study guide for module 11a—solutions i · pdf filestudy guide for module...

Chemistry 1020, Module 11A Name

Study Guide for Module 11A—Solutions I

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Reading Assignment: Sections 4.1, 4.2, 4.3, 11.1, and 11.3 in Chemistry, 6th Edition by Zumdahl.

Guide for Your Lecturer:

1. Solutions: Basic Information2. Classification of Solutions; Classification of Common Solutes3. Concentration Units: Molarity

✔ 4. Concentration Units: Mole Fraction✔ 5. Concentration Units: Weight Percent✔ 6. Concentration Units: Molality✔ 7. Concentration Units: Converting from One to Another

8. An Overview of Factors Which Affect Solubility; Predicting Whether or Not a Given Substance Will LikelyDissolve in Water

9. More Detail about the Temperature Dependence of Solubility✔ 10. More Detail about the Pressure Dependence of Solubility (Henry's Law)

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HomeworkNote: ✔ indicates problems to be stressed on drill quizzes and hour exams.■Solutions: Basic Information1a) A. Define solution. (p. 27)

-----------------------------------------------------------------------------------------------------------------------------------------------B. Define solute. (p. 136)

-----------------------------------------------------------------------------------------------------------------------------------------------C. Define solvent. (p. 136)

-----------------------------------------------------------------------------------------------------------------------------------------------b) If glucose is dissolved in water,...

•What is the solute? What is the solvent?-----------------------------------------------------------------------------------------------------------------------------------------------

c) If 6 moles of H2 and 2 moles of N2 are mixed,...•What is the solute? What is the solvent?-----------------------------------------------------------------------------------------------------------------------------------------------

d) Define concentration. (Amount of solute contained in a given amount of solvent or solution.)

-----------------------------------------------------------------------------------------------------------------------------------------------e) Define solubility. (p. A39)

-----------------------------------------------------------------------------------------------------------------------------------------------f) Distinguish between dilute and concentrated solutions. (A dilute solution has less solute per given amount of

solution or solvent than does a concentrated one.)

-----------------------------------------------------------------------------------------------------------------------------------------------g) Distinguish among unsaturated, saturated, and supersaturated solutions. (p. 752) (A saturated solution contains

as much solute as possible at a particular temperature and pressure. In an unsaturated solution, moresolute can dissolve. A supersaturated solution contains more solute than it normally can at a particulartemperature due to a heating and cooling down process.)

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Xavier University of Louisiana 1

Chemistry 1020, Module 11A

■Solutions: Basic Information (continued)1h) Define electrolyte. (p. 136)

-----------------------------------------------------------------------------------------------------------------------------------------------i) Distinguish among nonelectrolytes, weak electrolytes, and strong electrolytes. (Electrolytes are substances

which break up into ions when dissolved in water. All molecules in a strong electrolyte break up, i.e.100% ions; a few molecules in a weak electrolyte break up, i.e. 5-10% ions; no molecules in anonelectrolyte break up, i.e. 0% ions.)

-----------------------------------------------------------------------------------------------------------------------------------------------j) What determines the degree to which a solution conducts electricity if a low voltage is applied? (p. 136)

-----------------------------------------------------------------------------------------------------------------------------------------------k) Distinguish between miscible and immiscible. ( “Immiscible” substances do not mix together. An example is

oil and water. Miscible substances do mix together.)

-----------------------------------------------------------------------------------------------------------------------------------------------l) Give an example of...

•a liquid which is miscible with water. •a liquid which is immiscible in water.

-----------------------------------------------------------------------------------------------------------------------------------------------m) S. If the solubility of NaCl in water at a temperature is 35 g/100 g of water, how much NaCl would

dissolve in 40 grams of water at this temperature?35 g NaCl100 g H2O =

x40 g H2O so x = 40 g H2O *

35 g NaCl100 g H2O = 14 g NaCl

----------------------------------------------------------------------------------------------------------------------------------------------A. How many grams of KNO3 would dissolve in 200 grams of water at a certain temperature if the

solubility of KNO3 at this temperature is 42 g/100 g of water?

----------------------------------------------------------------------------------------------------------------------------------------------B. If the solubility of NaCl in water is 39 g/100 g of water at a certain temperature, how much NaCl

would dissolve in 60 grams of water at this temperature?

