Solutions and Concentration - Solution Composition
Chemical reactions often take place in aqueous solutions. To perform stoichiometric calculations in such cases the amounts of chemicals present in solution – the concentration of solution - must be known.
Concentrationof a solution is a measurement stating the amount of a solute present in a known amount of solution:
Concentration = amount of solute / amount of solution
The terms solute and solution are usually used for liquid samples but they can be extended to gaseous and solid samples.
The most common units of concentration are given in Table I.1:
Common Units of Concentration | ||
Name | Symbol | Units |
molarity | moles solute / liters of solution | M |
molality | moles solute / kg solvent | m |
normality | number of equivalent weights of solute / liters of solution | N |
formality | number of formal weights of solute / liters of solution | F |
weight % | g solute / 100 g of solution | % w/w |
volume % | ml solute / 100 ml solution | % v/v |
weight-to-volume % | g solute / 100 ml solution | % w/v |
parts per million | g solute / 106 g solution | ppm |
parts per billion | g solute / 109 g solution | ppb |
Note: Another way of describing solution concentration is the mole fraction (xi)
Molarity (M)is defined as the number of moles of solute per liter of solution.
i.e by dissolving 0.1 mol NaOH in 1 l of H2O gives a solution that contains 0.1 mol Na+ and 0.1 mol of OH- in 1 l. The concentration of the solution is [Na+] = 0.1 M and [OH-] = 0.1 M.
Since molarity depends on the volume of the solution it changes slightly with temperature.
Another way of describing solution concentration is molality (m) which is the number of moles of solute per kilogram of solvent.
Molality is independent of temperature since it depends on mass.
In very dilute aqueous solutions the molarity (M) and molality (m) are nearly the same.
Example #1
A solution of 1M H2SO4 has density 1.04 g/cm3. Calculate the (%w/w) concentration of the solution.
Given | [H2SO4] = 1M d = 1.04 g/cm3 PH2O = 41 mmHg MW H2SO4 = 98 g/mole |
Asked for | (%w/w) = ? |
From the definition of (%w/w) = g solute / 100 g of solution (1)
The mass of solute (g solute) is unknown but it can be calculated.
Since [H2SO4] = 1M ⇒1000 cm3 of H2SO4 solution contain 1 mole “pure” H2SO4 (2)
The mass of 1 mole “pure” H2SO4 can be calculated as shown below:
mass (g) = mole* MW = 1 mole * 98 g/mole = 98 g (3)
From (2) and d = m/V = 1.04 g/cm3 the mass of 1000 cm3 of H2SO4can be calculated:
m = d* V = 1.04g/cm3 * 1000cm3 = 1040 g of H2SO4 solution (4)
From (2), (3) and (4):
Mass of 1040 g of H2SO4 solution contain 98 g of“pure”H2SO4
Mass of 100 gof H2SO4 solution contain x = ? g of“pure”H2SO4
x = 98 g“pure”H2SO4 * (100gof H2SO4 / 1040gof H2SO4) = 9.42 g of“pure”H2SO4
Therefore,(%w/w) = 9.42
Provided that the theory and the definitions of solution concentration units is understood a % solution calculator can be used.
Meant to be used in both the teaching and research laboratory, a % solution calculator can be utilized to perform a number of different calculations for preparing percent (%) solutions when starting with the solid or liquid material.