Solutions Chapter 13

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Solutions Chapter 13

Solutions Chapter 13

Tro, 2nd ed.

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SOLUTIONS & PHYSICAL PROPERTIES

A solution is a system in which one or more substances are homogeneously mixed or dissolved in another substance.

The solute is the component that is dissolved or is the least abundant component of the solution.

The solvent is the dissolving agent or the most abundant component in the solution.

If the solution has only one phase, it is homogeneous.

(A mixture which has two phases is heterogeneous.)

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Similar to table 13.1

Metals in metals = alloys

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Properties of True Solutions

The solute remains uniformly distributed throughout the solution and will not settle out with time.

The solute can generally be separated from the solvent by purely physical means such as evaporation, usually of the solvent.

The solute particles of a true solution are molecular or ionic in size.

The solution can be either colored or colorless but is always clear or transparent.

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solvent

solute

solution

Formation of a Solution

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Solvated Ions

When materials dissolve, the solvent molecules surround the solvent particles due to the solvent’s attractions for the solute. The process is called solvation. Solvated ions are effectively isolated from each other. (Also called hydrated ions if the solvent is water.)

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TERMS THAT DESCRIBE THE SOLUBILITY OF LIQUIDS:

Solubility (limit) is the maximum amount of solute that can be dissolved in a given amount of solvent at a given fixed T, at equilibrium.

Saturated solutions contain the maximum amount of solute, so any additional solute appears as a precipitate or a gas, or a separate liquid phase.

Unsaturated means that more solute can be added to the solution.

Supersaturated is a temporary condition where more solute has dissolved, but add just 1 crystal to this, many crystals will precipitate from solution.

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Supersaturated Solution

A supersaturated solution has more dissolved solute thanthe solvent can hold. When disturbed, all the solute abovethe saturation level comes out of solution.

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TERMS THAT DESCRIBE THE SOLUBILITY OF LIQUIDS

Miscible: liquids that are capable of mixing and forming solutions.

methanol and water

Immiscible: liquids that are insoluble in each other.

oil and water

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Solubilities of substances vary widely.

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See back of blue Periodic Table for rules and table.

14.2

Solubility of Various Common Ions in Cold Water

Similar to Table 7.2

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ElectrolytesElectrolytes: substances whose

aqueous solution is a conductor of electricity

Strong electrolytes: all the electrolyte units are dissociated into ions (salts, strong bases, strong acids)

Nonelectrolytes: none of the units are dissociated into ions (molecular solutes)

Weak electrolytes: a small percentage of the units are dissociated into ions (weak bases, weak acids)

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1. Natural inclination of the universe towards disorder so substances do mix

2. Strength of Force of Attraction between solute particles, between solvent particles and between solute & solvent (nature of solute and solvent)

3. Temperature4. Pressure

NOTE: Gases are always completely miscible!

Factors affecting solubility:

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Factors affecting solubility:Nature/Intermolecular Forces

Liquids:Similar liquid molecules will dissolve in each other - heptane in octane - both have only London forces involved at about the same strength, both nonpolar.Different IP forces - octane and water. H-bonding very strong in water, very different from London forces in octane; water would have to give up H-bonding for weaker force. THIS WON'T HAPPEN. The less dense liquid will rise and stay in separate phase on top of water.

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Polar compounds tend to be more soluble in polar solvents than nonpolar solvents. NaCl (sodium chloride) is an ionic compound which is:

SolventPolarity

•soluble in water•slightly soluble in ethanol•insoluble in ether and benzene

General rule is “like dissolves like”

Factors affecting solubility:Nature/Intermolecular Forces

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Dissolution of sodium chloride in water.

Polar water molecules are attracted to Na+ and Cl- ions in the salt or crystal, weakening the attraction between the ions.

As the attraction between the ions weakens, the ions move apart and become surrounded by water dipoles.

The hydrated ions slowly diffuse away from the crystal to become dissolved in solution.

