CHEM-001 Chapter 14 - Kids in Prison Program · Solutions Chapter 14 Hein and Arena ... 14...

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SolutionsChapter 14

Hein and Arena

Version 1.1

Eugene Passer

Chemistry Department

Bronx Community College

© John Wiley and Sons, Inc

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General Properties of

Solutions

3

• A solute is a compound that is dissolved

in a solvent.

• The solvent is the dissolving agent or the

most abundant component in the solution.

• A solution is a system in which one or

more solutes are homogeneously

dissolved in a solvent.

4

Solubility

(…and the dissolution process)

5

Solubility describes the amount of a substance

that will dissolve in a specified amount of

solvent.

6Na+ and Cl- ions hydrated by H2O molecules.15.2

Dissolution of sodium chloride in water.

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– very soluble ( greatest degree of dissolution)

– soluble

– moderately soluble

– slightly soluble

– Insoluble (least degree of dissolution)

• Terms that describe the extent of

solubility of a solute in a solvent:

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• Terms that describe the solubility of

liquids:

– immiscible: liquids that are insoluble in

each other.

oil and water

methyl alcohol and water

– miscible: liquids that are soluble in each

other.

9

Factors Related to

Solubility

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The Nature of theSolute and Solvent

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• The general rule for predicting solubility

is “like dissolves like”.

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• Polar compounds tend to be more

soluble in polar solvents than nonpolar

solvents.

• Nonpolar compounds tend to be more

soluble in nonpolar solvents than in

polar solvents.

13

Rate of

Dissolving Solids

14

– Particle size; smaller particles dissolve

faster due to greater surface area.

– Temperature; higher temperatures

generally increase the dissolution rate.

– Mixing the solution increases contact

between solvent molecules and the

solute surface.

• Factors that affect solubility are:

15

Solutions:

A Reaction Medium

16

sodium chloride reacts

with silver nitrate when dissolved

in water

NaCl(aq) + AgNO3(aq) → AgCl(s) +NaNO3(aq)

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Concentration

of Solutions

18

• A dilute solution contains a relatively

small amount of dissolved solute.

• A concentrated solution contains a

relatively large amount of dissolved

solute.

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Mass Percent Solution (wt/wt)

20

Mass percent expresses the concentration

of a solution as the percent of solute in a

given mass of solution.

g solute g solutemass percent = x 100 = x 100

g solute + g solvent g solution

21

What is the mass percent of sodium hydroxide in a

solution that is made by dissolving 8.00 g NaOH in

50.0 g H2O?

grams of solute (NaOH) = 8.0 g

grams of solvent (H2O) = 50.0 g

2

8.00 g NaOH x 100 = 13.8% NaOH solution

8.00 g NaOH + 50.0 g H O

g solutemass percent = x 100

g solute + g solvent

22

What masses of potassium chloride and water are

needed to make 250. g of 5.00% (by wt/wt) solution?

The percent expresses the mass of the solute.

250.g = total mass of solution

5.00 x 250. g = 12.5 KCl solute

100

g solutemass percent = x 100

g solution

mass percent x g solution = g solute

100

250.g – 12.5 g = 238 g H2O

Dissolving 12.5 g

KCl in 238 g H2O

gives a 5.00% KCl

solution.

23

A 34.0% (by wt/wt) sulfuric-acid solution has a

density of 1.25 g/mL. How many grams of H2SO4 are

contained in 1.00 L of this solution?

Grams of solution are

determined from the

solution density.

Step 1. Determine

grams of solution.

1.00 L = 1.00 x 103 mL

MassDensity =

VolumeDensity x Volume = Mass

1.25 g

1 mL

3(1.00 x 10 mL) = 1250 g (mass of solution)

24

Solve the mass percent equation for grams of solute.

[(g H2SO4)/(g solution)] x 100 = mass %

g H2SO4 = [34.0 % (g solution)]/100

g H2SO4 = [34.0%(1250g solution)]/100 = 425gH2SO4

A 34.0% (by wt/wt) sulfuric-acid solution has a

density of 1.25 g/mL. How many grams of H2SO4 are

contained in 1.00 L of this solution?

25

Mass/Volume Percent(wt/v)

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Mass /volume percent expresses the

concentration of solute as g solute

per ml solution.

g solutemass/volume percent = x 100

mL solution

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Solve the mass/volume equation for mL of solution.

g solutemL solution = x 100

m/v percent

A 3.0% (by wt/vol) H2O2 solution is commonly used

as a topical antiseptic to prevent infection. What

volume of this solution will contain 10. g of H2O2?

2 210 g H O

x 100 =3.0 m/v percent

mL solution = 330 mL

g solutemass/volume percent = x 100

mL solution

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Volume Percent (vol./vol.)

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volume of liquid in questionvolume percent = x 100

total volume of solution

The volume percent is the volume of a

liquid per total volume of solution.

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Molarity

31

Molarity of a solution is the number of

moles of solute per liter of solution.

number of moles of solute molesmolarity = M = =

liter of solution liter

32

What is the molarity of a solution containing 1.4 mol

of acetic acid (HC2H3O2) in 250. ml of solution?

It is necessary to convert 250. mL to L since

molarity = mol/L.

mol molThe conversion is: = M

mL L

5.6 mol = 5.6 M

L

1.4 mol

250. mL

1000 mL =

L

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volume = 600 mL

The data are:

How many grams of potassium hydroxide are required

to prepare 600. mL of 0.450 M KOH solution?

Convert: mL L mol g

0.450 molM =

L

The calculation is:0.450 mol

L

600 mL1 L

1000 mL

56.11 g KOH =

mol

15.1 g KOH

56.11 gmolar mass KOH =

mol

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0.325 L

Calculate the number of moles of nitric acid in 325 mL

of 16 M HNO3?

