Solutions
Solutions consist of a solute (the substance that gets dissolved) and a solvent (the substance that does the dissolving). Solutions are homogeneous.
A
solution can either be dilute or concentrated, depending on how much solute is
dissolved in the solvent – this is a qualitative measure of the concentration
of a solution.
We
can also get a quantitative measure of concentration:
- percentage concentration – m/V, m/m or m/V
- parts per million (ppm) or parts per billion (ppb)
- molar concentration (usually referred to as simply “concentration’)
Percentage Concentration
Sometimes
concentrations are expressed as percentages.
For instance, the label of a vinegar bottle shows 5% V/V acetic acid (meaning
there are 5 mL of acetic acid for every 100 mL of vinegar solution).
Percentage
concentration is expressed as m/V (mass by volume), m/m (mass by mass) or V/V
(volume by volume):
W/V ↝ csolution = msolute/Vsolution
× 100, where mass is in g and volume is in mL
V/V ↝ csolution = Vsolute/Vsolution
× 100, volume is in mL
ex. A salt solution is formed by mixing
1.50 g of sodium chloride in enough water to make 250 mL of solution. What is the m/V percentage concentration of
salt in the solution?
msolute = 1.50 g csolution = msolute/Vsolution
× 100
Vsolution = 250 mL = (1.50 g/250 mL) × 100
= 0.60%
∴ the salt solution is 0.60% W/V
Parts per Million (Parts per Billion)
This
is used when we want to express the concentration of a very dilute solution (ppm
– one part solute for every million parts of solution)
ppm ↝ c = msolute/Vsolution × 106,
where m is in g and volume is in mL
ppb
↝ c = msolute/Vsolution × 109,
where m is in g and volume is in mL
ex.
In a chemical analysis, 0.0022 g of oxygen was measured in 250 mL of pond
water. What is the concentration of
oxygen in ppm?
mO2
= 0.0022 g cO2 = mO2/ VH2O
×
106
VH2O
= 250 mL = (0.0022
g/250 mL) ×
106
cO2
= ? = 8.8
ppm
∴ the
concentration is 8.8 ppm.
Molar Concentration
If we wish to make up a solution of
known concentration, we need to know how much solute to measure.
n = cV, where n is the number of moles
(mol), c is concentration (mol/L) and V is volume (L)
ex. What amount of lithium bromide is
required to make 250 mL of 1.25 mol/L solution?
V = 250 mL n = cV
c = 1.25 mol/L = (1.25 mol/L)(0.25 L)
n = ? = 0.313
mol
∴ 0.31
mol of lithium bromide is required
Homework:
Practice, p. 126 # 1, 3
Practice, p. 128 # 5, 6
Practice, p. 130 # 8, 9
Practice, p. 131 # 12, 13
Practice, p. 133 # 16
Answers:
