SOLUTIONS

A solution is a homogeneous mixture of a solute dissolved in a solvent.

The solvent is generally in excess.

ExampleThe solution NaCl(aq) is sodium chloride NaCl(s) dissolved in water H2O(l)

The solute is NaCl(s) and the solvent is H2O(l)

Solute + solvent

Solutions: homogeneous mixtures

• Two components (at the least)-– Solute – the substance being dissolved– Solvent – the dissolving medium

• usually water – aqueous solution

can have multi-solute solutions - seawater

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Solubility experiment

Different solutions

• It’s possible to prepare solutions in all phases of matter.

Different solutions

However,

• The solutions in which the solvent is water are called “AQUEOUS SOLUTIONS.”

• Aqueous solutions are indicated as (aq) in reaction equations.

Ni(s) + HCl(aq) NiCl2(aq) + H2(g) NiCl2(s)dry

Aqueous solutions

Types of Solutions

• Are solutions made from only one solvent and one solute?

– By definition, there can only be one solvent– However, many solutes can be dissolved in a

solvent to create a solution• Air is an example of a solution with one “solvent” (nitrogen) and many “solutes” (oxygen, helium, argon, carbon dioxide, etc.)

Unsaturated solution- a solution that is able to dissolve more solute.

Saturated solution- a solution that cannot dissolve any more solute at the given conditions.

Supersaturated solution- a solution holding more dissolved solute than is specified by its solubility at a given temperature

Electrolyte and Non-electrolyte

• Electrolyte: a substance that conducts electricity when dissolved in water. – Acids, bases and soluble ionic

solutions are electrolytes.

• Non-electrolyte: a substance that does not conduct electricity when dissolved in water. – Molecular compounds and insoluble

ionic compounds are non-electrolytes.

Electrolytes

• Some solutes can dissociate into ions.

• Electric charge can be carried.

Types of solutes

Na+Cl-

Strong Electrolyte -100% dissociation,all ions in solution

high conductivity

Types of solutes

CH3COOH

CH3COO-H+

Weak Electrolyte -partial dissociation,molecules and ions in solution

slight conductivity

Types of solutes

sugar

Non-electrolyte -No dissociation,all molecules in solution

no conductivity

Strong ElectrolytesStrong acids: HNO3, H2SO4, HCl, HClO4

Strong bases: MOH (M = Na, K, Cs, Rb etc)

Salts: All salts dissolving in water are completely ionized.

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Concentration • the amount of solute dissolved in a

solvent at a given temperature

described as dilute if it has a low concentration of solute

described as concentrated if it has a high concentration of solute

described as supersaturated if contains more dissolved solute than normally possible

Units of Concentrations

amount of solute per amount of solvent or solution

Molarity (M) =

moles of solute(n)

volume in liters of solution (V)

moles = M x V(L)

Examples:

Example 1: What is the concentration when 5.2 moles of hydrosulfuric acid are dissolved in 500 mL of water?

Step one: Convert volume to liters, mass to moles. 500 mL = 0.5 L

Step two: Calculate concentration. C = 5.2 mol/0.5 L = 10mol/L

• Example 2: What is the volume when 9.0 moles are present in

5.6 mol/L hydrochloric acid?

• Example 3: How many moles are present in 450 mL of 1.5

mol/L calcium hydroxide?

• Example 4: What is the concentration of 5.6 g of magnesium

hydroxide dissolved in 550 mL?

• Example 5: What is the volume of a 0.100 mol/L solution that

contains 5.0 g of sodium chloride?

Answers2) 9.0/5.6=1.6071428= 1.6 L

3) 1.5x0.45= 0.675=0.68 mol

4) Mg(OH)2= 58.316 g/mol

5.6/58.316=0.0960285342 mol

0.0960285342/0.55 = 0.1745973349 M

0.17 M

5) NaCl=58.5 g/mol

5.0/58.5= 0.0854700855 mol

0.0854700855 mol/0.100= 0.8547008547L

0.85 L

HOW MUCH SOLUTE IS NEEDED FOR A SOLUTION OF A PARTICULAR

MOLARITY AND VOLUME?

• EXAMPLE

How much solute is required to make 300 mL of 0.8 M CaCl2?

ANSWER

(111.0 g) (0.8 mole) (0.3 L) = 26.64 g

mole L

CaCl2

CaCl2

CaCl2 solution

CaCl2

EXAMPLE: Prepare 1 L, 1M Na2SO4 solution.

PREPARING DILUTE SOLUTIONS FROM CONCENTRATED ONES

• Concentrated solution = stock solution• Use this equation to decide how much

stock solution you will need: M1V1=M2V2

M1 = concentration of stock solution

M2 = concentration you want your dilute solution to be

V1 = how much stock solution you will need

V2 = how much of the dilute solution you want to make

EXAMPLE

• How would you prepare 1000 mL of a 1 M solution of KCl from a 3 M stock of the solution of KCl ?– The concentrated solution is 3 M, and is M1.

– The volume of stock needed is unknown, ?, and is V1.

– The final concentration required is

1 M, and is M2.

– The final volume required is 1000 mL and is V2.

SUBSTITUTING INTO THE EQUATION:

M1 .V1 = M2 .V2

3 M (?) 1 M (1000 mL)

? = 333.33 mL

So, take 333.33 mL of the concentrated stock solution and BTV (Bring to Volume)1 L.

EXAMPLE

• How would you prepare 500 mL of a 1.2 M solution of NaNO3 from a 2 M stock of the solution of NaNO3 ?– The concentrated solution is 2 M, and is M1.

– The volume of stock needed is unknown, ?, and is V1.

– The final concentration required is

1.2 M, and is M2.

– The final volume required is 500 mL and is V2.

SUBSTITUTING INTO THE EQUATION:

M1 .V1 = M2 .V2

2 M (?) = 1.2 M (500 mL)

? = 300 mL

So, take 300 mL of the concentrated stock solution and BTV (Bring to Volume)500 mL.