----------------------------------------------------------------------------------------------------------------------------------------------C. How many grams of NaCl will dissolve in 150 g of water at a temperature if the solubility of NaCl at

that temperature is 52 g/100 g of water?

----------------------------------------------------------------------------------------------------------------------------------------------D. How many grams of water would be required to dissolve 20 g of NaCl at a temperature if the

solubility NaCl at that temperature is 45 g/100 of water.

----------------------------------------------------------------------------------------------------------------------------------------------E. How many grams of water would be required to dissolve 45 g of NaCl at a temperature if the

solubility NaCl at that temperature is 32 g/100 of water.

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2 Xavier University of Louisiana


■Classif ication of Solutions; Classif ication of Common Solutes (continued)2a) The following diagrams are intended to represent solutions of the given solute in water. Determine the composition

of each of the following solutions and then choose the terms which correctly describe each. |Solute | Diagram of solution |Words which correctly describe solute or solution D. | ✭■ | ✭■ ✭■

| | ■- ✭■ ✭■ solution: saturated

|Composition | ✭■ ✭+ unsaturated

|of solution: | ✭■ ■- ✭■

| | ✭+ ✭■ ✭■

| | ✭■ ✭■ ■- solute: nonelectrolyte

| | ■- ✭■ ✭■ weak electrolyte

| | ✭■(s) ✭+ strong electrolyte

-----------------------------------------------------------------------------------------------------------------------------------------------b) List two common nonelectrolytes. (urea NH2CONH2 and glucose C6H12O6)

- -

-----------------------------------------------------------------------------------------------------------------------------------------------c) List the chemical formulas and names of the following:

•Six common strong acids used in thiscourse (four on page 657, hydrobromic acid, hydroiodic acid). Others areweak.

- - -

- - -

-----------------------------------------------------------------------------------------------------------------------------------------------•Seven common weak acids. (hydrofluoric acid, citric acid [C3H5O(COOH)3], acetic acid, boric acid,formic acid, hydrosulfuric acid, and H2CO3 [called carbonic acid = CO2 in water])

- - - -

- - ------------------------------------------------------------------------------------------------------------------------------------------------

d) List the chemical formulas and names of nine strong bases. (Six Group IA hydroxides, strontium hydroxide,barium hydroxide, and calcium hydroxide)

- - -

- - -

- - -

-----------------------------------------------------------------------------------------------------------------------------------------------e) What does “R” stand for as in RCOOH and RNH2? (A specific combination of C and H atoms. For example:

In acetic acid, CH3COOH, “R” = CH3, or formic acid, HCOOH, “R” = H.)

-----------------------------------------------------------------------------------------------------------------------------------------------f) List the chemical formulas of two common weak bases. (NH3, compounds with general formula RNH2)

- -

-----------------------------------------------------------------------------------------------------------------------------------------------g) Which salts are strong electrolytes? (All which dissolve.)

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■Classif ication of Solutions; Classif ication of Common Solutes (continued)2h) In the following beakers, represent solutions of the indicated solute in water and then choose the term which

correctly describes each. Diagram of solution S. HF

H F

H

H

H

HF

F

F

F

F

H

H

F

H

F

H

F

H

F

H+

F-

Composition of solution:10 HF1 H+

1 F-

solute: nonelectrolyteweak electrolyte✔strong electrolyte

-----------------------------------------------------------------------------------------------------------------------------------------------A. HBr

Composition of solution:

solute: nonelectrolyteweak electrolytestrong electrolyte

-----------------------------------------------------------------------------------------------------------------------------------------------B. KOH



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■Classif ication of Solutions; Classif ication of Common Solutes (continued)2h) In the following beakers, represent solutions of the indicated solute in water and then choose the term which

correctly describes each.

C. HNO2



-----------------------------------------------------------------------------------------------------------------------------------------------D. NH3



-----------------------------------------------------------------------------------------------------------------------------------------------E. LiOH





■Concentration Units: Molarity3a) Define molarity. (p. 140)

-----------------------------------------------------------------------------------------------------------------------------------------------b) Define gram formula weight. (The sum of the atomic weights of the atoms in the formula of a substance

expressed in grams. It has the same numerical value as the quantity we have called “molecular weight.”)

