Advanced Chemistry Unit #3 Solutions, Molarity, Percent Solutions, Molality And Colligative Properties

Advanced ChemistryUnit #3

Solutions, Molarity, Percent Solutions, MolalityAnd

Colligative Properties

Solution - A homogeneous mixture of 2 or more substances.

Examples: Salt water, Orange juice, “Kool-aid”

All solutions are a mixture of the solute(s) and the solvent.

They are always listed with the lesser constituent (the solute) first.

Solute - The substance that is dissolved. It is the lesser constituent by volume. (Ex: The Salt in the Saltwater.)

Solvent - The substance that does the dissolving. It is the greater constituent by volume. (Ex: The Water in the Saltwater.)

There are 6-types of solutions based onthe phases of their components.

1. Gas - Gas O2 in air

2. Liquid - Gas Water vapor in the air

3. Gas - Liquid CO2 in H2O

4. Liquid - Liquid “2-stroke” fuel (oil in gas)

5. Solid - Liquid Salt in water

6. Solid - Solid Alloys (Cu in Ag, “Sterling Ag”)

Miscible - Describes two liquids that are mutually soluble in each other.

Examples:1. Water and alcohol are completely miscible

with each other.

2. Water and ether are slightly miscible witheach other.

3. Water and oil are not miscible with each other.

The General Rule with regards to Miscibility:

“Likes dissolve likes.”

1. Polar solvents dissolve polar compounds.

2. Non-polar solvents dissolve non-polar compounds.

The “Solution process” involves two actions:

1. Dissolving

2. Crystallizing

Equilibrium (as its relates to solvation)-

The physical state where the two opposing processes of dissolving and crystallizing occur at equal rates.

Le Châtelier’s principle -

When stress is applied to a system at equilibrium, the system shifts to relieve the stress.

Example:

As you add salt to water, it dissolves. However, as the concentration of salt approaches saturation, the rate of solvation slows.

Saturated solution -

A solution in which the dissolved and un-dissolved substances are at equilibrium. (The solvent can no longer dissolve any additional solute. It will simply fall to the bottom of the solution.)The “solubility” of a substance determines when a solution will become saturated.

Solubility -

The maximum amount of a solute that can dissolve in a given amount of solvent under specified conditions.

Examples at 20˚C :1. 222 g of Silver nitrate / 100 g of water

2. 3.89 g of Barium hydroxide / 100 g of water

3. 0.169 g of CO2 / 100 g of water (at SP)

4. 0.0043 g of O2 / 100 g of water (at SP)

Factors that affect the rate of solution:

1. Particle size

2. Agitation

3. Temperature (aka: the Heat of Solution)

4. Polarities of solute and solvent

5. Concentration of solution (solvation slows itnears saturation)

Heat of Solution -

The difference between the heat content of a solution and it’s components.

Heats of solution are either positive or negative.

Positive Heat of Solution -

Solute + solvent + heat solution (endothermic)

“Heating helps solvation”

Heat has a positive effect on solvation.

Negative Heat of Solution -

“Heating helps solvation”

Heat has a negative effect on solvation.

Solute + solvent solution + heat (exothermic)

Concentration units for solutions:

1. Molarity -

2. Molality -

The number of moles per liter of H2O

The number of moles of solute per kilogram of solvent.

4. Percent solutions -

3. Normality - The number of equivalents of a substance dissolved in a liter of solution.

Molarity examples:

The following examples all utilize the “DIMO” concept as learned in Academic Chemistry.

4 Steps to use the DIMO chart.4 Steps to use the DIMO chart.

1 ) Determine the chart mass of thesubstance you're working with.

1 ) Determine the chart mass of thesubstance you're working with.

2) Deal with the concentration of that substance.

3) Deal with the volume of the solution you're working with.

4) Use "DIMO" to solve.

How many grams of NaOH are in 1 liter of a 1M solution of the NaOH?

Example 1:

STEP 1: Determine the chart mass

Example 1 (con’t.):

STEP 1: 40 g/mol

STEP 2: Deal with the concentration


STEP 1: 40 g/mol

STEP 2: 1 mol/L

STEP 3: Deal with the volume


STEP 1: 40 g/mol

STEP 2: 1 mol/L

STEP 3: 1 mol of NaOH in sample

STEP 4: Use "DIMO" to solve

How many grams of NaOH are in 1 liter of a 1M solution of the NaOH?

