Transcript الشريحة 1

Density Calculations
In this section, you will learn how to find the
molarity of solution from two pieces of
information (density and percentage).
Usually the calculation is simple and can
be done using several procedures. Look at
the examples below:
1
Example
What volume of concentrated HCl (FW =
36.5g/mol, 32%, density = 1.1g/mL) are required
to prepare 500 mL of 2.0 M solution.
Solution
Always start with the density and find how many
grams of solute in each mL of solution.
2
Remember that only a percentage of the solution
is solute .
g HCl/ml = 1.1 x 0.32 g HCl / mL
The problem is now simple as it requires
conversion of grams HCl to mmol since the
molarity is mmol per mL
mmol HCl = 1.1x0.32 x103 mg HCl/(36.5 mg/mmol)
= 9.64 mmol
M = 9.64 mmol/mL = 9.64 M
3
Now, we can calculate the volume required
from the relation
MiVi (before dilution) = MfVf (after dilution)
9.64 x VmL= 2.0 x 500mL
VmL = 10.4 mL
This means that 10.4 mL of the concentrated HCl
should be added to distilled water and the
volume should then be adjusted to 500 mL
4
Example
How many mL of concentrated H2SO4 (FW = 98.1
g/mol, 94%, d = 1.831 g/mL) are required to
prepare 1 L of 0.100 M solution?
g H2SO4 / mL = 1.831 x 0.94 g /mL
Now find the mmol acid present
mmol H2SO4 = (1.831 x 0.94 x 103 mg) / (98.1
mg/mmol)
M = mmol/mL = [(1.831 x 0.94 x 103 mg) / (98.1
mg/mmol)]
/
mL
=
17.5
M
5
To find the volume required to prepare the solution
MiVi (before dilution) = MfVf (after dilution)
17.5 x VmL = 0.100 x 1000 mL
VmL = 5.71 mL which should be added to distilled
water and then adjusted to 1 L.
6
An Easy Short-Cut
M=
Density * percentage * 103
Formula Weight
The percentage is a fraction: (i.e. a 35% is
written as 0.35)
7
Analytical Versus Equilibrium
Concentration
When we prepare a solution by weighing a
specific amount of solute and dissolve it in
a specific volume of solution, we get a
solution with specific concentration. This
concentration is referred to as analytical
concentration. However, the
concentration in solution may be
different from the analytical
concentration, especially when partially
dissociating substances are used.
8
An example would be clear if we consider
preparing 0.1 M acetic acid (weak acid) by
dissolving 0.1 mol of the acid in 1 L solution.
Now, we have an analytical concentration of
acetic acid (HOAc) equals 0.1 M. But what is the
actual equilibrium concentration of HOAc?
We have
HOAc = H+ + OAcThe analytical concentration ( CHOAc ) = 0.1 M
CHOAc = [HOAc]undissociated + [OAc-]
The equilibrium concentration = [HOAc]undissociated.
9
For good electrolytes which are 100% dissociated
in water the analytical and equilibrium
concentrations can be calculated for the ions,
rather than the whole species.
For example, a 1.0 M CaCl2 in water results in 0
M CaCl2, 1.0 M Ca2+, and 2.0 M Cl- since all
calcium chloride dissociates in solution.
For species x we express the analytical concentration
as Cx and the equilibrium concentration as [x]. All
concentrations we use in calculations using
equilibrium constants are equilibrium
concentrations.
10
Dilution Problems
In many cases, a dilution step or steps are
involved in analytical procedures. One should
always remember that in any dilution the number
of mmoles of the initial (concentrated) solution is
equal to the number of mmoles of the diluted
solution. This means
MiVi (concentrated) = MfVf (dilute)
11
Example
Prepare 200 mL of 0.12 M KNO3 solution from 0.48 M
solution.
MiVi (concentrated) = MfVf (dilute)
0.48 x VmL = 0.12 x 200
VmL = (0.12 x 200)/0.48 = 50 mL
Therefore, 50 mL of 0.48 M KNO3 should be diluted to
200 mL to obtain 0.12 M solution
12
Example
A 5.0 g Mn sample was dissolved in 100 mL water.
If the percentage of Mn (At wt = 55 g/mol) in the
sample is about 5%. What volume is needed to
prepare 100 mL of approximately 3.0x10-3 M
solution.
Solution
First we find approximate mol Mn in the sample =
5.0 x (5/100) g Mn/(55 g/mol) = 4.5x10-3 mol
13
Molarity of Mn solution = (4.5x10-3 x 103 mmol)/100 mL =
4.5x10-2 M
The problem can now be solved easily using the dilution
relation
MiVi (concentrated) = MfVf (dilute)
4.5x10-2 x VmL = 3.0x10-3 x 100
VmL = (3.0x10-3 x 100)/4.5x10-2 = 6.7 mL
Therefore, about 6.7 mL of the Mn sample should be
diluted to obtain an approximate concentration of 3.0x103 M solution.
14
Example
What volume of 0.4 M Ba(OH)2 should be added to 50
mL of 0.30 M NaOH in order to obtain a solution that
is 0.5 M in OH-.
Solution
We have to be able to see that the mmol OH- coming
from Ba(OH)2 and NaOH will equal the number of
mmol of OH- in the final solution, which is
mmol OH- from Ba(OH)2 + mmol OH- from NaOH =
mmol OH- in final solution
15
The mmol OH- from Ba(OH)2 is molarity
of OH- times volume and so are other
terms. Molarity of OH- from Ba(OH)2 is
0.8 M (twice the concentration of
Ba(OH)2, and its volume is x mL. Now
performing the substitution we get
0.8 * x + 0.30 * 50 = 0.5 * (x + 50)
x = 33 mL
16
For Solid Solutes
% (w/w) = [weight solute (g)/weight sample (g)] x 100
ppt (w/w) = [weight solute (g)/weight sample (g)] x 1000
ppm (w/w) = [weight solute (g)/weight sample (g)] x 106
ppb (w/w) = [weight solute (g)/weight sample (g)] x 109
A ppm can be represented by several terms like the
one above, (mg solute/kg sample), ( g solute/106g
sample), etc..
17
If the solute is dissolved in solution we have
% (w/v) = [weight solute (g)/volume sample (mL)] x 100
ppt (w/v) = [weight solute (g)/volume sample (mL)] x
1000
ppm (w/v) = [weight solute (g)/volume sample (mL)] x
106
ppb (w/v) = [weight solute (g)/volume sample (mL)] x
109
Also a ppm can be expressed as above or as (g
solute/106 mL solution), (mg solute/L solution), or
(mg/mL), etc..
18
For Liquid Solutes
% (v/v) = [volume solute (mL)/volume sample (mL)] x
100
ppt (v/v) = [volume solute (mL)/volume sample (mL)] x
1000
ppm (v/v) = [volume solute (mL)/volume sample (mL)] x
106
ppb (v/v) = [volume solute (mL)/volume sample (mL)] x
109
A ppm can be expressed as above or as (mL/L), (mL/103
L), etc..
19
Example
A 2.6 g sample was analyzed and found to contain
3.6 mg zinc. Find the concentration of zinc in
ppm and ppb.
A ppm is microgram solute per gram sample,
therefore
Ppm Zn = 3.6 mg Zn/2.6 g sample = 1.4 ppm
A ppb is nanogram solute/gram sample, therefore
Ppb Zn = 3.6 x103 ng Zn/2.6 g sample = 1400 ppb
20