Solution

Chapter 15 Section 2
Concentrations of Solutions
Concentration of Solution
The ratio of solute to solvent
Units – molarity (M), molality (m), mole
fraction, % by mass, parts per million (ppm),
parts per billion (ppb)
% by Mass (m/m)
grams of solute in 100 g of solution
g of solute
% by mass 
100
g of solution
 Different from solubility. How?
 mass = density × volume
m/v = mass of solute in volume of solution
v/v = volume of solute in volume of
solution
Masses are additive
 mass of solute + mass of solvent = mass of
solution
Volumes are not additive, unless told to do
so
 volume of solute + volume of solvent ≠ volume
of solution
Example 1
12.5 g of NaCl is dissolved in 100. mL of
water. Find the % of NaCl in the solution.
Density of water = 1 g/mL
Example 2
How much water would you need to dissolve
35.6 g KNO3 to make a 25.5% m/m KNO3
solution?
Parts Per Million (ppm)
 Useful for very diluted solution
 ppm=mg of solute in 1000 000 mg of solution
= mg of solute in 1000 g of solution
= mg of solute in 1 kg of solution
mg of solute
g of solute
ppm 

1, 000, 000
kg of solution g of solution
*For a very dilute solution, g of solution = g of
solvent
% vs. ppm vs. ppb
g of solute
% (parts out of 100) 
100
g of solution
g of solute
ppm (parts out of million) 
1, 000, 000
g of solution
ppb (parts out of billion) 
g of solute
1,000,000,000
g of solution
For ppm and ppb, g of solution ≈ g of solvent
Example
A chemical analysis shows that there are 2.2
mg of lead in exactly 500 g of a water sample.
Convert this measurement to ppm.
Solution
Convert 500 g to kg:
1 kg
500 g 
 0.5 kg
1000 g
mg of solute
ppm 
kg of solution
2.2 mg
=
0.5 kg
= 4.4 ppm
Practice
1. Helium gas, 3.0x10-4 g, is dissolved in 200.0 g of
water. Express this concentration in ppm.
2. A sample of 300.0 g of drinking water is found to
contain 38 mg Pb. What is this concentration in
ppm?
3. A solution of lead sulfate contains 0.425 g of lead
sulfate in 100.0 g of water. What is this concentration
in ppm?
4. A 900.0 g sample of sea water is found to contain
6.7x10-3 g Zn. Express this concentration in ppm.
5. A 365.0 g sample of water contains 23 mg
Au. How much gold is present in the sample
in ppm?
6. A 650.0 g hard-water sample contains 101
mg Ca. What is this concentration in ppm?
7. An 870.0 g river water sample contains 2 mg
of cadmium. Express the concentration of
cadmium in ppm.
Answers Pg 461
1)
2)
3)
4)
5)
6)
7)
1.5 ppm
130 ppm
4250 ppm
7.4 ppm
63 ppm
155 ppm
2.3 ppm
Molarity (M)
mol Solute
molarity (M) 
L Solution
Example
What is the molarity of a potassium chloride,
KCl, solution that has a volume of 400.0 mL
and contains 85.0 g KCl?
Solution
msolute = 85.0 g KCl
Vsolution = 400.0 mL
MKCl = ?
Convert 85.0 g KCl to moles:
1 mol KCl
85.0 g KCl 
 1.14 mol KCl
74.55 g KCl
Convert 400.0 mL to L:
1L
400.0 mL 
 0.400 L
1000 mL
Calculate the molarity:
mol KCl 1.14 mol KCl
M=

 2.85 mol/L
L Solution
0.4000 L
Example
How many grams of solute are needed to
make 2.50 L of a 1.75 M solution of
Ba(NO3)2?
Solution to Example
1.75 mol
n
1.75 M=

