Chapter 14 - Solutions Part II

Chapter 14 Solutions

2

Molarity (M)

Molarity (M) is

• _{a concentration term for solutions}

• _{the moles of solute in 1 L of solution}

A common unit for solution concentration due to convenience.

Example:

To prepare a 1.0 M KCl solution, 1.0 mol of KCl is dissolved in enough water to

make 1.0 L of solution.

mol solute L solution molarity =

Molarity

4

Molarity Conversion Factors

• What is the molarity of 0.500 L of a NaOH solution if it contains 6.00 g of NaOH?

grams of NaOH moles of NaOH molarity

• What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO₃?

6

Example of Using Molarity in Calculations

• _{How many grams of KCl are needed to prepare 0.125 L of a}

0.720 M KCl solution?

1 mol of KCl = 74.55 g of KCl

liters of KCl solution moles of KCl grams of KCl

How many milliliters of a 2.00 M HNO₃ solution contain 24.0 g of HNO₃?

g of solution  moles of HNO₃  mL of HNO₃

= 190. mL of HNO₃ solution

Example - Molarity

M = moles/L

•

_{What is the molarity of a sodium hydroxide}

Example - Molarity

•

_{Calculate the volume of a 2.50 M sugar}

Example - Molarity

•

_{What volume of 3.0 M NaOH can be prepared}

Example – Molarity

•

_{What mass of solute is dissolved in 25.0 mL of}

12

Dilution

In a dilution,

• _{water is added} • _{volume increases}

Comparing Initial and Diluted Solutions

In the initial and diluted solution,

• _{the moles of solute are the same}

• _{the concentrations and volumes are related}

by the equation M₁V₁ = M₂V₂

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Example of Dilution Calculations

What is the final molarity of the solution when 0.180 L of 0.600 M KOH is diluted to 0.540 L?

STEP 1 Prepare a table of the initial and diluted volumes

and concentrations.

Initial Solution Diluted Solution

M₁ = 0.600 M M₂ = ?

Learning Check

What is the final volume, in milliliters, if 15.0 mL of a 1.80 M KOH solution is diluted to give a

Example - Dilutions

•

_{A concentrated acetic acid is 20.0 M. What is}

Example - Dilution

•

_{What volume in mL of a 0.20 M strontium}

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Molarity in Chemical Reactions

In a chemical reaction,

• _{the volume and molarity of a solution are used to determine}

the moles of a reactant or product

volume (L) x molarity ( mol ) = moles 1 L

• _{if molarity (mol/L) and moles are given, the volume (L) can be}

determined

Example of Using Molarity in a Chemical

Equation

How many milliliters of a 3.00 M HCl solution are needed to react with 4.85 g of CaCO₃?

2HCl(aq) + CaCO₃(s) CaCl₂(aq) + CO₂(g) + H₂O(l)

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Learning Check

How many milliliters of a 0.150 M Na₂S solution are needed to react with 18.5 mL of a 0.225 M NiCl₂ solution?

Learning Check

How many liters of H₂ gas at STP are produced when 6.25 g of Zn react with 20.0 mL of a 1.50 M HCl solution?

Example - Solution Stoichiometry

Example - Solution Stoichiometry

•

Antacids containing CaCO

react with stomach

acid according to the reaction:

CaCO

+ 2 HCl



CaCl

+ CO

+ H

O

•

_{The normal volume acid in the stomach is}

between 20 – 100 mL.