14.3

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Nonpolar compounds tend to be more soluble in nonpolar solvents than in polar solvents. Benzene is a nonpolar organic compound which is:

SolventPolarity

•insoluble in water•soluble in ether

Factors affecting solubility:Nature/Intermolecular Forces

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Classifying Solvents

SolventSolvent ClassClassStructuraStructura

ll

FeatureFeature

Water, HWater, H22OO polarpolar O-HO-H

Ethyl Alcohol, CEthyl Alcohol, C22HH55OHOH polarpolar O-HO-H

Acetone, CAcetone, C33HH66OO polarpolar C=OC=O

Benzene, CBenzene, C66HH66 nonpolnonpolarar

C-C & C-HC-C & C-H

Hexane, CHexane, C66HH1414 nonpolnonpolarar

C-C & C-HC-C & C-H

Diethyl Ether, CDiethyl Ether, C44HH1010OO nonpolnonpolarar

C-C, C-H C-C, C-H &&

C-OC-O

Similar to Table 13.2

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Practice with solutes and solvents

1. Which solvent will ethanol dissolve in more readily, water or octane? Why?

2. What about liquid C6H13OH? Water or octane?

3. What about solid glucose? (Draw it.) Water or octane?

4. What about solid I2? Will it dissolve more readily in water or octane?

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Practice with solutes and solvents

1. Water, because of the polarity of both with their –OH groups.

2. Octane, because most of the molecule is nonpolar like octane is.

3. Solids must also be "like" the solvent. Glucose dissolves in water because of extensive H-bonding w/ its many -OH groups. (See Lewis structure)

4. Solid I2 - held by nonpolar London forces; octane also London forces. I2 dissolves in octane better than water..

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Effect of Temperature on Solubility

For most solids dissolved in a liquid, an increase in temperature results in increased solubility.

The solubility of a gas in water usually decreases with temperature.

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decrease in solubility with increasing temperature

slight increase in solubility with temperature

large increase in solubility with temperature

FIGURE 13.4

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Effect of Pressure on Solubility

Small pressure changes have little effect on the solubility of solids in liquids or liquids in liquids

Small pressure changes have a great effect on the solubility of gases in liquids- The solubility of a gas in a liquid is directly proportional to the pressure of that gas above the liquid

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Rate of Dissolving Solids: 4 factors

1. Particle SizeA solid can dissolve only at the surface that is in contact with the solvent.Smaller crystals have a larger surface to volume ratio than large crystals.Smaller crystals dissolve faster than larger crystals.

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Rate of Dissolving Solids: 4 factors

2. TemperatureIn most cases, the rate of dissolving of

a solid increases with temperature.This occurs because solvent molecules

strike the surface of the solid more frequently and harder, causing the solid to dissolve more rapidly.

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3. Concentration of the Solution

As solution concentration

increases, the rate of dissolving decreases.

Δc

Δt

The rate of dissolving is at a maximum when solute and solvent are

first mixed.Δt

Δc

Rate of Dissolving Solids: 4 factors

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Rate of Dissolving Solids: 4 factors

4. Agitation or stirring.When a solid is first put into water, it comes in contact only with water. The rate of dissolving is a maximum.As the solid dissolves, the amount of dissolved solute around the solid increases and the rate of dissolving decreases.Stirring distributes the dissolved solute throughout the water; more water is in contact with the solid causing it to dissolve more rapidly.

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Concentration of Solutions

The concentration of a solution expresses the amount of solute dissolved in a given quantity of solvent or solution.

The terms dilute and concentrated are qualitative expressions of the amount of solute present in a solute.

You will need to learn the mathematical representations of concentration: mass-percent, mass/volume percent, Molarity, etc.

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Mass Percent Solution

Mass percent expresses the concentration of solution as the percent of solute in a given mass of solution.

%-mass = (g solute/(g solute + g solvent)) * 100

= (g solute/g solution) * 100

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Example of Mass Percent

How do you prepare 425 g of a 2.40%-wt solution of sodium acetate?

Solve for grams of solute, then grams of water:(X g/425 g) x 100 = 2.40% x = 10.2 g NaC2H3O2

425 – 10.2 = 414.8 g H2OYou try: What is the mass-percent of a solution

that has 13.6 g of NaCl in 250.0 g of solution?

What is the mass-percent of a solution that has 15.0 g of ethanol in 35.0 g of water?

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Mass/Volume Percent (m/v)

Mass /volume percent expresses the concentration as grams of solute of solute per 100 ml solution.

%-mass/vol = g solute/mL solution * 100

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A 3.00 %-mass/vol H2O2 solution is commonly used as a topical antiseptic to prevent infection. What volume of this solution will contain 10.0 g of H2O2?

Solve the mass/volume equation for grams of solute.