-----------------------------------------------------------------------------------------------------------------------------------------------c) Define formality. (Formality is a concentration unit = the number of gram formula weights of a substance

per liter of solution. The value is the same as for molarity.)

----------------------------------------------------------------------------------------------------------------------------------------------d) Work the following problems. (pp. 141-5 and Module 4 in Chemistry 1010) NOTE: Starting with this problem

and continuing throughout the course, on the first line of each problem you will be requested to provide thefollowing information about all solutes

•Name or formula (which ever is not given).•Type of compound--i.e. "acid," "base," "salt," or "not acid, base or salt."•Strength--i.e. "strong," "weak,", or "nonelectroyte."

S. How many grams of sodium perchlorate are needed to prepare 155 milliliters of a 0.32 M solution of thesubstance?•Formula: NaClO4 •Type: salt •Strength: strong

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••NaClO4 M = (mol solute)/liter solnNa 23.0 = 23.0 M = (g/mw)/LCl 35.5 = 35.5 M*L = g/mw4"O" 4*16.0 = 64.0 mw*M*L = g

122.5 g/mol g = (122.5 g/mol)(0.32 mol/L)(155 mL)(1 L/1000 mL) =g = 6.076 g which rounds to 6.1 g

----------------------------------------------------------------------------------------------------------------------------------------------A. How many grams of sodium chloride must be used to prepare 502 ml of a 0.25 F solution?

•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------B. How many milliliters of 0.54 M sodium chloride could be prepared from 42 grams of sodium chloride?


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■Concentration Units: Molarity (continued)3d) Work the following problems. (Section 4.3 in text and Module 4 in Chemistry 1010)

C. How many grams of potassium sulfate must be added to water to obtain 412 ml of a 0.10 F solution of thesubstance?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------D.What is the molarity of a solution prepared by adding 132 grams of rubidium chloride to enough water to obtain

355 milliliters of solution?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------E. What is the molarity of a solution prepared by adding 13.2 grams of lithium acetate to enough water to obtain

135 milliliters of solution?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------F. How many grams of sodium iodate must be added to water to obtain 187 milliliters of a 0.23 M solution of the

substance?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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✔✔✔■Concentration Units: Mole Fraction4a) Define mole fraction. (p. 208)

----------------------------------------------------------------------------------------------------------------------------------------------b) Perform the calculation indicated.

S1. 8.0 moles of H2 and 6.0 moles of N2 are mixed. What is the mole fraction of nitrogen in the mixture?•Formula: N2 •Type: Not a, b, or s •Strength: nonelectrolyte

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

XN2 = moles N2total moles =

XN2 = 6.0

6.0 + 8.0 = 6.0

14.0 =

RXN2 = 0.42857 = 0.43

----------------------------------------------------------------------------------------------------------------------------------------------S2. How many grams of sodium hydroxide must be added to 75 mL of water in order to obtain a solution with a

mole fraction of 0.094 in sodium hydroxide?•Formula: NaOH •Type: base •Strength: strong electrolyte••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

XNaOH = moles NaOH

total moles =

XNaOH = mol NaOH = 0.392 = 0.906 mol NaOH mol NaOH + mol H2O

mol H2O = (75 mL)(1g/1mL)(1 mol/18.0 g) = 4.17 mol mol NaOH = 0.433 mol NaOH

0.094 = mol NaOH g NaOH= (0.433 mol)(40.0 g/mol) mol NaOH + 4.17

R0.094 mol NaOH + 0.392 = mol NaOH = 17 g NaOH

----------------------------------------------------------------------------------------------------------------------------------------------A. 1.2 moles of potassium chloride are dissolved in 5.0 moles of water. What is the mole fraction of

potassium chloride in the solution?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------B. How many moles of nitrogen must be added to 12 moles of hydrogen in order to obtain a mixture whose

mole fraction of nitrogen is 0.25?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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✔✔✔■Concentration Units: Mole Fraction (continued)4b) C. What is the mole fraction of sodium chloride in a solution prepared by adding 5.3 grams of the salt to

98 grams of water?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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D. How many of grams of urea (gfw=60.0) must be added to 125 milliliters of water in order to obtain a solutionwith a mole fraction of 0.034 in urea?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------E. What is the mole fraction of urea (gfw = 60.0) in a solution prepared by adding 15 grams of urea to 82 grams of

water?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------F. How many of grams of glucose (gfw=180.0) must be added to 55 milliliters of water in order to obtain a

solution with a mole fraction of 0.032 in glucose?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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✔✔✔■Concentration Units: Weight Percent5a) Define weight percent. (p. 512)

----------------------------------------------------------------------------------------------------------------------------------------------b) Perform the calculation indicated.