Example - Solution Stoichiometry

•

Given that 25.0 mL of a 0.100 M MgBr

Example - Solution Stoichiometry

•

_{What volume in mL of a 0.75 M sodium}

carbonate solution is required to completely

react with 15.0 mL of 6.00 M hydrochloric

acid? Given:

Properties of Different Types of Mixtures

Moles of Particles

The number of moles of particles in water depends on the type of solute

• _{nonelectrolytes dissolve as molecules}

1 mol of nonelectrolyte = 1 mol of particles

• _{strong electrolytes dissolve as ions}

1 mol of a strong electrolyte = 2 to 4 mol of particles

Examples:

Nonelectrolyte: 1 mol of C₂H₆O₂(l) = 1 mol of C₂H₆O₂(aq) Strong electrolyte: 1 mol of KCl (s) =

1 mol of K+_{(aq) + 1 mol of Cl}_{(aq) =2 mol of particles(aq)}

Molality

The molality, abbreviation m, is a concentration unit that gives

• _{number of moles of solute particles} • _{1 kg (1000 g) of solvent}

Molality (m) = moles of solute particles kilograms of solution

Example of Calculating Molality

What is the molality of a solution containing 80.0 g of the nonelectrolyte glucose (C₆H₁₂O₆, molar mass

180.16) in 0.250 kg of water?

STEP 1 Given 80.0 g of glucose; 0.250 kg of water

Need molality (m) of glucose solution

STEP 2 Plan grams of glucose moles of glucose moles of glucose particles molality

Example of Calculating Molality (continued)

STEP 3 Equalities/Conversion Factors

1 mol of glucose = 180.16 g of glucose

1 mol glucose and 180.16 g glucose 180.16 g glucose 1 mol glucose

1 mol of glucose = 1 mol of particles

Example of Calculating Molality (continued)

STEP 4 Set Up Problem

Moles of glucose particles =

80.0 g glucose x 1 mol glucose x 1 mol particles 180.16 g glucose 1 mol glucose

= 0.444 mol of glucose particles

Molality (m) = 0.444 mol particles = 0.178 m

0.250 kg water

Learning Check

What is the molality of particles in a solution of 125 g of MgCl₂, a strong electrolyte, dissolved in 0.500 kg of

water?

Solution

STEP 1 Given 125 g of MgCl; 0.500 kg of water

Need molality (m) of particles in water

STEP 2 Plan grams of MgCl₂ moles of MgCl₂ moles of particles molality

STEP 3 Equalities/Conversion Factors

1 mol of MgCl₂ = 95.21 g of MgCl₂ 1 mol MgCl₂ and 95.21 g MgCl₂

95.21 g MgCl₂ 1 mol MgCl₂

Solution (continued)

STEP 3 Equalities/Conversion Factors (continued)

1 mol of MgCl₂(s)

= 1 mol of Mg2+_{(aq) + 2 mol of Cl}_(aq)

= 3 mol of particles(aq)

1 mol of MgCl₂ = 1 mol of particles

Solution (continued)

STEP 4 Set Up Problem

Moles of glucose particles =

125 g MgCl₂ x 1 mol MgCl₂ x 3 mol particles 95.21 g MgCl₂ 1 mol MgCl₂

= 3.94 mol of particles

Molality (m) = 0.394 mol particles = 7.88 m

0.500 kg water

Freezing Point and Boiling Point Change

When a solute is added to water,

• _{the number of particles produced in water (molality) changes}

the physical properties of water

• _{freezing point is lower than pure water}

For a 1 molal solution, the freezing point depression (T_f ) is 1.86 °C

• _{boiling point is higher than pure water}

Effect of Solute Particles

Osmosis

In osmosis,

• _{water (solvent) flows from the}

lower solute concentration into the higher solute

concentration

• _{the level of the solution with}

the higher solute concentration rises

• _{the concentrations of the two}

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Example of Osmosis

A semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch is a colloid and cannot

pass through the membrane, but water can. What happens?

semipermeable membrane

10% starch 4% starch

Example of Osmosis (continued)

• _{The 10% starch solution is diluted by the flow of water out of}

the 4% solution, and its volume increases.

• _{The 4% starch solution loses water, and its volume}

decreases.

• _{Eventually, the water flow between the two becomes equal}

_.

7% starch

H₂O

7% starch

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Osmotic Pressure

Osmotic pressure

• _{is produced by the solute particles dissolved in a solution} • _{is the pressure that prevents the flow of additional water}

into the more concentrated solution

• _{increases as the number of dissolved particles in the}