Concentrations of Solutions

Solution: A solution is a homogeneous mixture of two or more substances. A

minor species in a solution is called solute and the major species is the solvent.
Standard solution: A solution of precisely known concentration is termed a
standard solution.
Examples: 1 M Na2CO3, 1 M HCl, 2 M NaCl etc.
Concentration is a general measurement unit stating the amount of solute
present in a known volume or mass of solution or solvent.
𝐀𝐦𝐨𝐮𝐧𝐭 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞
Concentration =
𝐀𝐦𝐨𝐮𝐧𝐭 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
The ideal standard solution for a titrimetric method will
➢ be sufficiently stable so that it is necessary to determine its concentration
only once;
➢ react rapidly with the analyte so that the time required between additions
of reagent is minimized;
➢ react more or less completely with the analyte so that satisfactory end
points are realized;
➢ undergo a selective reaction with the analyte that can be described by a
balanced equation. 2
Few reagents completely meet these ideals.
Two basic methods are used to establish the concentration of such solutions.
The first is the direct method in which a carefully determined mass of a
primary standard is dissolved in a suitable solvent and diluted to a known
volume in a volumetric flask.
The second is by standardization in which the titrant to be standardized is
used to titrate (1) a known mass of a primary standard, (2) a known mass of a
secondary standard, or (3) a measured volume of another standard solution.
In standardization, the concentration of a volumetric solution is determined
by titrating it against a carefully measured quantity of a primary or secondary
standard or an exactly known volume of another standard solution.
A titrant that is standardized is sometimes referred to as a secondary
standard solution.
The concentration of a secondary-standard solution is subject to a larger
uncertainty than the concentration of a primary-standard solution.

3
Primary Standard Substances:
A primary standard substance is a highly purified compound that serves as a
reference material in titrations and in other analytical methods.
The accuracy of a method critically depends on the properties of the primary
standard. Important requirements for a primary standard substance are the
following:
➢ High purity;
➢ Atmospheric stability;
➢ Absence of hydrate water so that the composition of the solid does not change
with variations in humidity;
➢ Modest cost;
➢ Reasonable solubility in the titration medium;
➢ Reasonably large molar mass so that the relative error associated with
weighing the standard is minimized.

Examples:

4
Very few compounds meet or even approach these criteria and only a limited
number of primary-standard substances are available commercially.
Secondary standard:
A secondary standard is a compound whose purity has been determined by
chemical analysis.
The secondary standard serves as the working standard material for titrations and
for many other analyses.
NaOH, HCl, CH3COOH, Na2S2O3, KMnO4 etc.

Units of concentration:
Molarity; Formality, Normality, ppm, ppb and weight percent.
Molarity is defined as the number of moles of solute dissolved per liter of solution.

𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 𝐦𝐨𝐥

Molarity (M) = =
𝐥𝐢𝐭𝐞𝐫𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐋
𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 𝐦𝐦𝐨𝐥
= =
𝐦𝐢𝐥𝐥𝐢𝐥𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐦𝐋
5
 M × L = mol; M × mL = mmol
For example, solutions are referred to as being 2.0 molar, or 2.0 M. The M
refers to molar, and the solution is said to have a molarity of 2.0 or 2.0 mol
dissolved per liter of solution.
It is important to recognize that molarity is the number of moles dissolved per
liter of solution and not per liter of solvent.
A mole (mol) is Avogadro’s number of particles (atoms, molecules, ions, or
anything else).
𝐠𝐫𝐚𝐦𝐬
Moles = 𝐠
𝐟𝐨𝐫𝐦𝐮𝐥𝐚 𝐰𝐞𝐢𝐠𝐡𝐭 (𝐦𝐨𝐥)

Grams = Moles × formula weight (g/mol)

𝐦𝐢𝐥𝐥𝐢𝐠𝐫𝐚𝐦𝐬
Millimoles = 𝐦𝐠
𝐟𝐨𝐫𝐦𝐮𝐥𝐚 𝐰𝐞𝐢𝐠𝐡𝐭 ( )
𝐦𝐦𝐨𝐥

Milligrams = millimoles × formula weight (mg/mmol)