Solution to Example 1:

There are 40 g of NaOH in 1 liter of a 1M solution of NaOH.

0.75 liters of a 0.5M solution contains how many grams of CaCl2?

Example 2:



STEP 1: 111 g/mol



STEP 1: 111 g/mol

STEP 2: 0.5 mol/L



STEP 1: 111 g/mol

STEP 2: 0.5 mol/L

STEP 3:0.375 mol of CaCl2 in sample


How many grams of CaCl2 are in 0.75 liter of a 0.5M solution of the CaCl2?


There are 41.625 g of CaCl2 in 0.75 liter of a 0.5M solution of CaCl2.

Notice in the previous examples we calculated the number of grams of a substance.

In the next example, we will determine how many moles of a substance are in a given volume of a solution.

Notice in the previous examples we calculated the number of grams of a substance.

In the next example, we will determine how many moles of a substance are in a given volume of a solution.

What is the Molarity of a solution created by dissolving 150 grams of NaI into 250 mL of distilled water?

Example 3:



STEP 1: 150 g/mol



STEP 1: 150 g/mol

STEP 2: 150 g / 0.25 L



STEP 1: 150 g/mol

STEP 2: 150 g / 0.25 L

STEP 3: 600 g / 1 L


What is the Molarity of a solution created by dissolving 150 grams of NaI into 250 mL of distilled water?


There are 4 moles of NaI in 1 liter of the solution of NaI.

Therefore the molarity is ____4M

600 g of NaI is 4 moles

Example #4:

A 10.0 mL sample of Potassium iodide solution was analyzed by adding an excess of Silver nitrate solution to produce Silver iodide crystals (which were filtered out of the solution), and Potassium nitrate.If 2.290 grams of Silver iodide were obtained, what is the Molarity of the original Potassium iodide solution ?

Clues:

1. Write a balanced equation.

2. Determine the theoretical and actual mole ratios of the involved substances.

3. Use the “4-Step method” to solve theMolarity problems.

Molality -

The number of moles of solute per kilogram of solvent.

Molality example #1:

5.67g of glucose are dissolved in 25.2g of water.

What is the Molality ?

Step #1: Determine the number of moles ofsolute.

Molecular weight of glucose =180.1572 g/mol

Use “DIMO” to determine # of moles.

5.67g

180.1572 g/mol= 0.0315 mol of glucose

Step #2: Determine the mass of the solvent.

Given 25.2g = Kg“X”0.0252

Step #3: Set up proportions to solve.

0.0315 mol glucose

0.0252 Kg solvent= X mol glucose

1 Kg solvent

X = 1.25m glucose


Toluene, C6H5CH3, is a liquid compound

similar to Benzene, C6H6. It is the starting

material for other substances, including

Trinitrotoluene.Find the molality of Toluene in a solution

that contains 35.6g of Toluene in 125g of

Benzene.


Fructose, C6H12O6, is a sugar found in

honey and fruits. The sweetest sugar, it is

nearly twice as sweet as sucrose.

How much water should be added to

1.75g of fructose to give a 0.125m solution of

Fructose ?

Step #1: Determine the number of moles ofsolute.

Molecular weight of Fructose =180.1572 g/mol

Use “DIMO” to determine # of moles.

1.75g

180.1572 g/mol= 0.00971 mol of Fructose

Step #2: Determine the mass of the solvent.

Given “X”g = Kg“X”

Step #3: Set up proportions to solve.

0.125 mol Fructose

1 Kg solvent=

0.00971 mol Fructose

“X” Kg solvent

X = 0.07768 Kg of solvent

Which equals g of solvent.“X”77.68

This problem requires an additional step to solve.

Step #4: Use the density of water to convertgrams to milliliters.

Percent solutions (con’t.)-

A. Mass percent of solute

Mass % Solute =Mass of solute

Mass of solutionx 100

Mass percent of solute example #1:

How would you prepare a 3.5% Sodium

chloride by mass solution ?

Notice: You are not given a mass or volume of the solution.

Clues:

1. Unless specified, assume 100g of solution.

2. Subtracting mass of solute from mass ofsolution yields mass of solvent.

3. Remember to convert percent to decimalequivalent when using algebra to

solve.

