Chapter# 4 Solution Chemistry. Solutions Solutions are homogeneous mixtures of two or more substances. The solvent is the substance in greatest quantity.

Chapter# 4Solution Chemistry

Solutions• Solutions are homogeneous mixtures of

two or more substances.

• The solvent is the substance in greatest quantity.

• Solutes are the other ingredients in the mixture.

• Solutions can exist in all states of matter.

Solution Examples

•Margarine

•Tap Water

•Steel

•18 Carat Gold

•Air

•Sterling Silver

What is the solvent in 18 ct gold ?

Composition of 18 carat gold: 75% gold, 12.5% silver,

12.5% copper.

Solution Examples


Gold

Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper.

Solution Examples

What is the solvent in 18 ct gold ?Gold

What are the solutes?


Solution Examples


Gold

What are the solutes?

Silver and Copper


Solution Examples

Some general properties of solutions include:

Solutions may be formed between solids, liquids or

gases.

They are homogenous in composition

They do not settle under gravity

They do not scatter light (Called the Tyndall Effect)Solute particles are too small to scatter light and therefore light will go right through a solution like is shown on the next slide.

Solution Properties

Tyndall Effect

Laser light reflected by a colloid. In a solution you would not see any red light.

Solutions

Aqueous Solutions

104.5o

Water is the dissolving medium

Some Properties of Water

• Water is “bent” or V-shaped.

• Water is a molecular compound.

• Water is a polar molecule.

• Hydration occurs when ionic compounds dissolve in water.

SOLUTION CONCENTRATIONThe ratio of the amount of solute to amount of solution, or

solvent is defined by the concentration.solute= solventsolutionConcentration solute =

There are various combinations of units that are used in these rations.

g solute

g solute

mL solute

g solution

mL solution

mL solution

=

=

=

% (w/w)

% (w/v)

% (v/v)

ppt (w/w)

ppt (w/v)

ppt (v/v)

ppm (w/w) ppb (w/w)

ppm (w/v) ppb (w/v)

ppm (v/v) ppb (v/v)

Ratio X 102 X 103 X 106 X 109

SAMPLE SOLUTION PROBLEMS1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g

H2O.

1. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution.

25.2 g NaCl


H2O.


25.2 g NaCl33.6g H2O + 25.2 g NaCl


H2O.


25.2 g NaCl33.6g H2O + 25.2 g NaCl

100


H2O.


25.2 g NaCl33.6g H2O + 25.2 g NaCl

100 = 43.0 % NaCl


H2O.


25.2 g NaCl33.6g H2O + 25.2 g NaCl

100 = 43.0 % NaCl

44.6 g NaCl100 g solution


H2O.


25.2 g NaCl33.6g H2O + 25.2 g NaCl

100 = 43.0 % NaCl


333 g solution


H2O.


25.2 g NaCl33.6g H2O + 25.2 g NaCl

100 = 43.0 % NaCl


333 g solution


H2O.


25.2 g NaCl33.6g H2O + 25.2 g NaCl

100 = 43.0 % NaCl


333 g solution= 149 g NaCl


H2O.


25.2 g NaCl33.6g H2O + 25.2 g NaCl

100 = 43.0 % NaCl



Mass of water?


H2O.


25.2 g NaCl33.6g H2O + 25.2 g NaCl

100 = 43.0 % NaCl



Mass of water? 333 g solution – 149 g NaCl = 184 g H2O


H2O.


25.2 g NaCl33.6g H2O + 25.2 g NaCl

100 = 43.0 % NaCl



Mass of water? 333 g solution – 149 g NaCl = 184 g H2O

SAMPLE SOLUTION PROBLEMS3. How many grams of NaCl are required to dissolve in 88.2 g

of water to make a 29.2% (w/w) solution.

4. A sugar solution is 35.2%(w/v) find the mass of sugar contained in a 432 mL sample of this sugar solution.

SOLUTION CONCENTRATIONThe solution concentration can also be defined using moles.

The most common example is molarity (M).

The molarity of a solution is defined as:

“The number of moles of solute in 1 L of solution”

and is given the formula:Moles solute

Molarity (M) = Liters solution

MOLARITY SAMPLE PROBLEMS

1. A student dissolves 25.8 g of NaCl in a 250 mL volumetric flask. Calculate the molarity of this solution. (picture of volumetric flask is on the next slide)

2. Find the mass of HCl required to form 2.00 L of a 0.500 M solution of HCl.

3. A student evaporates the water form a 333 mL sample of a 0.136 M solution of NaCl. What mass of salt remains?

In the lab we would use a piece of glassware called a volumetric flask to prepare this solution.

