CONCENTRATION
C = n / V — mol/L also written M
V must be in LITRES — divide mL ÷ 1000
VARIABLE KEY
C
concentration (mol/L = M)
EXAMPLE
0.5 mol NaCl in 2 L
C = 0.5 / 2 = 0.25 mol/L
Make 250 mL of 0.30 M CuSO₄ (M=159.6 g/mol)
n = 0.30×0.250 = 0.075 mol
m = 0.075×159.6 = 11.97 g
DILUTION
C₁V₁ = C₂V₂ — moles of solute = constant
C₁
initial (stock) concentration
C₂
final (diluted) concentration
V₂ = V₁ + Vwater added
V₂ is total volume, not volume of water alone!
DILUTION EXAMPLE
100 mL of 2.0 M → diluted to 500 mL
C₂ = (2.0 × 0.100) / 0.500
C₂ = 0.40 M
FIND STOCK VOLUME
Need 300 mL of 0.15 M H₂SO₄ from 6.0 M stock
V₁ = C₂V₂ / C₁ = 0.15×0.300 / 6.0
V₁ = 7.5 mL of stock → dilute to 300 mL
PREPARING SOLUTIONS
- Calculate mass: m = n × M
- Weigh solute precisely
- Dissolve in small vol. solvent
- Transfer to volumetric flask
- Fill to graduation mark
- Add acid TO water, never reverse
SOLUBILITY RULES (ionic compounds)
| Rule | Ion | Exception |
|---|
| ALWAYS soluble | Group 1, NH₄⁺ | None |
| ALWAYS soluble | NO₃⁻ | None |
| Usually soluble | CH₃COO⁻ | None (most) |
| Usually soluble | Cl⁻, Br⁻, I⁻ | Ag⁺, Pb²⁺, Hg₂²⁺ |
| Usually soluble | SO₄²⁻ | Ba²⁺, Pb²⁺, Ca²⁺ |
| Usually insoluble | OH⁻ | Gr 1, Ba²⁺ |
| Usually insoluble | CO₃²⁻, PO₄³⁻ | Group 1, 2 |
| Usually insoluble | S²⁻ | Group 1, 2 |
PRECIPITATION TEST
Mix two solutions: if both ions form an insoluble compound → precipitate forms
AgNO₃(aq) + NaCl(aq) → AgCl↓ + NaNO₃(aq)
AgCl is insoluble (Ag⁺ with Cl⁻) → white precipitate
Ba(NO₃)₂(aq) + Na₂SO₄(aq) → BaSO₄↓ + NaNO₃(aq)
BaSO₄ insoluble (Ba²⁺ with SO₄²⁻) → white precipitate
SATURATION STATES
Unsaturated
more can dissolve
Saturated
at max dissolved
Supersaturated
unstable excess
ELECTROLYTES
Strong
fully dissociates → ions
Nonelectrolyte
no ions formed
STRONG ACIDS (fully dissociate)
- HCl — hydrochloric
- HNO₃ — nitric
- H₂SO₄ — sulfuric
- HBr — hydrobromic
- HI — hydroiodic
- HClO₄ — perchloric
STRONG BASES
- NaOH, KOH
- Ca(OH)₂, Ba(OH)₂
- Most soluble ionic salts
WEAK & NONELECTROLYTES
Nonelec.
C₆H₁₂O₆ (glucose), C₂H₅OH
SOLUTION STOICHIOMETRY
mol = C (mol/L) × V (L)
Then use mole ratios as usual
ROADMAP
Step 1
n = C × V (in litres)
Step 2
mole ratio from equation
Step 3
m = n × M or C = n/V
30 mL of 2.0 M HCl + NaOH (1:1)
n(HCl) = 2.0×0.030 = 0.060 mol
n(NaOH) = 0.060 mol
m(NaOH) = 0.060×40 = 2.4 g