Percent (%) concentration
% (w/v) concentration:
mass of solute in grams contained by 100 mL solution
% (w/w) concentration:
mass of solute in grams contained in 100 g of solution
% (v/v) concentration:
volume of solute in mL in 100 mL solution
parts per million (ppm) are used to express the
concentrations of very dilute solution
A solution whose solute concentration is 1 ppm
contains 1 g of solute for each million (106) grams of
solution or, equivalently, 1 mg of solute per kilogram
of solution
A 2.5-g sample of groundwater was found to contain
5.4 g of Zn2+. What is the concentration of Zn2+ in
parts per million?
Mole Fraction, Molarity, and Molality are
concentration expressions based on the number of
moles of one or more components of the solution
The symbol X is used for mole fraction, with a
subscript to indicate the component of interest. The
mole fraction of HCl is represented as XHCl.
The sum of the mole fractions of all components of
a solution must equal 1
A solution of hydrochloric acid contains 36 percent
HCl by mass. Calculate the mole fraction of HCl in
the solution.
Molarity of a solute in a solution is defined as the
number of moles of solute per liters of solution
A 2.00L solution of hydrochloric acid contains 36
g of HCl. Calculate the molarity of HCl in the
solution.
0.99 mol
= 0.495 = 0.50 M
MHCl = ---------------2.00 L
The molality of a solution, denoted m, is defined as
the number of moles of solute per kilogram of solvent
A solution of hydrochloric acid contains 36 percent
HCl by mass. Calculate the molality of HCl in the
solution. (c) What additional information would you
need to calculate the molarity of the solution?
Mass of water = 100 gSolu – 36 g HCl = 64 g
(a) Calculate the mole fraction of NaOCl in a
commercial bleach solution containing 3.62
mass percent NaOCl in water.
3.62 gNaOCl
3.62 gNaOCl
 3.62 gNaOCl 
100 gsolution
100 gsolution
3.62 gNOCl  96.38 gWate
1molNaOCl
3.62 gNaOCl
74.44 gNaOCl
1molNaOCl
1molH 2O
3.62 gNOCl
 96.38 gWate
74.44 gNaOCl
18 gH 2O
0.048628molNaOCl
0.048628molNOCl  5.3544molWate
0.009000149
=0.00900
(b) What is the molality of a solution made by
dissolving 36.5 g of naphthalene, C10H8, in 420 g of
toluene, C7H8? (b) 0.678 m
1molC10 H 8
36.5 gC10 H 8
128.17 gC10 H 8 =0.28477 / .420
m
1kg
420 gSolvent 3
10 g
0.678024
Given that the density of a solution of 5.0 g of toluene
and 225 g of benzene is 0.876 g/mL, calculate (a) the
molarity of the solution; (b) the mass percentage of
solute.
Calculate the molality of a solution that
contains 25 g of H2SO4 dissolved in 80. g of
H 2O
Calculate the molality of a 10.0%
H3PO4 solution in water
Calculate the molality of a solution that contains
51.2 g of naphthalene, C10H—, in 500. mL of
carbon tetrachloride. The density of CCl4 is 1.60
g/mL
If 8.32 grams of methanol, CH3OH, are
dissolved in 10.3 grams of water, what is the
mole fraction of methanol in the solution?
What is the molarity of 2500. mL of a solution
that contains 160. grams of NH4NO3?
11-1Acid and base Reactions
100. mL of 0.100 M HCl solution and 100. mL of
0.100 M NaOH are mixed. What is the molarity of the
salt in the resulting solution? Assume that the volumes
1. Balance Chemical Equation
are additives.
