Download - 1 Some Definitions A solution is a _______________ mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT.

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Page 1: 1 Some Definitions A solution is a _______________ mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT.

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Some DefinitionsA solution is a

_______________ mixture of 2 or more substances in a single phase.

One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

Page 2: 1 Some Definitions A solution is a _______________ mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT.

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Parts of a Solution• SOLUTE – the

part of a solution that is being dissolved (usually the lesser amount)

• SOLVENT – the part of a solution that dissolves the solute (usually the greater amount)

• Solute + Solvent = Solution

Solute Solvent Example

solid solid

solid liquid

gas solid

liquid liquid

gas liquid

gas gas

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Definitions

Solutions can be classified as saturated or unsaturated.

A saturated solution contains the maximum quantity of solute that dissolves at that temperature.

An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature

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Definitions

SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved

Supersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways:

1. Warm the solvent so that it will dissolve more, then cool the solution

2. Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

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SupersaturatedSodium Acetate

• One application of a supersaturated solution is the sodium acetate “heat pack.”

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Homework

• Read section 8.5• Pages 392-386

• Page 395 #1-3• Page 397 #1-4,6

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IONIC COMPOUNDSCompounds in Aqueous Solution

Many reactions involve ionic compounds, especially reactions in water — aqueous solutions.

KMnO4 in water K+(aq) + MnO4-(aq)

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How do we know ions are present in aqueous solutions?

The solutions _________________________

They are called ELECTROLYTES

HCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

Aqueous Solutions

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Aqueous Solutions

Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes.Examples include:

sugarethanolethylene glycol

Examples include:sugarethanolethylene glycol

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Laboratory Concentration of

Solutions

Laboratory Concentration of

SolutionsThe amount of solute in a solution

is given by its concentration.

Molarity (C) = moles solute(n)litres of solution(V)

Units for molarity are mol/L also written as mol L-1 and M

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PROBLEM: Dissolve 5.00g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the molar concentration.

PROBLEM: Dissolve 5.00g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the molar concentration.

Step 1: Calculate moles of NiCl2•6H2O

5.00 g • 1 mol

237.7 g = 0.0210 mol

0.0210 mol0.250 L

= 0.0841 M

Step 2: Calculate Molarity

[NiCl2•6 H2O ] = 0.0841 M

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Try this problem

25.0 g of NaCl is dissolved in 5000.0 mL of water. Find the molar concentration of the resulting solution.

C = n/V

25.0 g NaCl

58.5 g/mol NaCl = 0.427 mol NaCl

C = n/V0.427 mol NaCl

5.000 L= 0.0854 M NaCl

Convert volume to L mlL divide by 1000

Convert mass of solute to moles

5000.0 mL = 5.000L

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Step 1: Change mL to L.250 mL = 0.250 LStep 2: Calculate moles using n = CV.Moles = (0.0500 mol/L) (0.250 L) = 0.0125 molesStep 3: Convert moles to mass.

(0.0125 mol)(90.00 g/mol) = 1.13 g

USING MOLARITYUSING MOLARITY

Moles(n) = C•V

What mass of oxalic acid, H2C2O4, is

required to make 250. mL of a 0.0500 Msolution?

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Learning Check

How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?

1) 12 g2) 48 g3) 300 g

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Using Molarity Cont.

What volume of solution would contain 42.6g of sucrose with concentration of 0.525M?

V = n/CConvert mass to moles

Moles = 42.6g / 342 g/mol = 0.1246 moles

V = n/CV= 0.1246 mol/0.525M = 0.237L

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Practice

Page 400 #1-4