SOLUTION PREPARATION

SOLUTION PREPARATION

VOLUMETRIC FLASK

SOLUTION DILUTION

Often we will want to make a dilute solution from a more

concentrated one. To determine how to do this we use the formula :

C1V1 = C2V2

Where:C1 = concentration of more concentrated solutionV1 = volume required of more concentrated solutionC2 = concentration of more dilute solutionV2 = volume of more dilute solution

We can use any units in this equation but they must be the same on both sides.

DILUTION PROBLEM

How would one prepare 50.0 mL of a 3.00 M solution of NaOH using a 7.10 M stock solution?

C1V1 = C2V2

(7.10 M)V1 = (3.00 M) (50.0 mL)

This means that you add 21.1 mL of the concentrated stock solution to a 50.0 mL volumetric flask and add water until the bottom of the meniscus touches the line on the volumetric flask.

DILUTION PROBLEM

How would one prepare 50.0 mL of a 3.00 M solution of NaOH using a 7.10 M stock solution?

(7.10 M)V1(7.10 M) (7.10 M)

C1V1 = C2V2

(7.10 M)V1 = (3.00 M) (50.0 mL)

V1 = 21.1 mL

This means that you add 21.1 mL of the concentrated stock solution to a 50.0 mL volumetric flask and add water until the bottom of the meniscus touches the line on the volumetric flask.

(3.00 M) (50.0 mL)=

Reaction Driving ForcesFive Driving Forces Favor Chemical Change

1. Formation of a solid 2. Formation of water 3. Transfer of electrons 4. Formation of a gas5. Formation of a weak electrolyte

Types of Aqueous SolutionsSolutions are homogeneous mixtures of a solute and a solvent.

• The solute is the solution component in the smallest amount while the solvent is the larger component of a solution.

• Solutes whose solutions conduct electricity are called electrolytes

• Solutes whose solutions do not conduct electricity are called nonelectrolytes

• Electrolytes are solutes that form ions when they dissolve. Ionic solutes or acids usually form solutions that conduct electricity.

Solution Conductivity

Strong electrolyte Weak electrolyte Nonelectrolyte

Solution FormationWater is one of the best solvents known. It is able to dissolve ionic solutes, such as sodium chloride, to produce solutions that conduct electricity. Molecules, containing a positive and negative regions, are called polar. Water is an example of a polar molecule and can dissolve ionic solutes by the positive region of water attracting to the negative ion of an ionic solute thus separating the crystal lattice in to a solution of solvated ions.

Acids undergo characteristic double replacement reactions with oxides, hydroxides, carbonates and bicarbonates.

2HCl (aq) + CuO (s) CuCl2 (aq) + H2O (l)

2HCl (aq) + Ca(OH)2 (aq) CaCl2 (aq) + 2H2O (l)

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

2HC l (aq) + Sr(HCO3)2 (aq) SrCl2 (aq) + 2H2O (l) + 2CO2 (g)

Acid-Base Reactions

Bases undergo a double replacement reaction with acids called neutralization:

NaOH (aq) + HCl (aq) H2O (l) + NaC l (aq)

In words this well known reaction is often described as: “acid plus base = salt plus water”

We previously discussed this reaction when describing types of reactions.

Acid-Base Reactions

We have discussed the double replacement reactions and ionic equations before. Since the acids and bases undergo double replacement reactions called neutralization reactions, then they can have ionic equations too.

Molecular equation:

HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)

Total ionic equation:

H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) Na+ (aq) + Cl- (aq) + H2O (l)

Net ionic equation:

H+ (aq) + OH- (aq) H2O (l)

Acid-Base Reactions

Another property of acids is their reaction with certain metals to produce hydrogen gas, H2 (g).

Zn (s) + 2HC l (aq) H2 (g) + ZnCl2 (aq)

This is an example of a single replacement reaction and is a redox reaction.

Total ionic equation:

Zn (s) + 2H+ (aq) + 2Cl- (aq) H2 (g) + Zn2+ (aq) + 2Cl- (aq)

Net ionic equation:

Zn (s) + 2H+ (aq) H2 (g) + Zn2+ (aq)

Acid-Base Reactions

Solution StoichiometryConsider the following balanced equation:

CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq)

1. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride.




