# Preparing Solutions

Review of Formulas
kg = 103 g
mg = 10-3 g
μg = 10-6 g
eg. 1.6 kg = 1.6 x 103 g
6 mg = 6 x 10-3 g
0.6 mg = 6 x 10-4 g
Making
Molar
Solutions
From
Liquids
(More accurately, from
stock solutions)
What are molar solutions?
A molar solution is one that expresses
“concentration” in moles per volume
Molar solutions are prepared using:
• a balance to weigh solids (in grams)
• a pipette to measure small liquid volumes (μL/mL)
• a volumetric flask to measure large volumes (L)
and for mixing
Molar Volume is measured in mol/L,
∴ we can use the equation c = n/V
• mol/L can be abbreviated as M or [ ]
Calculations with molar solutions
Q: How many moles of NaCl are required to make
7.5 L of a 0.10 M solution?
M=n/L, n = 0.10 M x 7.5 L = 0.75 mol
# mol NaCl = 7.5 L x 0.10 mol NaCl = 0.75 mol
1L
But in the lab we weigh grams not moles, so …
Q: How many grams of NaCl are required to make
7.5 L of a 0.10 M solution?
# g NaCl =
7.5 L x 0.10 mol NaCl x 58.44 g NaCl =43.83 g
1L
1 mol NaCl
Practice Questions
1. How many grams of nitric acid are present in 1.0 L of a
63 g
1.0 M HNO3 solution?
2. Calculate the number of grams needed to produce 1.00
L of these solutions: a) 1.00 M KNO3
101 g
g M KClO3
b) 1.85 M H2SO4 181
c) 0.67
82 g
3. Calculate the # of grams needed to produce each:
a) 0.20 L of 1.5 M KCl b) 0.160 L of 0.300 M HCl
a) 22 g b) 1.75 g
c) 0.20 L of 0.09 mol/L AgNO3
d) 250 mL of 3.1 mol/L BaCl2
c) 3 g d) 0.16 kg
4. Give the molarity of a solution containing 10 g of each
solute in 2.5 L of solution: a)H2SO4 b)Ca(OH)2
5. Describe how 100 mL of a 0.10 mol/L a) 0.041 mol/L
b) 0.054 mol/L
C, V, n, m  if these variables are in the
the following steps
e.g. Given c and V. Find m.
Find # of moles (c=n/v)
Convert to grams (n= m/M)
Preparing Standard Solutions
 This is a volumetric flask
It is used for preparing solutions
(but not for storing them)
A standard solution is one with an accurate,
known concentration. This is also known as a
stock solution.
• These are used as reactant solutions
• They usually have a higher concentration
than is needed for creating solutions and
therefore must be diluted
Preparing Standard Solutions
After diluting a solution, the concentration of the
solution changes. To calculate the new molar
concentration, we use the equation:
c1V1 = c2V2
C = concentration; V = volume
1 = initial (concentrated); 2 = final (diluted)
Don’t forget the equation for molar
concentration! (c = n/V)
Identify each volume to two decimal places
(values tell you how much you have expelled)
4.48 - 4.50
4.86 - 4.87
5.00
The Dilution Formula
E.g. if we have 1 L of 3 M HCl, what is M if we
dilute acid to 6 L?
M1 = 3 mol/L, V1 = 1 L, V2 = 6 L
M1V1 = M2V2, M1V1/V2 = M2
M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M
V1 = 1 L
M1 = 3 M
M1V1 = 3 mol
V2 = 6 L
M2 = 0.5 M
M2V2 = 3 mol
Examples
Q – What volume of 0.5 M HCl can be prepared from 1 L
of 12 M HCl?
M1 = 12 mol/L, V1 = 1 L, M2 = 0.5 L
M1V1 = M2V2, M1V1/M2 = V2
V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L
Q – 1 L of a 3 M HCl solution is added to 0.5 L of a 2 M
HCl solution. What is the final concentration of HCl?
(hint: first calculate total number of moles and total
number of L)
# mol = (3 mol/L)(1 L) + (2 mol/L)(0.5 L)
= 3 mol + 1 mol = 4 mol
#L
= 1 L + 0.5 L = 1.5 L
# mol/L = 4 mol / 1.5 L = 2.67 mol/L
Dilution problems
1. How many mL of a 14 M stock solution must
be used to make 250 mL of a 1.75 M solution?
2. You have 200 mL of 6.0 M HF. What
concentration results if this is diluted to a
total volume of 1 L?
3. 100 mL of 6.0 M CuSO4 must be diluted to
what final volume so that the resulting
solution is 1.5 M?
1. M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL
V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M)
V1 = 0.03125 L = 31.25 mL
2. M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 L
M2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L)
M2 = 1.2 M
3. M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ?
V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M)
V2 = 0.4 L or 400 mL
Dilution problems (cont’d)
4. What concentration results from mixing 400
mL of 2.0 M HCl with 600 mL of 3.0 M HCl?
5. What is the concentration of NaCl when 3 L
of 0.5 M NaCl are mixed with 2 L of 0.2 M
NaCl?
6. What is the concentration of NaCl when 3 L of
0.5 M NaCl are mixed with 2 L of water?
Mixing two solutions together, need to find
new concentration
1. Calculate the amount of moles in the final
solution (n=m/M) n1+n2=n(final)
2. Calculate the final volume after mixing the
two solutions together
3. From #1 & 2  calculate final
concentration
4. # mol = (2.0 mol/L)(0.4 L) + (3.0 mol/L)(0.6 L)
= 0.8 mol + 1.8 mol = 2.6 mol
# L = 0.4 L + 0.6 L
# mol/L = 2.6 mol / 1 L = 2.6 mol/L
5. # mol = (0.5 mol/L)(3 L) + (0.2 mol/L)(2 L)
= 1.5 mol + 0.4 mol = 1.9 mol
# mol/L = 1.9 mol / 5 L = 0.38 mol/L
6. # mol = (0.5 mol/L)(3 L) + (0 mol/L)(2 L)
= 1.5 mol + 0 mol = 1.5 mol
# mol/L = 1.5 mol / 5 L = 0.3 mol/L
Or, using M1V1 = M2V2,
M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L
Dilution problems (cont’d)
7. Water is added to 4 L of 6 M antifreeze
until it is 1.5 M. What is the total volume
of the new solution?
8. There are 3 L of 0.2 M HF. 1.7 L of this is
poured out, what is the concentration of
the remaining HF?
7. M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ?
V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M)
V2 = 16 L
8. The concentration remains 0.2 M, both
volume and moles are removed when the
solution is poured out. Remember M is
mol/L. Just like the density of a copper
penny does not change if it is cut in half, the
concentration of a solution does not change
if it is cut in half.
Homework
• Read Table 8.7 and Table 8.8 thoroughly to
prepare for Friday’s lab
– MANDATORY!!
Arab people

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