Mass % Solute =Mass of solute

Mass of solutionx 100

3.5% = 0.035 =grams of NaCl

100g of solution

“X”3.5g

To prepare dissolve:

3.5g of NaCl into what volume of water ?

3.5g of NaCl into what volume of water ?

3.5g NaCl

(100g of solution - 3.5g NaCl) = mass of solvent

Mass of solvent = 96.5g of H20

Use the density of water to calculate the volume required.

H20 = 1g/mL

Therefore 96.5g of H2O = 96.5 mL of water

How would you prepare a 3.5% Sodium

chloride by mass solution ?

Dissolve 3.5g of NaCl into 96.5mL of water.


How would you prepare 425g of an aqueous solution that is 2.40% by mass Sodium acetate (NaC2H3O2).

Answer:

Dissolve 10.2g of Sodium acetate into 414.8g of water.


An experiment requires 35.0g of HCl (aq) that is 20.2% by mass.

1. How many grams of HCl is this ?

2. How many grams of water is this ?

3. What is the molarity of this solution ?

1. How many grams of HCl is this ?

20.2% of a total of 35g = 7.07g HCl

2. How many grams of water is this ?

35 total grams of solution

- 7.07g of HCl

27.93g of H2O

3. What is the molarity of this solution ?

Formula weight HCl = 36.4609 g/mol

7.07g HCl = 0.194 mol HCl

M =Mol HCl

L of H2O=

0.194 mol HCl

0.02793 L H2O= 6.95 M


What mass of solution containing 6.50% by mass of Sodium sulfate, contains 1.50g of Sodium sulfate ?

Answer:

23.1g of 6.50% Sodium sulfate solution contains 1.50g of Sodium sulfate.


How would you prepare 455g of an aqueous solution that is 4.45% by mass of Barium hydroxide ?

What is the mass of the water ?

Calculate the molarity of this solution.


B. Percent by volume (v/v) (usually a liquid - liquid)

Volume % Solute =Volume of solute

Volume of solutionx 100

Percent by volume (v/v) example #1:

What is the percent by volume of Ethanol

(C2H6O) in the final solution when 20.0 mL

EtOH are diluted to a volume of 250 mL ?



“X” = 20.0 mL EtOH250 mL solution

x 100

X = 8.00% EtOH (v/v)


Isopropyl alcohol can be purchased as a

90% Isopropyl (v/v) solution. If you dilute 25.0

mL of this solution to 100 mL with water,

what will the final percent (v/v) be ?



90% = “X” mL Isopropyl25.0 mL solution

x 100

X = 22.5 mL Isopropyl

Continued on next slide

22.5 mL Isopropyl

100 mL of solution= 22.5% Isopropyl (v/v)


A 750 mL bottle of Rum is 40% Ethanol

(v/v). How many mL of C2H6O can be distilled

from this solution (assume 100% efficiency of

the “still”) ? What is the volume of water that

will remain ?



40% = “X” mL Ethanol750 mL solution

x 100

X = 300 mL Ethanol

Therefore, 450 mL of water remain.


The density of an Ethanol is 0.789 g/mL. What is the Molarity of the solution in the previous example ?

Clues:

1. Using the density of EtOH, determine howmany grams of EtOH are in the

bottle.2. Convert grams of EtOH to moles; and, convert mL of water to Liters.

3. Set up a proportion to determine Molarity.

0.789 g

1 mL=

“X” g

300 mL

“X” = 236.7g EtOH

236.7g EtOH

How much H2O ?750 mL sol’n.

Convert grams toMoles.

Convert mL of water to Liters of water.

236.7g EtOH

750 mL sol’n.= Mol EtOH“X”5.14

L of sol’n“X”0.750

What is the molarity of EtOH in this solution ?

6.85 M


C. Percent mass/volume (m/v) (usually a solid-liquid)

% Mass / Volume =Mass of solute (g)

Solution volume (mL)x 100

Percent mass/volume (m/v) or (w/v) example #1:

A solution contains 4.3g of Iron (II) sulfate in 0.591 liters of solution. What is the percent (m/v) of the solution ?