S1. 12 grams of calcium chloride are dissolved in 123 grams of water. What is the weight percent of calciumchloride in the mixture?•Formula: CaCl2 •Type: salt •Strength: strong

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

%CaCl2 = g CaCl2total g * 100% =

%CaCl2 = 12 g CaCl2

123g water + 12 g CaCl2 * 100% =

%CaCl2 = 12 g CaCl2135 g soln * 100% =

R%CaCl2 = 8.89% CaCl2 = 8.9%CaCl2

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S2. How many grams of potassium sulfate must be added to 105 grams of water to obtain a solution which is 8.0 %potassium sulfate by weight?•Formula: K2SO4 •Type: salt •Strength: strong

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

%K2SO4 = g K2SO4

total g * 100% = 0.080 g K2SO4 + 8.40 = g K2SO4

%K2SO4 = g K2SO4 * 100 %= 8.40 = 0.920 g K2SO4g K2SO4 + g H2O

R8.0 % K2SO4 = g K2SO4 *100 % g K2SO4 = 9.1 g K2SO4

g K2SO4 + 105 g H2O

0.080 K2SO4 = g K2SO4 = g K2SO4 + 105 g H2O

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A. How many grams of a solution which is 35% sodium chloride by weight must be measured out to obtain12 grams of sodium chloride?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••



✔✔✔■Concentration Units: Weight Percent (continued)5b) B. What is the weight percent of sodium chloride in 62 grams of a mixture which contains 35 grams of

sodium chloride?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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C. How many grams of a solution which is 2.5% in glucose (gfw=180.0) must be taken in order to obtain 15grams of glucose?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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D. What is the weight percent of sulfuric acid in a solution prepared by adding 14 grams of sulfuric acid to432 milliliters of water?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------E. What is the weight percent of sodium chloride in a solution prepared by dissolving 32 grams of the substance in

72 grams of water?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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✔✔✔■Concentration Units: Weight Percent (continued)5b) F. How many grams of glucose (gfw = 180.0) must be added to 35 grams of water to obtain a solution which is

41% glucose by weight?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

---------------------------------------------------------------------------------------------------------------------------------------------✔✔✔■Concentration Units: Molali ty6a) Define molality. (p. 512)

----------------------------------------------------------------------------------------------------------------------------------------------b) Perform the indicated calculation.

S. How many grams of sodium sulfide must be combined with 50.0 grams of water to obtain a 0.20 molalsolution of the substance?•Formula: Na2S •Type: salt •Strength: strong

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Na2S | m = mol solute/kg solvent2Na 2*23.0 = 46.0 | m = mol Na2S/kg waterS 32.1 = 32.1 | m = (g/mw)/kg solvent

78.1 g/mol | g = (kg solvent)(mw)(m)| g = (0.0500 kg)(78.1 g/mol)(0.20 molal) =

kg water = (50.0 g H2O)(1 kg/1000 g) = | Rkg water = 0.0500 kg H2O | g = 0.781 g Na2S = 0.78 g Na2S

----------------------------------------------------------------------------------------------------------------------------------------------A. Calculate the molality of a solution prepared by dissolving 2.5 grams of a potassium chlorate in 512 grams of

water.•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------B. How many grams of sodium nitrate must be combined with 213 grams of water to obtain a 0.40 molal

solution of the substance?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

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✔✔✔■Concentration Units: Molality (continued)6b) C. Calculate the molality of a solution prepared by dissolving 4.5 grams of naphthalene (C10H8) in 50.0 grams

of benzene (C6H6)•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------D. How many grams of iodine must be combined with 159 grams of ethanol (C2H5OH) in order to obtain a

0.30 m solution of the substance?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------E. What is the molality of a solution prepared by adding 13 grams of sodium iodide to 835 grams of water?


