# Chemical Calculations for Solutions | General Chemistry 2

Chemical calculations for solutions are studied in this chapter: solubility and mole fraction, electrolytes, ionic strength, precipitation reactions, net ionic equation, molarity vs. molality, dilution, acid-base titrations

## Solutions and Solubility

Solution:

A homogeneous mixture consisting of a solvent (usually present in greater quantity) and one or more dissolved species called solute. A saturated solution is an homogeneous mixture that contains as much solute as possible ⇒ additional solute will remain undissolved

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

Solubility:

The maximum of solute that can be dissolved in a specified amount of 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:

The number of moles of a component divided by the total number of moles in a mixture. The mole fraction of a solute is calculated as follows:

xsolute = $\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{n}}_{\mathrm{solution}}}$ =

## Electrolytes

Electrolyte:

A substance that produces ions when dissolved in solution. Electrolytic solutions conduct electricity: mobile ions move and conduct an electric current

• Strong electrolytes are completely dissociated into ions (ex: NaCl)
• Weak electrolytes are partially dissociated into ions (ex: HF)
• Non-electrolytes dissolve to give non-conducting solution ⇒ they do not dissociate into ions (ex: CCl4)

Ionic strength I (in mol.L-1):

A measure of the electrical intensity of a solution containing ions

I = $\frac{1}{2}$

ci = concentration of ion i (in mol.L-1)
zi = charge of ion i

## Precipitation Reactions

Precipitation reaction:

A chemical reaction in which a precipitate is formed. A solid forms if ions of an insoluble salt are present

Ni2+ (aq) + S2- (aq) → NiS (s)
Aqueous solution of Ni2+ and S2- results in the formation of a precipitate: NiS

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)
Homogeneous mixture of AgNOand NaCl results in the formation of a precipitate: AgCl

Types of equations:

• Molecular equation: compounds are represented as if none of the reactants or products has dissociated
• Ionic equation: strong electrolytes are represented as ions
• Net ionic equation: an ionic equation from which spectator ions have been eliminated ⇒ only ions involved in the reaction are shown

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)   [molecular equation]
Ag+ (aq) + NO3- (aq) + Na+ (aq) + Cl- (aq) → AgCl (s) + Na+ (aq) + NO3- (aq)   [ionic equation]
Ag+ (aq) + Cl- (aq) → AgCl (s)   [net ionic equation]

How to determine the net ionic equation for a precipitation reaction:

1. Write and balance the molecular equation
2. Write the ionic equation by representing strong electrolytes into their constituent ions
3. Identify the ions that appear on both sides of the equation: these are the spectator ions
4. Write the net ionic equation removing spectator ions

## Molarity vs. Molality

Molarity M (in mol.L-1):

The number of moles of solute per liter of solution

M = $\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{V}}_{\mathrm{solution}}}$

nsolute = moles of solute (in mol)
Vsolution = volume of solution (in L)

Molality m (in mol.kg-1):

The number of moles of solute dissolved in 1 kg of solvent

m = $\frac{{\mathrm{n}}_{\mathrm{solute}}}{{\mathrm{m}}_{\mathrm{solution}}}$

nsolute = moles of solute (in mol)
msolution = masse of solution (in kg)

## Dilution of a Solution

Mole-volume relationship:

n = M x V

n = number of moles (in mol)
M = molarity (in mol.L-1)
V = volume (in L)

Dilution:

The process of preparing a less concentrated solution from a more concentrated one. The principle of a dilution is to decrease the concentration (molarity) of a solute in a solution by adding more solvent without changing the total number of moles of solute present in solution: moles of solute before dilution = moles of solute after dilution

Volume of solvent required to dilute a solution 10 times:

Solution 1: concentrated solution; solution 2: dilute solution

Diluting a solution 10 times means: M2 = $\frac{{M}_{1}}{10}$
The total number of moles of solute present in the solution does not change during dilution ⇒ moles of solute before dilution = moles of solute after dilution: n1 = n2
⇒ M1 x V1 = M2 x VM1 x V1 = $\frac{{M}_{1}}{10}$ x V2 ⇒ V2 = 10 V1

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

## Acid-Base Titrations

Titration:

A laboratory technique used to determine the unknown concentration of an acid or base using a neutralization reaction. A neutralization reaction is a reaction between an acid and a base resulting in the formation of water and a salt

Titration principle

A solution of known concentration (the titrant) is gradually added to a solution of unknown concentration (the analyte) in order to determe the unknown concentration. Indicators, substances which change color with pH, are used to identify the endpoint of the titration. At the endpoint of the titration, the number of moles of acid is equal to the number of moles of base

How to determine the concentration of an acid solution:

1. Add a strong base solution with known concentration to the acid solution
2. Stop adding the base exactly when all the acid has been neutralized: indicator should change color. This is called the endpoint of the titration
3. Determine the volume of base added
4. Determine the concentration of the acid solution
At the endpoint of the titration, moles of acid = moles of base
MaVa = MbVb with M = molarity (mol.L-1) and V = volume (L)