Chemical Calculations for Solutions | General Chemistry 2

Solution: homogeneous mixture composed of a solute and a solvent
Solute: material dissolved
Solvent: material into which the solute is dissolved (usually present in greater quantity)
 

An aqueous solution of NaCl is a homogeneous mixture composed of NaCl (the solute) dissolved in water (the solvent)

Saturated Solution: homogeneous mixture that contains the maximum amount of solute possible ⇒ additional solute remains undissolved
Solubility: maximum quantity of solute dissolvable in a solvent at a particular temperature. A solid is more soluble at higher temperatures and in a solvent with similar types of intermolecular forces
 

The solubility of NaCl in water at 25°C is 360 g per kg of water
⇒ 360 g of NaCl can be dissolved in 1 kg (1 L) of water
If more NaCl is added, it will remain undissolved

Mole fraction x: 

Molarity M (in mol.L^-1):

Molality m (in mol.kg^-1):

Electrolytes: substances that produce ions when dissolved in solution

Strong electrolytes ⇒ completely dissociate into ions (ex: NaCl)
Weak electrolytes ⇒ partially dissociate into ions (ex: HF)
Nonelectrolytes ⇒ do not dissociate into ions (ex: CCl₄)

Electrolyte solutions conduct electricity: mobile ions move and conduct an electric current

Ionic strength I (in mol.L^-1): measure of the electrical intensity of a solution containing ions

Mole-volume relationship:

Dilution Principle: decreasing the concentration (molarity) of a solute in a solution by adding more solvent

Solution 1 = concentrated solution; Solution 2 = diluted solution
The amount of solute does not change, only the amount of solvent:
n₁ = n₂ ⇒ M₁ x V₁ = M₂ x V₂

Diluting a solution 10 times means we want: M₂ = $\frac{M_1}{10}$

The amount of solute does not change:
n₁ = n₂ ⇒ M₁ x V₁ = M₂ x V₂ ⇒ M₁ x V₁ = $\frac{M_1}{10}$ x V₂ ⇒ V₂ = 10 V₁

To dilute a solution 10 times, we need to add 10 times the volume of solvent

When solutions of salts are mixed, a solid can form if ions of an insoluble salt are present. This is a precipitation reaction and the solid is called a precipitate. Molarity is usually used to calculate quantities in precipitation reactions.

Aqueous solution of Ni²⁺ and S^2- results in the formation of a precipitate: NiS
Ni²⁺ (aq) + S^2- (aq) → NiS (s)

Titration: lab technique used to determine the unknown concentration of an acid or base using a neutralization reaction

Chemical equation of neutralization reaction:
strong acid (pH < 7) + strong base (pH > 7) → salt + H₂O

Titration principle:

- Titrant: acid/base with a known concentration
- Analyte: acid/base solution being analyzed
- Indicators: substances which change color with pH

1) For an unknown concentration of strong acid, add strong base (with a known concentration)
2) Stop adding base exactly when all the acid had been neutralized: indicator should change color. This is called the endpoint of the titration
3) Determine the strong base volume added
4) At neutralization: moles of acid = moles of base
M_aV_a = M_bV_b with M = molarity (mol.L^-1) and V = volume (L)