HCl + NaOH  H2O + NaCl
Reaction Ratio: 1 mol 1 mol
1 mol 1 mol
0
10.0
m
mol
10.0
m
mol
Start:
10.0
-10.0
Change:
-10.0
After reaction: 0.00
0.00
10.0
For HCl: (100 mL)(0.100mol/L) = 10.0 mmol HCl
For NaOH = (100. mL)(0.100mol/L) = 10.0 mmol
For NaCl: moles of NaCl at the end of reaction =
moles of HCl at beginning = 10.0 m mol
10.0 mmol NaCl
MNaCl = ---------------------------- = 0.0500
(100+100) mL solution
11-1Acid and base Reactions
80. mL of 0.100 M HCl solution and 100. mL of 0.100
M NaOH are mixed. What is the molarity of the salt in
the resulting solution? Assume that the volumes are
1. Balance Chemical Equation
additives.
HCl + NaOH  H2O + NaCl
Reaction Ratio: 1 mol 1 mol
1 mol 1 mol
Start:
8.0 m mol 10.0 m mol
0
8.0 m mol
Change:
-8.0
-8.0
After reaction: 0.00
2.0
8.0
For HCl: (80 mL)(0.100mol/L) = 8.0 mmol HCl
For NaOH = (100. mL)(0.100mol/L) = 10.0 mmol
For NaCl: moles of NaCl at the end of reaction =
moles of HCl at beginning = 8.0 m mol
8.0 mmol NaCl
MNaCl = ---------------------------- = 0.044
(100+80) mL sol 2.0 mmol NaOH
MNaOH = ------------------------ = 0.011
180 mL solution
100. mL of 0.100 M HCl solution and 100. mL of 0.80
M NaOH are mixed. What is the molarity of the salt in
the resulting solution? Assume that the volumes are
1. Balance Chemical Equation
additives.
HCl + NaOH  H2O + NaCl
Reaction Ratio: 1 mol 1 mol
1 mol 1 mol
10.0 m mol 8.0 m mol
0
Start:
Change:
-8.0
+8.0
-8.0
After reaction: 2.0
0.0
8.0
For HCl: (100 mL)(0.100mol/L) = 10.0 mmol HCl
For NaOH = (100. mL)(0.80mol/L) = 8.0 mmol
For NaCl: moles of NaCl at the end of reaction =
moles of NaOH at beginning = 8.0 m mol
8.0 mmol NaCl
MNaCl = ---------------------------- = 0.040
(100+100) mL solution
2.0 mmol NaCl
MHCl = ---------------------------= 0.010
(100+100) mL solution
100. mL of 0.100 M HCl solution and 100. mL of 0.80
M NaOH are mixed. What is the molarity of the salt in
the resulting solution? Assume that the volumes are
1. Balance Chemical Equation
additives.
HCl + NaOH  H2O + NaCl
Reaction Ratio: 1 mol 1 mol
1 mol 1 mol
10.0 m mol 80. m mol
0
Start:
Change:
-10.0
+10.0
-10.0
After reaction: 0.00
70.
8.0
For HCl: (100 mL)(0.100mol/L) = 10.0 mmol HCl
For NaOH = (100. mL)(0.80mol/L) = 80. mmol
For NaCl: moles of NaCl at the end of reaction =
moles of NaOH at beginning = 10. m mol
10. mmol NaCl
MNaCl = ---------------------------- = 0.050
(100+100) mL solution
70. mmol NaCl
MNaOH = ---------------------------- = 0.035
(100+100) mL solution
100. mL of 1.00 M H2SO4 solution is mixed with 200.
mL of 1.00 M KOH. What is the molarity of the salt in
the resulting solution? Assume that the volumes are
1. Balance Chemical Equation
additives.
H2SO4 + 2KOH  2H2O + K2SO4
Reaction Ratio: 1 mol 2 mol
2 mol 1 mol
Start:
100 m mol 200 m mol
0
100 m mol
Change:
-200
-100
After reaction: 0.00
0.0
100
Moles of H2SO4: (100 mL)(1.00mol/L) = 100 mmol
mols KOH = (200. mL)(1.00mol/L) = 200 mmol
moles of K2SO4 at the end of reaction = moles of
H2SO4 at beginning = 100 m mol
100 mmol K2SO4
M H2SO4 = --------------------= 0.333
(100+200) mL sol
Titration
Titration is a process in which a solution of one
reactant, titrant, is carfully added