Consider the summation:
+ 0.50 (+ 0.02)
+4.10 (+ 0.03)
-1.97 (+ 0.05)
2.63 (+ ?)
s y  sa2  sb2  sc2  ...
Sy = + 0.06
3-5. Estimating Random Error

Consider the following operation:
4.10(0.02)  0.0050(0.0001)
 0.010406( ?)
1.97(0.04)
2 2
2
2 2
2
   sb0.0001
 sc   0.04 
0s.a02
    
        ... 
yy
4a.10
   b 0.0050
 c   1.97 
sy
 0.02897 s y  0.02897 0.010406
y
=
y
ssyy
Try this one
14.3(0.2)  11.6(0.2) 0.050(0.001)
820(10)  1030(5) 42.3(0.4)
3-5. Estimating Random Error

For exponents
For
ya
x
uncert ainty in a is Sa
 sa 
 x 
y
a
sy
3-5. Estimating Random Error

Logarithms
antilogs
For
For
y  log a
y  antilog a
uncertainty in a is Sa
uncertainty in a is Sa
 sa 
s y  0.434 
a
sy
y
 2.303sa
Question
y  log a
Calculate the absolute
uncertainty in a is Sa
standard deviation for a
 sa 
the pH of a solutions
s y  0.434 
whose hydronium ion
a
concentration is
 0.02 
-4
s y  0.434

2.00 (+ 0.02) x 10
 2.00 

pH = 3.6990 + ?
Question

y  antilog a
Calculate the absolute
value for the hydronium uncertainty in a is Sa
ion concentration for a
sy
 2.303sa
solution that has a pH of
y
7.02 (+ 0.02)
[H+] = 0.954992 (+ ?) x 10-7
s y  2.303sa  y
Suppose
You determine the density of some mineral by
measuring its mass

4.635 + 0.002 g
And then measured its volume

1.13 + 0.05 ml
g
m ass( g )
 4.1018
d
ml
volum e(m l)
What is its uncertainty?
The minute paper
Please answer each question in 1 or 2
sentences
1)
2)
What was the most useful or meaningful thing
you learned during this session?
What question(s) remain uppermost in your
mind as we end this session?