----------------------------------------------------------------------------------------------------------------------------------------------F. How many grams of iodine must be added to 52 grams of carbon tetrachloride to produce a solution which is

0.47 m in iodine?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■Concentration Units: Converting from One to Another7. Perform the conversion between concentration units indicated.

Algorithm for solving problems which convert from one type of concentration unit to another:❶Divide your paper into two columns.❷Write the given concentration value and its definition in the left column under the heading “Given”.❸If the given concentration unit is %, assume you have “100 grams of solution”. If the given

concentration unit is anything else, assume you have “1” of the unit in the denominator of theconcentration unit. In either case, write your assumption on paper. (Note: You will get the sameanswer if you assume ANY amount of the substance in the denominator of the concentration unit. Theassumptions here are merely the easiest ones to use.)

❹Write down the amount of solute associated with the amount of solvent/solution you assumed in step 3.(Note: This equals the value of the given concentration. The units are the same as those in thenumerator of the given concentration.)

❺Write the desired concentration units and its definition on the right under the heading “Desired”.❻Write any other information you are given (such as density of the solution) in the “Given” column.❼Use the data in your “Given” column to calculate the quantities needed to obtain the concentration unit

you desire.S1. What is the mole fraction of sodium chloride in a solution which is 25% sodium chloride by weight?

•Formula: NaCl •Type: salt •Strength: strong••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Given: |Desired:❷ Concentration = 25% NaCl |❺ mole fraction NaCl = ?

% NaCl = g NaCltotal g * 100% | mole fraction NaCl =

mol NaCltotal mol

❸ Assume have 100 grams soln |❼ Need moles Na and moles H2O

❹ Therefore, have | mol NaCl = 25 g NaCl * 1 mol NaCl

58.5 g NaCl =

25 grams NaCl | = 0.427 mole NaCl

75 grams H2O | mole H2O = 75 g H2O * 1 mol H2O

18.0 g H2O =

and know that | = 4.17 mol H2O

gfw NaCl = 35.5 + 23.0 = 58.5 g/mol | XNaCl = 0.427 mol NaCl

4.17 + 0.427 mol total =

| Rgfw H2O = 2*1.0 + 16.0 = 18.0 g/mol | XNaCl = 0.09295 = 0.093

----------------------------------------------------------------------------------------------------------------------------------------------S2. What is the molarity of a solution of H2SO4 that is 4.3% H2SO4 if the specific gravity of the solution is 1.21?

•Formula: H2SO4 •Type: acid •Strength: strong

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••Given: |Desired:❷ Concentration = 4.3% H2SO4 |❺ molarity of H2SO4 = ?

% H2SO4 = g H2SO4

total g * 100% | molarity of H2SO4 = mol H2SO4

L sol’n❸ Assume have 100 grams soln |❼ Need moles H2SO4 and liters sol’n

❹ Therefore, have | mol H2SO4 = 4.3 g H2SO4 *1 mol H2SO4 =4.3 grams H2SO4 | 98.1 g

| = 0.0438 mol H2SO4and know that | L sol’n = 100 g * 1 mL * 1 L = 0.0826 L

l 1.21 g 1000mLgfw H2SO4= 2*1.0 + 4*16.0 + 32.1 l

= 98.1 g/mol | M of H2SO4 = 0.0438 mol H2SO4| 0.0826 L sol’n

❻ specific gravity = 1.21 | Rtherefore, density = 1.21 g/mL | = 0.53 M

----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■Concentration Units: Converting from One to Another (continued)7. A. What is the weight percent of calcium acetate in a solution whose mole fraction of calcium acetate is 0.40?


----------------------------------------------------------------------------------------------------------------------------------------------B. What is the weight % of sodium sulfate in a 0.500 M solution whose specific gravity1 is 1.12?


----------------------------------------------------------------------------------------------------------------------------------------------C. What is the molarity of a solution whose concentration is 0.200 m and whose specific gravity1 is 1.10? GFW = 153.


----------------------------------------------------------------------------------------------------------------------------------------------

1Specific gravity of X = density of X

density of H2O . Therefore, at room temperature where density of H2O = 1.00 g/mL, the

specific gravity of a substance has the same value as its density.