0.100 moles AgNO3

L solution




0.100 moles AgNO3

L solution 103 mLL solution




0.100 moles AgNO3


moles AgNO3

moles AgClmoles AgCl143.45 g AgCl 33.2 mL




0.100 moles AgNO3


moles AgNO3





0.100 moles AgNO3


moles AgNO3


= 0.476 g AgCl




2. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and 200.0 mL of a 0.200 M solution of calcium chloride solution.

3. Find the volume of the excess reactant.

Acid-Base Reactions• Bronsted-Lowry acids are proton (H+)

donors.

• Bronsted-Lowry bases are proton acceptors.

• Free hydrogen ions don’t exist in water because they strongly associate with a water molecule to create a hydronium ion (H3O+).

Acid-Base Reactions

• A neutralization reaction takes place when an acid reacts with a base and produces a solution of a salt and water.

• A salt is made up of the cation characteristic of the base and the anion characteristic of the acid.

• Example: HCl + NaOH ---> NaCl + H2O

Strong Acids and Bases• A strong acid or strong base is completely

ionized in aqueous solution.

• HCl, HBr, HI, HNO3, HClO4 and H2SO4 are all strong acids. All other acids are assumed to be weak acids.

• A weak acid or weak base only partially ionize in aqueous solution.

• Amphiprotic substances can behave as either a proton acceptor or a proton donor.

Types of Equations• Molecular Equations have reactants and

products written as undissociated molecules.

HCl + NaOH ---> NaCl + H2O

• Overall Ionic Equations show all the species, both ionic and molecular present in the reaction.

H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O

Continued• Strong acids and strong bases are written

as the corresponding ions in an overall ionic equation.

• Net Ion Equations describe the actual reaction taking place.

H+ + OH- ----> H2O• The Na+ and Cl- ions are spectator ion is

this reaction, because they are unchanged by reaction taking place.

Precipitation Reactions• Reactions in which a solid product

forms from the reactants in solution.

• Solubility guidelines for ionic compounds allows the prediction of the formation of solid products.

Solubility GuidelinesAll compounds containing the following ions are

soluble:

Cations:Group I ions (alkali metals) and NH4+

Anions:NO3- and CH3COO- (acetate)

Compounds containing the following anions are soluble except as noted:Group 17 ions (halides), except with Ag+, Cu+, Hg2

2+, Pb2+, SO42-

All other compounds are insoluble except the following group 2 (alkaline earth) hydroxides: Ba(OH)2, Ca(OH)2, and Sr(OH)2

Precipitation

Precipitation is the formation of a solid when two solutions are combined.

Precipitate FormationDoes a precipitate form when sodium chloride is mixed with silver nitrate? If so what is the precipitate?

NaCl Na+ + Cl-

AgNO3 Ag+ + NO3-

(Salt water solution)

(Silver nitrate solution)

Net Ionic Equations• Soluble ionic compounds are called strong

electrolytes and completely ionize in aqueous solution.

• Write the balanced net ionic equation when sodium sulfate reacts with barium acetate.

Types of Solutions• Unsaturated solution does not contain the

maximum concentration

• A saturated solution contains the maximum concentration of solute that can dissolve in it.

• A supersaturated solution contains more than the quantity of a solute that is predicted to be soluble in a given volume of solution at a given temperature.

Solubility Curves of Various Solutes

Supersaturated Solution

Sodium acetate precipitates from a supersaturated solution.

By forming a solution at a high temperature then slowly cooling it we can form supersaturated solutions that contain more solute than in a saturated solution.

These kinds of solutions are very unstable and tend to separate out the excess solute with the slightest disturbance.

http://www.youtube.com/watch?v=uy6eKm8IRdI&NR=1

http://www.youtube.com/watch?v=aC-KOYQsIvU&feature=related

Supersaturated Solutions









TitrationTitration is an experimental procedure to determine the concentration of an unknown acid or base.

The figure on the left shows the glassware for a titration experiment. A buret clamp holds the buret to a ring stand and below the buret is a flask containing the solution to be titrated, which includes an indicator. The purpose of the indicator is to indicate the point of neutralization by a color change.

Key Titration Terms• A titration is a volumetric method used to

determine the concentration of an unknown solution by reacting it with a standard solution.

• A standard solution is a solution of known concentration.

• The equivalence point in a titration is reached when enough standard solution has been added to completely react with the unknown solution.

• The end point in a titration is reached when the indicator changes color.