% Mass / Volume =Mass of solute (g)

Solution volume (mL)x 100

% Mass / Volume =4.30 g

591 mLx 100

% (m/v) = 0.728% (m/v) Iron (II) sulfate

What is the molarity of the above solution ?

Molarity = M = # of moles of solute

Liter of solution

Iron (II) sulfate = FeSO4 = 151.9046 g/mol

Therefore: 4.3 g FeSO4 = mol FeSO4“X”0.0283

0.0283 mol FeSO4

0.591 Liters of sol’n

Molarity = 0.0479 M

What is the molality of the previous solution ? (Assume addition of the 4.3g of solute does not effect the volume of solvent.)

Molality = m =Moles of solute

Kg of solvent

Moles of solute

Kg of solvent=

0.0283 mol0.591 Kg

= 0.0479 m


How many liters of 2.00% glucose (w/v) solution can you prepare with 40.0 grams of Glucose ?

2.00% = 0.0200 =40.0g

“X” mL

“X” = 2000 mL = 2.00 L

For example #3, see homework problem involving “toothpaste”.


How many grams of Chloride are in 5.00 liters of seawater if the molal concentration of Chloride is 0.568m ?(The density of this seawater = 1.024 g/mL.)

What is the molarity of this solution ?

What is the percent (m/v) of this solution ?

Equivalent -

The quantity of acid in an acid base reaction that yields 1 mol of H+ ions.

or

The quantity of base that reacts with 1 mol H+ ions.This definition depends on the chemical reaction you are dealing with.

1 equivalent of acid

reacts with

1 equivalent of base.

Equivalent example #1:

HCl + NaOH NaCl + H2O

1 mol of HCl supplies 1 mol of H+ ions.Therefore, 1 mol of HCl = 1 equivalent of HCl

1 mol of NaOH reacts with 1 mol of H+ ions.Therefore, 1 mol of NaOH = 1 equivalent of NaOH.

AND


H2SO4 + NaOH

Na2SO4 + H2O

1 mol of H2SO4 supplies 2 mol of H+ ions.

Therefore, 1 mol of H2SO4 = 2 equivalent of

H2SO41 mol of NaOH reacts with 1 mol of H+ ions.Therefore, 1 mol of NaOH = 1 equivalent of NaOH.

AND

22

BUT

In this reaction:

1 mol of H2SO4 = 2 equivalent of H2SO4

And this requires 2 mol of NaOH,

(remember: 1 mol of NaOH = 1 equivalent of

NaOH)

Therefore 2 mol NaOH = 2 equivalents.2 equiv. H2SO4 reacts with 2 equiv. NaOH; which

in lowest terms is:

1 equiv. H2SO4 reacts with 1 equiv. NaOH.


H2SO4 + NaOH

NaHSO4 + H2O

1 mol of H2SO4 supplies 1 mol of H+ ions.

Therefore, 1 mol of H2SO4 = 1 equivalent of

H2SO41 mol of NaOH reacts with 1 mol of H+ ions.Therefore, 1 mol of NaOH = 1 equivalent of NaOH.

AND

AND

In this reaction:

1 mol of H2SO4 = 1 equivalent of H2SO4

And this requires 1 mol of NaOH,

(remember: 1 mol of NaOH = 1 equivalent of

NaOH)

1 equiv. H2SO4 reacts with 1 equiv. NaOH; which

in lowest terms is:

1 equiv. H2SO4 reacts with 1 equiv. NaOH.

Equivalent mass -

The mass of one equivalent.

You can calculate this from the mol:equivalent relationship.

1 mol H2SO4 = 2 equivalents H2SO4

And1 mol H2SO4 = 98.1 g/mol

98.1 g H2SO4

1 mol H2SO4

x 1 mol H2SO4

2 equiv. H2SO4

=98.1 g H2SO4

2 equiv. H2SO4

Therefore: 49.05 g/equiv

Normality -

The number of equivalents of a solute dissolved in a liter of solution.

N = # of equivalents

Liter of solution

N = Molarity x number of ionizable H+ or OH-

AND

Normality example #1:

What is the N of:

A. 2M HCl = 2M x 1 = 2 N HCl

B. 0.1M CH3COOH = 0.1M x 1 = 0.1 N CH3COOH

C. 0.3M H3PO4 = 0.3M x 3 = 0.9 N H3PO4

D. 0.1M NaOH = 0.1M x 1 = 0.1 N NaOH

E. 1.0M Ca(OH)2 = 1M x 2 = 2.0 N Ca(OH)2


What is N of 20.0g NaOH dissolved in 1.0 L of solution ?