6
 Preparation of Molar Solutions
Problem: How would you prepare 500.0 mL of a 0.20 M solution of NaOH
from pure, solid NaOH?
Volume = 500 mL, Concentration = 0.20 M, Weight of NaOH =?
grams to weigh = LD × MD × FWSOL
where LD refers to the liters that are desired, MD to desired molarity, and
FWSOL to the formula weight of solute.
𝟏𝐋
LD = 500 mL  = 0.5 L
𝟏𝐨𝐨𝐨 𝐦𝐋
MD = 0.20 mol/L
FWSOL = 40 g/mol
𝐦𝐨𝐥 𝒈
grams to weigh (NaOH) = 0.5 L  0.20  40 = 4.0 g
𝐋 𝒎𝒐𝒍
Tell how you would prepare each of the following.
(a) 500.0 mL of a 0.10-M solution of KOH from pure solid KOH (MW = 56.11)?
(b) 250.0 mL of a 0.15-M solution of NaCl from pure solid NaCl (MW= 58.88)?
(c) 100.0 mL of a 2.0-M solution of glucose from pure solid glucose (C6H12O6)? 7
Example: What is the molarity of a solution that has 4.5 mol of solute dissolved
in 300.0 mL of solution?
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞 𝟒.𝟓 𝐦𝐨𝐥
Molarity (M) = = = 15 M
𝐥𝐢𝐭𝐞𝐫𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝟑𝟎𝟎.𝟎 𝐦𝐋 × 𝟏𝐋
𝟏𝟎𝟎𝟎 𝐦𝐋

Example: What is the molarity of a solution of NaOH (MW = 40) that has 0.491
g dissolved in 400.0 mL of solution?

𝐠𝐫𝐚𝐦𝐬 𝟏 𝐦𝐨𝐥 𝐍𝐚𝐎𝐇

Moles = 𝐠
𝐌𝐖 (𝐦𝐨𝐥)
= 𝟎. 𝟒𝟗𝟏 𝐠 ×
𝟒𝟎 𝐠 𝐍𝐚𝐎𝐇
= 0.0123 mol

𝟎.𝟎𝟏𝟐𝟑 𝐦𝐨𝐥
∴ Molarity = 𝟏𝐋 = 0.03068 M = 0.0307 M
𝟒𝟎𝟎.𝟎 𝐦𝐋 × 𝟏𝟎𝟎𝟎 𝐦𝐋

Problem: A solution is prepared by dissolving 1.26 g AgNO3 (FW = 169.9) in a

250-mL volumetric flask and diluting it to volume. Calculate the molarity of
the silver nitrate solution. How many millimoles AgNO3 were dissolved?
Solution:
8
Problem
1. What is the molarity of a solution of Na2CO3 (106 g/mol) that has 2.650 g
dissolved in 250.0 mL of solution?
2. How many grams Na2SO4 (142 g/mol) should be weighed out to prepare 500
mL of a 0.100 M solution?
3. What weight of potassium permanganate KMnO4 (158.04 g/mol) is needed
to make a 0.15 molar solution in 425 mL of H2O?
4. Calculate the molarity of each of the following solutions: (a) 10.0 g H2SO4 in
250 mL of solution, (b) 6.00 g NaOH in 500 mL of solution, (c) 25.0 g AgNO3
(169.87 g/mol) in 1.00 L of solution.
5. Calculate the number of grams in 250 mL of each of the following solutions:
(a) 0.100 M NaHCO3 (84.007 g/mol), (b) 0.250 M Fe(NH4)2(SO4)2·6H2O (392.14
g/mol), (c) 0.0167 M K2Cr2O7 (294.12 g/mol).
6. Calculate the grams of each substance required to prepare the following
solutions: (a) 250 mL of 0.100 M KOH (56.1056 g/mol), (b) 1.00 L of 0.25 M
C6H12O6, (180.156 g/mol) (c) 500 mL of 0.0500 M CuSO4 (159.61 g/mol).
9
6. What mass of sodium iodide (149.89 g mol−1 ) is contained in 250 mL of a
0.500 M solution?
7. Calculate the molar concentrations of all the cations and anions in a solution
prepared by mixing 10.0 mL each of the following solutions: 0.100 M
Mn(NO3)2, 0.100 M KNO3, and 0.100 M K2SO4.
8. Calculate the concentration of potassium ion, K+ in grams per liter after
mixing 100 mL of 0.250 M KCl and 200 mL of 0.100 M K2S04
Formality (F):
The number of moles of solute, regardless of chemical form, per liter of
solution.
There is no difference between a substance’s molarity and formality if it
dissolves without dissociating into ions. The molar concentration of a solution
of glucose, for example, is the same as its formality.
For substances that ionize in solution, such as NaCl, molarity and formality
are different.
For example, dissolving 0.1 mol of NaCl in 1 L of water gives a solution
containing 0.1 mol of Na+ and 0.1 mol of Cl–. 10
The molarity of NaCl, therefore, is zero since there is essentially no un-
dissociated NaCl in solution. The solution, instead, is 0.1 M in Na+ and 0.1 M
in Cl– (0.1 mole/ L CaCl2 = M and F?).
The formality of NaCl, however, is 0.1 F because it represents the total
amount of NaCl in solution.
When we state that a solution is 0.1 M NaCl we understand it to consist of
Na+ and Cl– ions.
The unit of formality is used only when it provides a clearer description of
solution chemistry.
Normality:
Normality is defined as the number of equivalents of solute dissolved per
liter of solution.
𝐍𝐨. 𝐨𝐟 𝐞𝐪𝐮𝐢𝐯𝐚𝐥𝐞𝐧𝐭𝐬 𝐞𝐪
Normality (N) = =
𝐥𝐢𝐭𝐞𝐫𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧 𝐋