%-mass/vol = (g solute/mL solution) * 100

mL solution = (g solute/%-mass/vol) * 100

= (10.0 g/3.00%) * 100 = 333 mL

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Volume Percent

Solutions that are formulated from liquids are often expressed as volume percent with respect to the solute.

The volume percent is the volume of a liquid in 100 ml of solution.

%-vol = (solute vol/solution vol) * 100

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Volumes are not necessarily additive.

A bottle of rubbing alcohol reads 70% by volume.

The alcohol solution could be prepared by mixing 70 mL of alcohol with water to make a total volume of 100 mL of solution.

30 m L of water could not be added to 70 mL of alcohol because the volumes are not necessarily additive.

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Practice

Find both the mass-percent and volume percent of a solution that has 10.0 g of ethanol (D = 0.7893 g/mL) and 90.0 g of water (D = 0.9987 g/mL). Assume volumes are additive.

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Molarity

Molarity of a solution is the number of moles of solute per liter of solution.

Molarity = M = moles solute/liter of solution

= mol/LSometimes n represents moles, then

M = n/V.

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Molarity Calculations

How many moles of NaOH are present in 25.0mL of 0.555 M NaOH?

M = mol/vol, therefore mol = M * V (must be

in Liters)0.555 mol/L * (25.0mL * 1L/103mL)

= 0.0138 mol NaOH

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Molarity Calculations

What is the Molarity of a solution that has 13.6 g of NaCl in 250.0 mL of solution?

What is the volume of 1.25 M HCl solution that will provide 0.4414 moles of HCl?

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Solubility g/100g grams of solute/100 grams of solventNormality N molarity * (# of H+ or OH-)

Yes, you have to know all of these!

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NORMALITY: EXAMPLES

1.0 M HCl * 1 H+ = 1.0 N1.0 M H2SO4 * 2 H+ = 2.0 N

1.0 M NaOH * 1 OH- = 1.0 N1.0 M Ba(OH)2 * 2 OH- =

2.0 N

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Dilution Problems

If a solution is diluted by adding pure solvent:

- the volume of the solution increases.

- the number of moles of solute remain the same.

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Dilution Problems

When the moles of a solute in a solution before and after dilution are the same, the moles before and after dilution may be set equal to each other.

Remember M = n/V, or n = M * V. If V changes, M will change to keep moles, n, constant.

Dilution Equation: M1V1 = M2V2

MEMORIZE THIS!!!

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Practice

How would you make 500.0 mL of a 0.500 M solution of NaOH from 6.00 M NaOH?

0.500 M * 500.0 mL = 6.00 M * V2

V2 = 41.7 mL

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Practice Making Solutions

How would you make 250.0ml of a 0.555 M standard NaOH solution?

Use two methods: (1) Weigh out dry chemical NaOH solid and dissolve it.

(2) Dilute a more concentrated solution of 6.00 M NaOH.

A. Weighing a solid and using M = n/V. Find n first, then grams of solid.

250.0mL(1L/103mL)(0.555 mol/L) = 0.13875 mol NaOH

0.13875 mol * 39.997 g/mol = 5.55 g NaOH Get a vol. flask and add water about halfway up. Add

solid. Mix til dissolved, then dilute to mark on flask.

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Practice Making Solutions

B. Dilution: Given 6.00 M NaOH. Use the Dilution Equation, M1V1=M2V2

6.00 mol/L * V1 = 0.555 mol/L * 0.2500 L

V1 = O.O231 L or 23.1 mL

Obtain 23.1 mL in a graduated cylinder. Get a vol. flask and add water about halfway up. Add the correct amount of 6.00 M NaOH. Mix until dissolved, then dilute to mark on flask.

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Preparation of a 1 molar solution

14.7

Start with some water in the flask!

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Molarity, Dissociation andSolution Inventories

When strong electrolytes dissolve, all the solute particles dissociate into ions.

By knowing the formula of the compound and the molarity of the solution, it is easy to determine the molarity of the dissociated ions simply multiply the salt concentration by the number of ions.

It’s like doing stoichiometry!