The picture on the left shows the tip of a buret, with air bubble, which is not good, and also shows the stop-cock. Note the position of the stop-cock is in the “off” position. This picture shows the color of the phenolphthalein indicator at the end-point. In this experiment a 23.00 mL aliquot of 0.1000 M NaOH titrant is added to 5.00 mL of an unknown HCL solution. The acid solution in the beaker starts out clear and becomes pink when all of the HCL has been consumed.

NaOH + HCl NaCl + HOH

Titration

How can we calculate the concentration of acid in the beaker?

Titration


Normal procedure, yes, a conversion. Steps 1-4, again!

Titration

Titration


0.100 mole NaOHL NaOH solution


Titration


0.100 mole NaOHL NaOH solution mL solution

10-3 L solution 23.00 mL soln mole HClmole NaOH0.00500L

0.460 M HCl


IndicatorsIndicators are weak organic (carbon containing) acids of

various colors depending on the formula of the acid.

Below is a generic acid.

HA H+ + A- colorless pink

1. Describe the color change when a strong acid is added?






Less pink






2. Describe the color change when a strong base is added?

Less pink


various colors depending on the formula of the acid. Below is

a generic acid.




Less pink

Darker pink







3. Describe the color change when the pH is lowered?

Less pink

Darker pink








Less pink

Darker pink

Less pink








4. Describe the color change when the pH is raised?

Less pink

Darker pink

Less pink








4. Describe the color change when the pH is raised?

Less pink

Darker pink

Less pink

Darker pink

Color versus pH of Many Different indicators

How can we make an indicator?

How can we make an indicator?

Step One

Red Cabbage

Step Two

Cook the Cabbage

Step Three

Filter the Juice

http://images.google.com/imgres?imgurl=http://www.coffeemakersetc.com/images/Paper_Filters.jpg&imgrefurl=http://www.coffeemakersetc.com/bunn-coffee-filters-20111-series-gallon-urns-c-109_29_35.html&usg=__XztN17uTxz0ZX46CLKI9DDpKRBQ=&h=500&w=500&sz=43&hl=en&start=2&um=1&tbnid=J4lede1Vryg6gM:&tbnh=130&tbnw=130&prev=/images?q=Filter&um=1&hl=en&sa=G

What color is the juice after filtering?

What color is the juice after filtering? The color of pH 6, 7, or

8

Colors of cabbage juice at various pH values

CH#4 REVIEW

Titraction of HCl and NaHO

To the left is a plot that shows the pH of an HCl solution as a function of the added volume of 0.011 M NaOH. Which of the following plots would correspond to the same titration but using 0.022 M NaOH?

A) B) C)

Titraction of HCl and NaHO

Consider the following arguments for each answer and vote again:

A. The shape of both titration curves is the same, but the pH for titration with the stronger base should be higher at every point on the curve.

B. The titration curve stays the same up to the equivalence point, but the pH will be higher when there is excess strong base.

C. The curve will be shifted to the left because only half the volume of 0.022 M NaOH will be required to reach the equivalence point.

Dilution of Acidic Acid

An acetic acid (HAc) solution of pH 3.0 is diluted by a factor of 10 with water. What is the new pH of the solution?

A) < 4.0 B) 4.0 C) > 4.0

Dilution of Acidic Acid


A. Dilution of an acetic acid solution will drive the HAc/Ac- equilibrium toward further ionization of HAc. Therefore, the final pH will be between 3 and 4.

B. Dilution of an acidic solution with pH 3.0 by a factor of 10 will result in a decrease in the H3O+ concentration from 0.001 M to 0.0001 M, giving a pH of 4.0.

C. Dilution of a weak acid solution with pH 3.0 will give a less acidic solution than the dilution of a strong acid solution with the same pH.

pH of HCl,H2SO4,and NaOH Solutions

Which of the following three solutions would have the highest pH?

A) 10-3 M NaOH B) 10-6 M H2SO4 C) 10-12 M HCl

pH of HCl,H2SO4, and NaOH Solutions


A. Of the three solutions, only the NaOH solution is basic, so its pH must be the highest.

B. Although H2SO4 is a strong acid, it dissociates in water to form the base SO4

2-, making its pH higher than that of the other two solutions.

C. Since HCl dissociates completely in water, the concentration of H3O+ is 10‑12 M for this solution. Therefore, the pH of the solution is 12.

THE END

Chapter# 4 Solution Chemistry. Solutions Solutions are homogeneous mixtures of two or more substances. The solvent is the substance in greatest quantity.

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