20.0g NaOHx

1 equiv. NaOH

40.0g=

0.5 equiv.


Liter of solution=

0.5 equiv.

1 L= 0.5 N


What is N of 4.90g H2SO4 dissolved in 500 mL of solution ?

4.90g H2SO4x1 equiv. H2SO4

49.0g=

0.10 equiv.


Liter of solution=

0.10 equiv.

0.5 L= 0.2 N


What is N of 15.0g HCl dissolved in 0.400 L of solution ?

15.0g HClx

1 equiv. HCl

36.5g=

0.411 equiv.


Liter of solution=

0.411 equiv.

0.400 L= 1.03 N


What is N of 80.0g NaOH dissolved in 1.5 L of solution ?

80.0g NaOHx

1 equiv. NaOH

40.0g=

2.0 equiv.


Liter of solution=

2.0 equiv.

1.5 L= 1.33 N


What is N of 196g H3PO4 dissolved in 2.0 L of solution ?

196g H3PO4 x1 equiv. H3PO4

32.7g=

5.99 equiv.


Liter of solution=

5.99 equiv.

2.0 L= 3 N


How many equivalents are in 2.5 L of 0.60 N H2SO4 ?


Liter of solution=

“X” equiv.

2.5 L0.60 N =

“X” = 1.5 equivalents


How many equivalents are in 0.55 L of 1.8 N NaOH ?


Liter of solution=

“X” equiv.

0.55 L1.80 N =



How many equivalents are in 1.60 L of 0.5 N H3PO4 ?


Liter of solution=

“X” equiv.

1.60 L0.50 N =



How many equivalents are in 250 mL of 0.28 N H2SO4 ?


Liter of solution=

“X” equiv.

0.250 L0.28 N =


Dilutions with Normality:

What if you wished to dilute a more concentrated Normal solution to a specific concentration. How would you do it ?

NiVi = NfVf

Normal Dilutions example #1:

A lab requires 500 mL of 0.20 N Sulfuric acid. You have a significant volume of 4.0 N

H2SO4.How do you prepare the desired solution ?

NiVi = NfVf

0.20 N x 0.500 L = 4.0 N x “X”

“X” = 0.025 L

Dilute 25 mL of 4.0 N Sulfuric acid to 500 mL.

Normal Dilutions example #2:

A lab requires 870 mL of 2.0 N Potassium hydroxide. You have a significant volume of 3.0 N KOH.

How do you prepare the desired solution ?

NiVi = NfVf

2.0 N x 0.870 L = 3.0 N x “X”

“X” = 0.58 L

Dilute 580 mL of 3.0 N Sulfuric acid to 870 mL.

Titrations by Normality:

You can determine the Normality of an unknown solution by adding a specific volume of a solution of known Normality until the mixture is neutral.

Nacid x Volumeacid = Nbase x Volumebase

Both “volumes” must be the same unit.

Titrations by Normality example #1:

If 17 mL of 2.0 N HCl neutralizes 50 mL

of Ca(OH)2, what is the Normality of the base ?


2.0 N x 17 mL = “X” N Ca(OH)2 x 50 mL

“X” = 0.68 N Ca(OH)2


A 25.0 mL sample of Ba(OH)2 solution

was neutralized by 45.3 mL of 0.150 N HCl.

What is the Normality of the Ba(OH)2 ?


0.150 N x 45.3 mL = “X” N Ba(OH)2 x 25 mL

“X” = 0.272 N Ca(OH)2

What is the molarity of the above

determined Ba(OH)2 solution ?

N = M x ionizable H+ or OH-

0.272 N = “X” x 2

“X” = 0.136 M Ba(OH)2


An antacid tablet contains Magnesium hydroxide and an inert binding agent. The mass of the Magnesium hydroxide in the tablet was determined by titration with HCl. If one tablet required 39.1 mL of 0.2056 N HCl for neutralization; how many equivalents of the base are in the tablet ? How many grams is this ?

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Advanced Chemistry Unit #3 Solutions, Molarity, Percent Solutions, Molality And Colligative Properties