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Find the solution inventory for each ofthese strong electrolytes

1. 0.25 M MgBr2

2. 0.33 M Na2CO3

3. 0.0750 M Fe2(SO4)3

1. MgBr2(aq) → Mg2+(aq) + 2 Br-

(aq)

I: 0.25 M 0 0R: -0.25 M +0.25 +0.50F: 0 0.25 M0.50 M So we say: [Mg2+] = 0.25 M, [Br-]

= 0.50 M

2. Na2CO3(aq) → 2 Na+(aq) + CO3

2-(aq)

I: 0.33 M 0 0R: -0.33 +0.66 +0.33F: 0 0.66 M 0.33 M[Na+] = 0.66 M,

[CO32-] = 0.33 M

3. [Fe3+] = 0.150 M, [SO42-] = 0.225 M

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STOICHIOMETRY WITH AQUEOUS SOLUTIONS:

A. GAS-FORMING: use regular stoichiometry and concentration informationB. PRECIPITATION: use solubility rules (see examples on following pages)C. NEUTRALIZATION: use concentration information and stoichiometry steps. (Do NOT use dilution equation! Use stoichiometry.)

acid + base salt + waterTAKE GOOD NOTES – THE TEXT ASSUMES

YOU KNOW HOW TO DO THIS!

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STOICHIOMETRY OF SOLUTION PRECIPITATION REACTIONS:

How many grams of silver will precipitate if excess copper is added to 500.0 ml OF 0.100 M AgNO3?

1: Balanced chemical equationCu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)

What you know & what you want to know500.0 ml OF 0.100 M AgNO3. Want grams Ag.

2: 500.0 ml * 1 L * 0.100 moles = 0.0500 mol AgNO3

103 ml 1.00 L3: 0.0500 mol AgNO3 * 2 mol Ag/2 mol AgNO3 =

0.0500 mol Ag4: 0.0500 mol Ag * 107.9 g/mol = 5.40 g Ag

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Group Practice: Stoichiometry/PrecipitationWhat was the concentration in Molarity of

a silver nitrate solution, if 25.00 mL of it required 23.31 mL of 0.3161 M NaCl solution to completely precipitate all the silver as silver chloride?NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)

23.31mL 25.00mL.3161 M ??M

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C. NEUTRALIZATION

Acid-base reactions are analyzed by titration. This is volumetric analysis using a standardized solution to find mass or concentration of another substance.

Burettes - contain the titrant, usually standardized

Flask - contains analyte, the substance we want to analyze

Endpoint - when the indicator changes color to say reaction is done

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C. NEUTRALIZATIONStandardized solution: exact concentration of a reagent

known. Two methods to standardize – (1) use a pure dry solid or (2)

use a purchased aqueous standard.Method 1: Use 0.250 grams of sodium carbonate to

determine exact concentration of hydrochloric acid. (NOTE: this is also a gas-producing reaction.) It took 25.76 mL of HCl to reach "endpoint" - where the chemical indicator changed color indicating a pH change.Na2CO3(s) + 2 HCl(aq) 2 NaCl(aq) + H2O(l) + CO3(g)

Moles solid base = 0.250 g/(106.0 g/mol) = 0.0023585 molMoles acid = mol base (2 HCl/1 Na2CO3) = 0.004717 molMolarity of acid = mol/vol = 0.004717 mol/0.02576 L =

0.183 M

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C. NEUTRALIZATION

Method 2: Use a purchased standardized solution of 0.100 M HCl to standardize a NaOH solution. Exactly 25.00 mL of base took 32.56 mL of acid.HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)

Moles acid = 0.100mol/L x 0.03256 L = 0.003256 mol

Moles base = mol acid (1/1) = 0.003256 mol baseMolarity of base = mol/vol =

0.003256mol/0.02500L = 0.0130 M

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STEPS TO PERFORMING CALCS FOR ACID-BASE REACTIONS:

1. List species as reactants, decide what products will form, and balance the equation.

2. Calculate moles of reactants. (Use molarity and volume.)

3. Calculate moles of desired product or other reactant using mole/mole ratio.

4. Convert to volume or molarity or grams, as required.

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Example of steps

A 1.034 g sample of clover has oxalic acid extracted from it into a small volume of water. The endpoint of titration is 34.47 mL of 0.100 M NaOH. What is the mass-percent of oxalic acid in clover?