The number of equivalents (eq), n, is based on a reaction unit, which is that

part of a chemical species involved in a reaction.
11
 H2SO4 + 2NaOH Na2SO4 + H2O
For the above reaction we find that n = 2 for H2SO4 and n = 1 for
NaOH.
Na2CO3 + HCl NaHCO3 + H2O
Na2CO3 + 2HCl 2NaCl + H2CO3
NaOH + HCl NaCl + H2O
It is clear from these reactions that determining of number of
equivalents for a chemical species requires an understanding of
how it reacts.

n = 1 for Fe3+ and n = 2 for Sn2+

12
 𝐠
𝐠 𝐅𝐖 (𝐦𝐨𝐥)
eq = 𝐠 (eq wt = equivalent weight); eq wt =
𝐞𝐪 𝐰𝐭 (𝐞𝐪) 𝐞𝐪/𝐦𝐨𝐥
𝐠
𝟗𝟖 (𝐦𝐨𝐥)
For H2SO4 : eq wt = 𝐞𝐪 = 49 (g/eq)
𝟐( )
𝐦𝐨𝐥

𝐠
𝐠
𝐞𝐪 𝐰𝐭 (𝐞𝐪)
Normality (N) =
𝐋
Example:
What is the normality of a solution of oxalic acid dihydrate
(H2C2O4·2H2O; formula weight, 126.07 g per mole) if it is to be used
as in the following reaction and 0.4920 g of it is dissolved in 250.0
mL of solution?
H2C2O4 + 2NaOH Na2C2O4 + H2O
13
Solution: In the reaction given, two hydrogen atoms are lost per
molecule of oxalic acid, i.e., number of reacting unit per mole of
oxalic acid is 2.
𝐠
𝟏𝟐𝟔.𝟎𝟕 ( )
𝐦𝐨𝐥
eq wt of oxalic acid = 𝐞𝐪 = 63.035 (g/eq)
𝟐( )
𝐦𝐨𝐥
𝟎.𝟒𝟗𝟐𝟎 𝐠
𝐠
𝟔𝟑.𝟎𝟑𝟓( ) 𝟎.𝟎𝟑𝟏𝟐𝟐 𝐞𝐪
𝐞𝐪
Normality = 𝟏𝐋 = = 0.03122 N
𝟐𝟓𝟎.𝟎 𝐦𝐋 × 𝐋
𝟏𝟎𝟎𝟎 𝐦𝐋

Relation between normality and molarity: N = n × M

Example: Calculate the equivalent weight and normality for a

solution of 6.0 M H3PO4 given the following reactions:

