Solutions and Molarity

Describe the types of solutions.Define the vocabulary words.List and explain the factors that influence solubility and the rate at which a solute dissolves in a solvent.Explain and calculate molarity.

Types of Solutions Gas in a gas – Gases mix freely and will always form a

solution unless they react (Example: air) Solid in a solid – Alloy: a homogeneous mixture of metals

(Example: brass is a mixture of copper and zinc) Liquid in a liquid Miscible – when liquids can be mixed together to form a

solution (Example: ethylene glycol and water form antifreeze)

Immiscible – when liquids cannot be mixed (Example: oil and water)

“Like dissolves like” – polar dissolves polar Polar and nonpolar (vinegar and oil) are immiscible

Solid in a liquid

Energy is required to separate particles of a solid (Endothermic)

Energy is released when solute particles and solvent particles are attached (Exothermic)

Goes back and forth: Dynamic equilibrium

Differences are responsible for different solubilities

Gases in liquids – when gas is attracted to solvent particles---release energy

Free---move toward entropy

Examples: CO2 in soda and O2 in seawater

Entropy (S) – disorder or randomness

Systems tend to go from a state of order (low entropy) to a state of maximum disorder (high entropy)

Solubility - quantity of solute that will dissolve in a specific amount of solvent at a certain temperature.(pressure must also be specified for gases).

Solubility Curve

-soluble vs. insoluble,-saturated (on or above) vs.unsaturated (below)

- solubility should NOT be confused with the rate at which a substance dissolves

Reading the curve:

At 30°C approximately 10g of KClO3 will dissolve in 100g of water

Refer to Solubility Curves

At 10 °C, 135 grams of KI will dissolve At 50 °C, 85 grams of KNO3 will dissolve

At 30 °C, 42 grams of NH4Cl will dissolve

At 75 °C, 50 grams of KCl will dissolve

Factors Influencing the Rate at which a Solute Dissolves in a

Solvent

1. Agitation – stirring; a surface phenomenon

2. Particle size 3. Temperature – the only factor that

affects both the rate of solution and the solubility

Solid in a liquid

Energy is required to separate particles of a solid (Endothermic)

Energy is released when solute particles and solvent particles are attached (Exothermic)

Goes back and forth: Dynamic equilibrium

Differences are responsible for different solubilities

Gases in liquids – when gas is attracted to solvent particles---release energy

Free---move toward entropy

Examples: CO2 in soda and O2 in seawater

Entropy (S) – disorder or randomness

Systems tend to go from a state of order (low entropy) to a state of maximum disorder (high entropy)

Temperature Effects on Solubility

In general, an increase in temperature increases the solubility of solids in liquids

In general, an increase in temperature decreases the solubility of gases in liquids (gases escape)

Pressure Effects Pressure increases the solubility of gases

in liquids (nail being hammered into wood) Henry’s Law – At a given temperature, the

solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid (Page 506)

=S1 S2

P1 P2

S = solubility

P = pressure

Concentration of Solutions Dilute – contains relatively little solute Concentrated – contains relatively large

amount of solute

Vocabulary Words

Solubility: the amount of a substance that dissolves in a given quantity of solvent at specified conditions of temperature and pressure to produce a saturated solution

Solute: dissolving particles

Solvent: the dissolving medium in a solution (Example- water)

Saturated – a solution that holds as much solid as it normally can at a given temperature (If more solid is added, it will not dissolve)

Unsaturated – a solution which has not yet reached the limit of solubility at a given temperature

Supersaturated – rare solution that contains MORE dissolved solute than it can normally hold at a given temperature

(Crystallization)

Suspension: if the particles are so large that they settle out unless the mixture is constantly stirred or agitated.

Colloid: particles that are intermediate in size between those in solutions and suspensions. AKA- colloidal suspensions

colloid

suspension

Molarity The number of moles of solute dissolved per

liter (1000mL) of solution Water: 1 mL = 1 g; 1000 mL = 1 Kg

Moles of solute = liters of solution x molarity Problem: What is the molarity of the solution

obtained by dissolving 90g of glucose (C6H12O6) in 1000 grams of water?

molarity (M) =mol of solute

liters/Kg of solution

Answer 1 mole of glucose = 180 g 90g ÷ 180 g = .5 mole 1000 g = 1000 mL .5M per 1000 mL water Problems: How many grams are needed

to make a molar solution of

a. 1M glucose

b. 2M glucose

c. .5M glucose

Answer

1 mole of glucose = 180 grams a. 1M = 180 grams b. 2M = 2 x 180 g = 360 grams c. .5M = .5 x 180 g = 90 grams

Preparing Molar Solutions The solute will take up some of the available

space in the volumetric flask. Steps 1. The solute should be added to some of the

solvent and dissolved. 2. Then solvent is added to the 1L mark on the

volumetric flask. If these steps are not followed, the total volume

of the mixture is likely to exceed the desired volume.

A volumetric pipet measures volumes even more accurately.

Making Dilutions

Dilution reduces the moles of solute per unit volume, however, the total moles of solute in solution does not change.

Moles of solute = molarity (M) x liters of solution (V)

Moles of solute = M1 x V1 = M2 x V2

Problem: How many milliliters of a stock solution of 2.00M MgSO4 would you need to prepare 100.0 mL of 0.400 M MgSO4?

Answer

0.400M x 100.0 mL ÷ 2.00M = 20.0 mL Thus, 20.0 mL of the initial solution must

be diluted by adding enough water to raise the volume to 100.0 mL

OR

0.400M is 1/5th of 2.00M

1/5th of 100mL is 20 mL

Percent Solutions

If both the solute and solvent are liquids, a convenient way to make a solution is to measure volumes.

Example: 20 mL of pure alcohol is diluted with water to a total volume of 100 mL – The concentration of alcohol is 20% (v/v)

A commonly used relationship for solutions of solids dissolved in liquids is percent (mass/volume).

Problems

1. What is the percent by volume of ethanol (C2H6O), or ethyl alcohol, in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water?

2. How many grams of glucose (C6H12O6) would you need to prepare

2.0 L of 2.8% glucose (m/v) solution?

Answers

1. 85/250 x 100% = 34% ethanol (v/v) 2. In a 2.8% solution, each 100 mL of

solution contains 2.8 grams of glucose

100mL is 1/10th of a liter, so you need 28 grams per liter

2 L x 28 g = 56 grams of glucose