Chapter 13 Solutions Some Definitions A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase. One part is usually regarded as.

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Presentation transcript:

Chapter 13 Solutions

Some Definitions A solution is a HOMOGENEOUS mixture of 2 or more substances in a single phase. One part is usually regarded as the SOLVENT and the others as SOLUTES.

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 liquid gassolid liquid gasliquid gas

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

Definitions SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved. SUPERSATURATED SOLUTIONS are unstable. SUPERSATURATED SOLUTIONS are only temporary, and usually accomplished in one of two ways:

HOW TO MAKE SUPERSATURATED SOLUTIONS 1.Warm the solvent so that it will dissolve more, then cool the solution – undisturbed 2.Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.”One application of a supersaturated solution is the sodium acetate “heat pack.”

More Definitions SUSPENSIONS – The particles are so large that they settle out unless the mixture is constantly stirred.SUSPENSIONS – The particles are so large that they settle out unless the mixture is constantly stirred. COLLOID – Contains particles that are intermediate in size between those of a solution and a suspensionCOLLOID – Contains particles that are intermediate in size between those of a solution and a suspension Flour suspended in water

Factors Affecting Dissolving Rate Increasing the surface area –Increasing the surface area – Making the solute into smaller pieces gives more contact between the solute and the solvent.

Factors Affecting Dissolving Rate Agitating the solution – Stirring or shaking increases contact between solute and solvent

Factors Affecting Dissolving Rate Heating the solution – Increasing the temperature increases the KE making the particles of the solute move faster and increases the collisions between the solute and the solvent.

Concentration of Solute Molarity – The number of moles of solute dissolved in 1 liter of solution. Molarity (M) = moles solute liters of solution

1.0 L of water was used to make 1.0 L of solution. Notice the water left over.

PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl 2 6H 2 O Step 2: Calculate Molarity NiCl 2 6 H 2 O [NiCl 2 6 H 2 O ] = M

Learning Check How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? a.12 grams b.48 grams c.300 grams

Two Other Concentration Units grams solute grams solution MOLALITY (m) % by mass = % by mass m of solution= mol solute kilograms solvent 100

Calculating Concentrations Calculate molality Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate m & % of ethylene glycol (by mass). Calculate weight %

Learning Check A solution contains 15 g Na 2 CO 3 and 235 g of H 2 O? What is the % by mass of the solution? 1) 15% Na 2 CO 3 2) 6.4% Na 2 CO 3 3) 6.0% Na 2 CO 3

Preparing Solutions Weigh out a solid solute and dissolve in a given quantity of solvent.Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is less concentrated.Dilute a concentrated solution to give one that is less concentrated.

Diluting a Concentrated Solution V 1 M 1 =V 2 M 2 V = Volume M = Molarity

Calculations How many mL of a 6.00 M solution are needed to make 500. mL of 2.50 M solution? M 1 V 1 = M 2 V 2 V 1 = M 2 V 2 M 1 V 1 = 2.50M x 500. mL 6.00 M V 1 = 208 mL

THE END