14
Molality (m) is concentration expressed as moles of substance per
kilogram of solvent (not total solution).
𝐦𝐨𝐥 𝐬𝐨𝐥𝐮𝐭𝐞
Molality (m) =
𝐊𝐠 𝐬𝐨𝐥𝐯𝐞𝐧𝐭
Percent Concentration: Chemists frequently express concentrations in
terms of percent (parts per hundred). Three common methods are
generally employed to express percent composition.
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞
weight percent (w/w) = × 100
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
For example, nitric acid is sold as a 70% (w/w) solution, meaning that the
reagent contains 70 g of HNO3 per 100 g of solution
15
 𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞
volume percent (v/v) = × 100
𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐭𝐨𝐭𝐚𝐥 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
For example, a 5% (v/v) aqueous solution of methanol usually describes a
solution prepared by diluting 5.0 mL of pure methanol with enough water
to give 100 mL.
𝐦𝐚𝐬𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞, 𝐠
weight/volume percent (w/v) = × 100%
𝐯𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧, 𝐦𝐋
For example, 5% (w/v) aqueous silver nitrate often refers to a solution
prepared by dissolving 5 g of silver nitrate in sufficient water to give 100
mL of solution.
Always specify the type of percent when reporting concentrations in this
way.
Parts per Million (ppm) and Parts per Billion (ppb)
When a solution contains a very small concentration of solute,
concentration is often expressed in parts per million (ppm) or parts per
billon (ppb).
16
Whereas percent concentration is the number of “parts”—grams or
milliliters—in 100 parts (100 mL) of solution, parts per million is the
number of “parts” in 1,000,000 parts of solution. The “parts” may be
expressed in either mass or volume units as long as the same unit is used
for both the numerator and denominator.

When water is the solvent (for very dilute solutions), the density of the
solution is close to the density of pure water, which is 1.0 g/mL at room
temperature. Then, we frequently equate 1 g of water with 1 mL of water,
although this equivalence is only approximate. Therefore, 1 ppm
corresponds to 1 mg/L or 1 μg/mL and 1 ppb is 1 μg/L or 1 ng/mL.
17
 mass solute (mg) mass solute (𝛍g)
ppm = =
volume solution (L) volume solution (mL)
mass solute (𝛍g) mass solute (ng)
ppb = =
volume solution (L) volume solution (mL)

Problem: What is the concentration in parts per million of DDT in the

tissues of a seabird that contains 50. mg of DDT in 1,900 g of tissue?
Solution:

a) Calculate the molar concentrations of 1.00 ppm solutions each of Li+

(FW = 6.94 g/mol) and Pb2+ (FW =207 g/mol).
b) What weight of Pb(NO3)2 (FW = 283.2 g/mol) will have to be dissolved
in 1 liter of water to prepare a 100 ppm Pb2+ solution? [CLi = 1.44  10−4
mol/L; CPb = 4.83  10−6 mol/L; 0.137 g Pb(NO3)2]
18
1. What is the molar concentration (molarity, M) of a solution made up of
6.239 g of pure (>99.9%) lithium hydroxide (FW = 24.02) in 857 mL of
H2O? What is the molar concentration of the lithium (Li+) and hydroxide
(OH−)?
2. A 0.456 g sample of an ore is analyzed for chromium and found to
contain 0.560 mg Cr2O3 (FW = 151.99 g/mol). Express the concentration
of Cr2O3 in the sample as (a) percent and (b) parts per million. (0.123%,
1.23 ×103 ppm)
3. What is the parts per million (ppm) of a solution made up of 12.3 mg
lactose in 750 mL water? What is the parts per billion (ppb)? What is the
lactose concentration expressed in percentage (%)?
4. A 2.6 g sample of plant tissue was analyzed and found to contain 3.6 μg
zinc. What is the concentration of zinc in the plant in ppm and ppb?
5. A 25.0-μL serum sample was analyzed for glucose content and found to
contain 26.7 μg. Calculate the concentration of glucose in μg/mL and in
mg/dL. 19
Density: The mass (or weight) per unit volume of a material at a given
temperature.
grams per cubic centimeter (g/cc or g/cm3)
kilograms per cubic meter (kg/m3)
pounds per cubic foot (lb/ft3)
pounds per cubic inch (lb/cu in or lb/in3)
Specific gravity: The ratio of the density of a material at a given
temperature to the density of an equal volume of water at the same
temperature.
Specific gravity is dimensionless and so is not tied to any particular system
of units
Since the density of water is approximately 1.00 g/mL (4 oC), we use
density and specific gravity interchangeably.
𝝆𝑯𝑪𝒍
SG =  HCl = SG × water
𝝆𝒘𝒂𝒕𝒆𝒓
 HCl = 1.19 × 1.00 g/mL = 1.19 g/mL
20
Problem: Calculate the molar concentration of HCl (36.46 g/mol) in a
solution that has a specific gravity of 1.19 and is 37.0% HCl (w/w). How
many milliliters of this reagent should be diluted to 1.000 L to make 0.100
M HCl?
Solution:
g HCl 𝟏.𝟏𝟗 𝐊𝐠 𝐫𝐞𝐚𝐠𝐞𝐧𝐭 𝟏𝟎𝟑 𝐠 𝐫𝐞𝐚𝐠𝐞𝐧𝐭 𝟑𝟕.𝟎 𝐠 𝐇𝐂𝐥
= × ×
𝐋−𝐫𝐞𝐚𝐠𝐞𝐧𝐭 𝐋 𝐫𝐞𝐚𝐠𝐞𝐧𝐭 𝟏 𝐊𝐠 𝐫𝐞𝐚𝐠𝐞𝐧𝐭 𝟏𝟎𝟎 𝐠 𝐫𝐞𝐚𝐠𝐞𝐧𝐭
= 4.4 × 102 (g/L-reagent)
𝐠
(𝟒.𝟒 ×𝟏𝟎𝟐 𝐠)/𝟑𝟔.𝟒𝟔(𝐦𝐨𝐥) 𝟏𝟐.𝟏 𝐦𝐨𝐥
Molar concentration = = = 12.1 M
𝐋 𝐫𝐞𝐚𝐠𝐞𝐧𝐭 𝐋