Substitute the data given in the problem and solve:

moles = 316 mol HNO =

1 L

3 5.2 mol HNO

moles = liters x MUse the equation:

Convert: mL L

1 L(325 mL) = 0.325 L

1000 mL

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2 3

1 L =

0.250 mol K CO

2 3mass K CO = 16.0 g

The conversion is:

What volume of 0.250 M solution can be prepared

from 16.0 g of potassium carbonate?

2 3 2 3Convert: g K CO mol K CO L solution

0.250 molM =

L2 3

138.2 gmolar mass K CO =

mol

The data are:

2 316.0 g K CO 2 3

2 3

1 mol K CO

138.2 g K CO

0.463 L

36

How many mL of 2.00 M HCL will react with 28.0 g

NaOH?

Convert: g NaOH mol NaOH

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

Step 1 Write and balance the equation for the

reaction.

Step 2 Find the number of moles of NaOH in 28.0 g

NaOH.

0.700 mol NaOH 28.0 g NaOH1 mol

=40.0 g

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How many mL of 2.00 M HCL will react with 28.0 g

NaOH?

Step 3 Solve for moles and volume of HCl.

0.700 mol NaOH1 mol HCl

1 mol NaOH

1 L HCl =

2.00 mol HCl

0.350 L HCl

Convert: mol NaOH mol HCl L HCl mL HCl

350 mL HCl0.350 L HCl1000 mL

=1 L

1

38

How many mL of 0.5000 M H2SO4 will be required to

completely react with 31.60 g Al(OH)3? The formula

weight of Al(OH)3 is 78.01 g/mol.

Answer

1215 mL

3H2SO4(aq) + 2Al(OH)3(aq) → Al2(SO4)3 (aq) + 6H2O(aq)

39Note: volumes are not always additive.

40

Introduction to pH and pOH(Please see pages 363 - 365 of Ch. 15)

41

pH is the negative logarithm of the hydrogen

ion concentration.

pH = -log[H+]

pOH = -log[OH-]

pH + pOH = 14

[ ] = molarity

pOH is the negative logarithm of the hydroxide

ion concentration.

42

Calculation of pH and pOH

43

What is the pH of a solution with a [H+] of 1.0 x 10-11?

pH = - log(1.0 x 10-11)

pH = 11.00

pH = - log[H+]

44

What is the pOH of a solution with a [OH-] of

1.0 x 10-3 and what is the pH of this solution?

pOH = - log[1.0 x 10-3] = 3.00

pH + pOH = 14.00

pH = 14.00 – pOH = 14.00 - 3.00 = 11.00

pOH = -log[OH-]

45

What is the pH of a 0.250 M HClO4 solution?

pH = - log(0.250) = 0.60

pH = - log[H+]

0.250 mol HClO4 x

L

1 mol. H+

1 mol HClO4

= 0.250 M H+

HClO4 H+ + ClO4-2

46

What is the pH of a 0.250 M Ca(OH)2 solution?

pOH = - log(0.500) = 0.30

pH + pOH = 14.00

pOH = - log[OH-]

0.250 mol Ca(OH)2 x

L

2 mol. OH+

1 mol Ca(OH)2

= 0.500 M. OH-

Ca(OH)2 Ca+2 + 2OH-

pH = 14.00 - 0.30 = 13.70

pH + 0.30 = 14.00

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Dilution and Concentration Problems

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• If a solution is diluted by adding pure

solvent:

– the volume of the solution increases.

– the number of moles of solute remain the

same.

– the ratio of mol./L gets smaller.

– the molarity decreases.

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• If a solution is concentrated by

removing the solvent via evaporation:

– the volume of the solution decreases.

– the number of moles of solute remain the

same.

– the ratio of mol./L gets larger.

– the molarity increases.

50

Calculate the molarity of a sodium hydroxide solution

that is prepared by mixing 100. mL of 0.20 M NaOH

with 150. mL of water. Assume volumes are additive.

Step 1 Calculate the moles of NaOH in the original

solution.

0.020 mol NaOH 0.100 L0.20 mol NaOH

=1 L

molM =

Lmol = L M

0.020 mol is the number of moles of NaOH in the

original 100. mL of the 0.20 M NaOH solution.

51

Calculate the molarity of a sodium hydroxide solution

that is prepared by mixing 100. mL of 0.20 M NaOH

with 150. mL of water. Assume volumes are additive.

Step 2 Solve for the new molarity.

New solution volume = 100. mL + 150. mL = 250. mL

0.020 mol NaOH

0.250 L= 0.080 M NaOH

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53

Quiz 10

1. How many moles of CO2 are present in 9.55L at 45°C and 752

torr? [5 pts.]

2. How many grams of CO2 are present in 9.55L at 45°C and 752

torr? See question 1 above. [5 pts.]

3. How many grams of BaCl2 are present in 75.0 g of a 12.0 % by

mass solution of BaCl2? [5 pts.]

4. How many milliliters of a 0.250M solution of H2SO4 would be

needed to completely react with 25.5 mL of 0.750 M NaOH?

Given:

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

[5 pts.]

54

Midterm 3 Topic’s Review

1. Electron Configurations

2. Valence Shells and Valence electrons.

3. Full Octet

4. Periodic Trends as per Rule of Thumb

a. atomic radii

b. ionic radii

c. electronegativity

d. ionization energy

5. Bond Classifications: ionic or covalent

6. Lewis Structures and VSEPR

7. Molecular Shapes and Polarity

8. PV = nRT (i.e. STP and non-STP conditions)

9. % by Mass

10. Molarity