1L
2.5 L
1 L  n  (1.75 mol)  (2.5 L)
n=4.38 mol
261.3 g Ba(NO3 ) 2
4.38 mol Ba(NO3 ) 2 
1 mol Ba(NO3 ) 2
 1144 g Ba(NO3 ) 2
Practice
1. Vinegar contains 5.0 g of acetic acid, CH3COOH in 100.0
mL of solution. Calculate the molarity of acetic acid in
vinegar
2. If 18.25 g HCl is dissolved in enough water to make 500.0
mL of solution, what is the molarity of the HCl solution?
3. If 20.0 g H2SO4 is dissolved in enough water to make 250.0
mL of solution, what is the molarity of the sulfuric acid
solution?
4. A solution of AgNO3 contains 29.66 g of solute in 100.0 mL
of solution. What is the molarity of the solution?
5. A solution of barium hydroxide, Ba(OH)2
contains 4.285 g of barium hydroxide in 100.0
mL of solution. What is the molarity of the
solution?
6. What mass of KBr is present in 25 mL of
0.85 M solution of potassium bromide?
7. If all the water in 430.0 mL of a 0.45 M NaCl
solution evaporates, what mass of NaCl will
remain?
Answers
1)
2)
3)
4)
5)
6)
7)
0.83 M
1.001 M
0.816 M
1.75 M
0.2501 M
2.5 g KBr
11 g NaCl
Molarity (M)
moles of solute
molarity (M) =
liters of solution
% Concentration (By Mass)
mass solute
% by mass=
100
mass solution
Masssolution = dsolution x vsolution
Example
What mass of lithium nitrate, LiNO3, would
have to be dissolved in 30.0 g of water in
order to make an 18.0% solution?
Solution to Example
msolute
 100
msolution

msolute
100=18
msolute  30.0
msolute
 0.18
msolute  30.0
msolute  0.18( msolute  30.0)
msolute  0.18msolute  5.4
 0.18msolute  0.18msolute
0.82msolute  5.4
msolute 
5.4
 6.59( g )
0.82
PRACTICE
1. A student wants to make a 5.00% of
rubidium chloride, RbCl using 0.377 g of
the substance. What mass of water will
be needed to make the solution?
2. A chemist dissolves 3.50 g of potassium
iodate and 0.23 g of potassium hydroxide
in 805.05 g of water. What is the
percentage concentration of each
solution?
3. What is the percentage concentration of
75.0 g of ethanol dissolved in 500.0 g of
water?
Answers
1. 7.16 g H2O
2. 0.430% potassium iodate
0.0284% potassium hydroxide
3. 13.0% ethanol
Molality (m)
mol Solute
molality =
kg Solvent
Do NOT capitalize “m”.
Example
A NaCl (molar mass = 58 g/mol) solution is
made by dissolving 29 g of NaCl in 200 g of
water. What is the molality of the solution?
Mole Fraction
mol of solute
mole fraction,  solute 
totoal mol of solution
mol of solvent
 solvent 
totoal mol of solution
 solute   solvent  1
Example
What is the mole fraction of sulfur dioxide
in an industrial exhaust gas containing
128.0 g of sulfur dioxide dissolved in every
1500. g of carbon dioxide?
Solution Stoichiometry
Mass, Volume, Concentration of Substance A
Moles of Substance A
Moles of Substance B
Mass, Volume, Concentration of Substance B
Example
What volume, in mL, of a 0.500 M solution of
copper (II) sulfate, CuSO4, is needed to react with
an excess of aluminum to provide 11.0 g of
copper?
3CuSO4(aq) + 2Al(s) → 3Cu(s) + Al2(SO4)3(aq)
Solution to Example
CuSO4 solution = 0.500 M
mCu = 11.0 g
Vsolution = ?
1 mol Cu 3 mol CuSO4
11.0 g Cu 

 0.173 mol CuSO4
63.55 g Cu
3 mol Cu
0.500 mol CuSO 4 0.173 mol CuSO 4
0.500 M =

1 L Solution
V
(0.500)  V=(0.173) 1 L
0.173 L
1L
V=
 0.346 L 
 346 mL
0.500
1000 mL
Practice Pg 467
1) Commercial hydrochloric acid, HCl, is 12.0 molar.
Calculate the mass of HCl in 250.0 mL of the
solution.
2) An excess of zinc is added to 125 mL of 0.100 M
HCl solution. What mass of zinc chloride is
formed?
Zn + 2HCl → ZnCl2 + H2
3) Yellow CdS pigment is prepared by reacting
ammonium sulfide with cadmium nitrate. What
mass of CdS can be prepared by mixing 2.50 L of
a 1.25 M Cd(NO3)2 solution with an excess of
(NH4)2S?
Cd(NO3)2(aq) + (NH4)2S(aq) → CdS(aq) + 2NH4NO3(aq)
Answers
1) 109 g HCl
2) 0.852 g ZnCl2
3) 451 g CdS