Solutions (Chapter 12)

Solutions (Chapter 12)

Taylor, Tommy, Kayla

Homogeneous mixture - mixture of two or more substances that has a uniform appearance throughout ex: salt water

Heterogeneous mixture - can be separated physically and the different components are visibly distinguishable from one another

ex: chocolate chip cookie

http://merryann12.hubpages.com/hub/Heterogeneous-and-Homogeneous-Mixtures

Solute - substance being dissolved ex: Kool-Aid mix

Solvent - dissolves another substance ex: water

Solution - a homogeneous mixture of two or more substances.ex: Kool-Aid

https://www.google.com/search?q=kool+aid&oe=utf-8&aq=t&

http://www.chem4kids.com/files/matter_solution.html

Visual

Salt/ Solute

Water/ Solvent

Salt Water/ Solution

google.com

Suspension - a mixture in which large particles can be evenly distributed and the components will settle out

ex: muddy water

Colloid - a mixture in which small particles can be mixed so they remain evenly distributed without settling out ex: milk

http://www.800mainstreet.com/9/0009-001-mix-solut.html

Dilute - a solution containing a relatively small quantity of solute as compared with the amount of solvent ex: bleach

Concentrated - a solution that contains a large amount of solute relative to the amount that could dissolve ex: alcohol

http://www2.ucdsb.on.ca/tiss/stretton/chem2/acid02.html

Unsaturated - the solute concentration is lower than its equilibrium solubility

Saturated - a point of maximum concentration in which no more solute can be dissolved

Supersaturated - increase of concentration beyond saturation

http://shikagami21.blogspot.com/2007/08/saturated-unsaturated-and.html

Solubility CurvesSolute below the line indicates the solutionis unsaturated at a certain temperature

Solute above the line shows all of the solute has dissolved and is supersaturated

If the amount of solute is on line then the solution is saturated

Ex: At 500C and 100g, NaNo3 is saturated

At 200C and 100g, NaNo3

is supersaturated

At 200C and 50g, NaNo3 is unsaturated

http://gcserevision101.files.wordpress.com/2009/02/solubility-curves.jpg

Stirring A SolutionStirring a solution helps disperse the solute particles and bring fresh solvent into contact with the solute surface. This increases the contact between the solvent and solute.

google.com

Surface Area

• The dissolution process occurs at the surface of the solute, so the larger the surface area of the solute the faster it dissolves.

google.com

Temperature of a Solution

• As the temp. of a solvent increases the solvent molecules move faster. This causes collisions between the solvent and solute molecules. This helps separate solute molecules and disperse them among the solvent molecules.

google.com

Concentration

When you increase the concentration of the solute, the rate of solution formation will also increase. They are directly related.

MolarityThe concentration of a solution

expressed as the number of moles of solute dissolved in each liter of solution

moles of solute ormoles

Molarity= Liters of solutionL

Molarity CalculationsA saline (salt water) solution contains

0.70 g of NaCl per 100 ml of solution. What is its molarity?

First calculate the number of moles NaCl in the solution. You must use dimensional analysis to convert from grams to moles.

0.70 g NaCl x 1 mole NaCl = 0.01 mole

58.44 g NaCl

Molarity CalculationsWhere do the numbers come from?

0.70 g NaCl x 1 mole NaCl = 0.01 mol

58.44 g NaCl

Grams of solute given in problem

Formula mass of NaCl from periodic table

Formula mass correlates with grams in one mole

Molarity CalculationsNow that you have obtained the

number of moles solute, use this number in the molarity equation:

M= 0.01 moles NaCl0.1L

To find this value, convert from ml (given in the problem) to liters.

M= 0.1

Molarity Practice1. How many grams of CaCl2 are

needed to make 125 ml of a 1.5 M solution of CaCl2?

2. How many moles of solute are present in 750 ml of a 0.10 M Na2SO4 solution?

Molarity Answers1. How many grams of CaCl2 are

needed to make 125 ml of a 1.5 M solution of CaCl2?

1.5M = moles of solute.125 L

.188 mol x 110.98 g CaCl2 = 20.9 g 1 mole CaCl2

Molarity Answers2. How many moles of solute are

present in 750 ml of a 0.10 M Na2SO4 solution?

.10 M = moles of solute = .075 moles .750 L Na2SO4

MolalityThe number of moles of solute

dissolved in each kilogram of solvent

moles of soluteor moles

Molality= Kilograms of solvent Kg

Molality is represented by "m "

Example Molality Problem:What is the molality of a solution if 100

g of NaCl is dissolved into 250 g of water?