Vconcd × Cconcd = Vdil × Cdil

𝟏.𝟎𝟎𝟎 𝐋 ×𝟎.𝟏𝟎𝟎 𝐌
Vconcd = (Vdil × Cdil )/Cconcd =
𝟏𝟐.𝟏 𝐌
= 0.00826 L = 8.26 mL

21
1. Calculate the molar concentration of HNO3 (63.0 g/mol) in a solution
that has a specific gravity of 1.42 and is 70.5% HNO3 (w/w).
2. Calculate the molar concentration of H2SO4 (98.0 g/mol) in a solution
that has a specific gravity of 1.84 and is 96.5% H2SO4 (w/w).
3. How many milliliters of concentrated sulfuric acid, 94.0% (g/100 g
solution), density 1.831 g/cm3, are required to prepare 1 liter of a 0.100
M solution?
Molar Analytical Concentration or Analytical concentration:
Molar analytical concentration or analytical concentration is the total
number of moles of a solute, regardless of its chemical state, in 1 L of
solution.
In other words, the molar analytical concentration specifies a recipe by
which the solution can be prepared regardless of what might happen to
the solute during the solution process.
If 1.0 mole or 98 g of H2SO4 is dissolved in water and diluting the acid to
exactly 1.0 L; analytical concentration CH2SO4 = 1.0 M 22
Molar Equilibrium Concentration
The molar equilibrium concentration, or just equilibrium concentration,
refers to the molar concentration of a particular species in a solution at
equilibrium.
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+ + OAc-
The analytical concentration ( CHOAc ) = 0.1 M
CHOAc = [HOAc]undissociated + [-OAc]
The equilibrium concentration = [HOAc]undissociated.
For species X we express the analytical concentration as cX and the
equilibrium concentration as [X].
23
Problem:
Calculate the molar concentrations of all the cations and anions in a
solution prepared by mixing 10.0 mL each of the following solutions: 0.100
M Mn(NO3)2, 0.100 M KNO3, and 0.100 M K2SO4.
Solution:
Amount Mn(NO3)2 = 10.0  0.100 M = 1.00 mmol
Mn2+ = 1.00 mmol; NO3− = 2  1.00 mmol
Similarly,
KNO3 : K+ = 1.00 mmol; NO3− = 1.00 mmol
K2SO4 : K+ = 2.00 mmol; SO42− = 1.00 mmol
Total: NO3− = (2.00 + 1.00) mmol; K+ = (1.00 + 2.00) mmol; SO42− = 1.00
mmol
Total volume of solution = (10.0 + 10.0 + 10.0) mL = 30.0 mL
𝟑.𝟎𝟎 𝒎𝒎𝒐𝒍
Concentration of NO3− = = 0.100 M
𝟑𝟎.𝟎 𝒎𝑳
24