CONCENTRATION OF SUBSTANCES
•Concentration is usually defined in terms of solute (solid) per volume of solvent (liquid).
•The unit of concentration depends on the units for the quantity of solute and quantity of solution.
•Solution is the result of a solute (solid) being dissolved in a solvent (liquid).
•A concentrated solution may be diluted by the addition of more solvent.
CONCENTRATION UNITS
•There are many ways to express concentration units:
1. Grams per litre (g/L).
2. Parts per million (ppm)- Micrograms per gram (µg/g)
OR milligrams per kilogram (mg/kg)
OR milligrams per litre (mg/L).
3. Parts per billion (ppb)- Micrograms per kilogram (µg/kg)
OR micrograms per Litre (µg/L)
4. Percentage by mass (%w/w) or (m/m)
5. Percentage mass / volume (%w/v) or (m/v)
6. Percentage by volume (%v/v).
•Concentration is usually defined in terms of solute (solid) per volume of solvent (liquid).
•The unit of concentration depends on the units for the quantity of solute and quantity of solution.
•Solution is the result of a solute (solid) being dissolved in a solvent (liquid).
•A concentrated solution may be diluted by the addition of more solvent.
CONCENTRATION UNITS
•There are many ways to express concentration units:
1. Grams per litre (g/L).
2. Parts per million (ppm)- Micrograms per gram (µg/g)
OR milligrams per kilogram (mg/kg)
OR milligrams per litre (mg/L).
3. Parts per billion (ppb)- Micrograms per kilogram (µg/kg)
OR micrograms per Litre (µg/L)
4. Percentage by mass (%w/w) or (m/m)
5. Percentage mass / volume (%w/v) or (m/v)
6. Percentage by volume (%v/v).
CONCENTRATION OF SOLUTIONS
•Molar concentration, or molarity, is the amount of solute (moles) present 1.00 Litre of solution.
•Symbol for Concentration is C.
•Unit for Concentration is mol L-1 or M.
•Molar Concentration is defined as:
•Concentration = quantity of solute (n)
volume of solution (V)
C = n
V
Sample Problem:
•What mass of sodium chloride, NaCl, would be needed to prepare a 500 mL solution with a concentration of 0.08 mol L^-1 ?
Solution:
(1) Calculate n (NaCl):
n (NaCl) = C x V
= 0.08 x 0.500 = 0.04 mol
(2) Calculate m (NaCl):
m (NaCl) = n x M
= 0.04 x (23 + 35.5)
= 0.04 x 58.5 = 2.34 g
DILUTION OF SOLUTIONS
•When a solution is diluted by the addition of more solvent (eg: water); the number of moles (n) of the solute remains the same.
•If n1 = initial amount of substance of solute (before dilution) and n2 = final amount of substance of solute (after dilution); then
n1 = n2
•Because n = C x V then we can rewrite the above expression as:
C1 x V1 = C2 X V2
Sample Problem:
•What volume of 18 M hydrochloric acid is needed to prepare 250 mL of a 2.0 M hydrochloric acid solution?
Solution:
(1) List Data:
C1 = 18 M C2 = 2.0 M
V1 = ??? V2 = 250 mL = 0.250 L
(2) Calculate Volume:
C1 x V1 = C2 x V2
18 x V1 = 2.0 x 0.250
V1 = 2.0 x 0.250 / 18
V1 = 0.0278 L = 27.8 mL
Volume of 18 M hydrochloric acid required was 27.8 mL