First calculate the number of moles NaCl in the solution. You must use dimensional analysis to convert from grams to moles.

100g NaCl x 1 mole NaCl = 1.71 moles NaCl

58.44g NaCl

Molality CalculationsWhere do the numbers come from?

100g NaCl x 1 mole NaCl = 1.71 mol NaCl

58.44g NaCl

Grams of solute given in problem

Formula mass of NaCl from the periodic table

Formula mass correlates with grams in one mole

Molality CalculationsNow that you have obtained the

number of moles solute, use this number in the molality equation:

m = 1.71 moles NaCl.25 Kg

m = 6.84

To find this value, convert from g (given in the problem) to Kg.

Molality Practice1. A solution was prepared by

dissolving 8.55 g of sucrose (table sugar, C12H22011) in 62.5 g of water. Find the molality of this solution.

2. A solution of iodine, I2, in carbon tetrachloride,CCl4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a 0.240 m solution of iodine in CCl4 if 50.0 g of CCl4 is used?

Molality Answers1. A solution was prepared by

dissolving 8.55 g of sucrose (table sugar, C12H22011) in 62.5 g of water. Find the molality of this solution.

8.55g C12H22011 x 1 mol C12H22011 = 0.0250 mol C12H22011

342.34 g C12H22011

m = 0.0250 mol C12H22011 = 0.400 m C12H220110.0625 kg H2O

Molality Answers2. A solution of iodine, I2, in carbon

tetrachloride,CCl4, is used when iodine is needed for certain chemical tests. How much iodine must be added to prepare a 0.240 m solution of iodine in CCl4 if 50.0 g of CCl4 is used?

50.0 g CCl4 x 1 kg = 0.05 kg CCl4 1000 g CCl4

0.240 m = X X= 0.012 moles I2

0.05 kg CCl40.012 moles I2 x 253.8 g I2 = 3.05 g I2

1 mole I2

DilutionThe concentration of a solution can be

decreased by adding water.

M1 x V1 = M2 x V2 Original molarity

Original volume

Molarity after dilution

Volume after dilution

Volume should always be expressed in liters

Dilution Equation:

Dilution CalculationsHow much concentrated 6.0 M

hydrochloric acid is needed to prepare 50.0 ml of a 1.0 M solution?

First use the information given in the question in order to substitute into the dilution equation.

6.0 M x V1 = 1.0 M x 0.05 L

Dilution CalculationsNow, solve for the missing value

algebraically.

6.0 M x V1 = 1.0 M x 0.05 L6.0V1 = 0.05

6.0 6.0 V1 = 0.0083

L

Dilution Practice1. To what volume should 8.3 ml of 5.0

M nitric acid be diluted to prepare a 1.0 M solution?

2. To how much water should 12.5 ml of 3.0 M hydrochloric acid be added to produce a 1.0 M solution?

Dilution Answers1. To what volume should 8.3 ml of 5.0

M nitric acid be diluted to prepare a 1.0 M solution?

5.0 M x 0.0083 L = 1.0 M x V2 0.0415 = 1.0V2

1.0 1.0V2= 0.042

L

Dilution Answers2. To how much water should 12.5 ml

of 3.0 M hydrochloric acid be added to produce a 1.0 M solution?

3.0 M x 0.0125 L = 1.0 M x V20.0375 = 1.0V2

1.0 1.0

V2 = 0.038 L

0.038 L - 0.0125 L = 0.026 L of H2O

Further Explanation2. To how much water should 12.5 ml

of 3.0 M hydrochloric acid be added to produce a 1.0 M solution?

3.0 M x 0.0125 L = 1.0 M x V2

Molarity of solute given in problem

Volume of solute given in problem

Final solute molarity given in problem

Unknown

2. To how much water should 12.5 ml of 3.0 M hydrochloric acid be added to produce a 1.0 M solution?

3.0 M x 0.0125 L = 1.0 M x V20.0375 = 1.0V2

1.0 1.0

V2 = 0.038 L

Further Explanation

Solve algebraically

3.0 M x 0.0125 L = 1.0 M x V20.0375 = 1.0V2

1.0 1.0

V2 = 0.038 L

0.038 L - 0.0125 L = 0.026 L of H2O

Further Explanation

Subtract the final volume and the original volume