1.) H2C2O4(aq) + 2 NaOH(aq) 2 H20(l) + Na2C2O4(aq)

2 & 3.)0.03447 L *(0.100mol NaOH/L)(1 mol acid/2 mol

base) = 0.0017235 mol acid

4.) 0.0017235 mol (90.04g/mol) = 0.15518 g(0.15518 g/1.034g)*100 = 15.01 %-mass oxalic acid

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Colligative Properties of Solutions

When a nonvolatile solute is added to a solvent, four physical properties of the solvent, called Colligative Properties, will change:Freezing Point Depression

Boiling Point Elevation

Vapor Pressure Lowering

Osmotic Pressure

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Colligative Properties of Solutions

Colligative properties of a solution depend only on the number of solute particles in a solution and not on the nature of those particles.

You already know something about these! Consider?Why does salt water boil at a higher temp? Why do we throw salt on snow and ice on roads?

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Each solvent shows a characteristic:

freezing point depression constant

boiling point elevation constant

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A liquid boils when its vapor pressure equals atmospheric pressure.A solution will have a lower vapor pressure. and consequently a higher boiling point.

Vapor Pressure Curve of Pure Water and Water Solution: Boiling Point Elevation

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Vapor Pressure Curve of Pure Water and Water Solution: Freezing Point Depression

A liquid freezes when its vapor pressure equals the vapor pressure of its solid.

water vapor pressure curve

ice vapor pressure curve

A solution will have a lower vapor pressure. and consequently a lower freezing point.

solution vapor pressure curve

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Molality

The freezing point depression and the boiling point elevation are directly proportional to the number of moles of solute per kilogram of solvent, in a concentration unit called molality.

molality = moles solute/kg of solvent = molal

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What is the molality (molal) of a solution prepared by dissolving 2.70 g

CH3OH in 25.0 g H2O?

The concentration unit of molality is moles solute/kg solvent, so convert mass of methanol to moles and mass of water to kg:

2.70 g/32.042g/mol = 0.084264 mol CH3OH

25.0 g/1000g/kg = 0.0250 kg of watermolality = 0.084264 mol/0.0250 kg

= 3.37 molal

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Symbols and equations used in the calculation of colligative properties for FP and BP

m = molality (molal)ΔTf = freezing point depression in oC

ΔTb = boiling point elevation in oC

Kf = freezing point depression constant

Kb = boiling point elevation constant

ΔTf = Kf * molal new Tf = solvent Tf – ΔTf

ΔTb = Kb * molal new Tb = solvent Tb + ΔTb

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A solution is made by dissolving 100. 0 g of ethylene glycol (CH2OHCH2OH) in 200. g of water. What is the freezing point of the solution?The calculation takes two steps:1. Determine the freezing point depression.2. Subtract the depression value from the

freezing point of the pure solvent.Find the molality and the ΔTf:

100.0 g/62.068 g/mol = 1.611 mol EGmolality = 1.611 mol/0.2000 kg = 8.056 molal

ΔTf = Kf * molal new Tf = solvent Tf – ΔTf

ΔTf = 1.86oC/molal * 8.056 molal = 15.0oCnew Tf = solvent Tf – ΔTf = 0.0oC – 15.0oC = -

15.0oC

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Practice

Calculate both the new freezing point and the new boiling point for both 1.0 molal and 2.0 molal glucose solutions.

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Osmosis and Osmotic Pressure

Osmosis is the diffusion of water, either from a dilute solution or from pure water, through a semipermeable membrane into a solution of higher concentration.

Semipermeable membrane: A semipermeable membrane allows the passage of water (solvent) molecules through it in either direction, but it prevents the passage of larger solute molecules.

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Some Definitions

Osmotic pressure is the pressure that must be exerted to prevent osmosis of solvent particles through a semipermeable membrane that is separating two solutions of different solute concentrations

Isotonic: Having the same concentration of solute particles as the blood

Hypotonic: Having a lower concentration of solute particles than blood plasma

Hypertonic: Having a higher concentration of solute particles than blood plasma

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water passes through the cellophane

In osmosis,the net transfer of solvent is always from the more concentrated to the less concentrated solution.

14.9

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Hemolysis & CrenationHemolysis & Crenation

normal red bloodcell in an isotonic

solution

red blood cell inhypotonic solution – water flows into

the cell –eventually causing

the cell to burst

red blood cell inhypertonic solution – water flows out

of the cell –eventually causingthe cell to distort

and shrink

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COLLIGATIVE PROPERTIES

Know the definitions, try the homework. Be able to do the freezing point depression and boiling point elevation.

There is an equation for vapor pressure lowering, but you don’t want to know it!

The osmotic pressure will not be a major deal on any quiz or test, but you need to understand it for biology.