✔✔✔■Concentration Units: Converting from One to Another (continued)7. D. What is the molality of a 25% solution of ethanol (C2H5OH)?


E. What is the molality of a solution of methanol (CH3OH) in water in which the mole fraction of methanol is 0.23?•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------F. What is the mole fraction of sodium chloride in a 1.8 M solution with a specific gravity1 of 1.17?

•Formula: •Type: •Strength:••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------






✔✔✔■Concentration Units: Converting from One to Another (continued)7. G. What is the molarity of a solution of lithium chloride in water in which the mole fraction of lithium chloride is

0.036? The specific gravity of the solution is 1.21.•Formula: •Type: •Strength: ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------H. What is the mole fraction of urea (gfw = 60.0) in a solution that is 0.83m?

•Formula: •Type: •Strength:••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

----------------------------------------------------------------------------------------------------------------------------------------------I. What is the molality of 0.95 M solution of potassium chloride with a specific gravity1 of 1.02?


----------------------------------------------------------------------------------------------------------------------------------------------






✔✔✔■Concentration Units: Converting from One to Another (continued)7. J. What is the weight percent of sodium bromide in a 0.56 m solution of sodium bromide?


----------------------------------------------------------------------------------------------------------------------------------------------K. What is the molarity of a solution of HCl that is 3.6% HCl if the specific gravity1 of the solution is 1.13?


----------------------------------------------------------------------------------------------------------------------------------------------






■An Overview of Factors Which Affect Solubil i ty; Predicting Whether or Not a Given SubstanceWil l Likely Dissolve in Water8a) List three factors which affect solubility and state how each generally does so. (Section 11.3 in text)

Factor | How it generally affects solubility |||

| | For solids in water:

Temperature || For gases in water:

| |||

|

b) Variation of solubili ty as the nature of the solvent and solute varies: Indicate the extent to whicheach of the substances will dissolve in water and explain your answer. (Section 11.3 in text)S. CH3CH2CH2OH

The "CH3CH2CH2" part of the molecule is nonpolar because "C" and "H" have about the sameelectronegativity. On the other hand, the "OH" part is polar because "O" has a mucher higherelectronegativity than the "C" or the "H" it is bonded to. The nonpolar "CH3CH2CH2" part wouldtend NOT to dissolve in H2O whereas the polar "OH" WOULD tend to dissolve in H2O.Whether or not the molecule overall is soluble in water depends on the relative size of the nonpolar andpolar parts. In this case, the nonpolar part is relatively small so this substance is probablymoderately soluble in water.

----------------------------------------------------------------------------------------------------------------------------------------------A. CH3CH2CH3

----------------------------------------------------------------------------------------------------------------------------------------------B. CH3CH2CH2CH2CH2CH2CH2CH2CH2OH

----------------------------------------------------------------------------------------------------------------------------------------------



■An Overview of Factors Which Affect Solubil i ty; Predicting Whether or Not a Given SubstanceWill Likely Dissolve in Water (continued)8b) Variation of solubili ty as the nature of the solvent and solute varies: Indicate the extent to which

each of the substances will dissolve in water and explain your answer.C. CH3CH2-O-CH2CH3

----------------------------------------------------------------------------------------------------------------------------------------------D. CH3CH2NH2

----------------------------------------------------------------------------------------------------------------------------------------------E. CH2 CH CH CH2

| | | |OH OH OH OH

----------------------------------------------------------------------------------------------------------------------------------------------F. CH3CH2OH

----------------------------------------------------------------------------------------------------------------------------------------------



■More Detail About the Temperature Dependence of Solubili ty9. Answer the following regarding the solubility curve below. (Section 11.3 in text)

|60_ |

S (grams |o 50_ |l of |u 40_ |b solute |i 30_ |l per |i 20_ |t 100 g |y 10_ |

of water) |0_ | . . . .

0 20 40 60 80temp (oC) -->

S. a) What is the solubility of NaCl at 20oC?≈ 40 g/100 g H2O

b) What happens to a solution with a concentration of KNO3 of 30 grams/100 grams of water if it is

cooled from 70oC to 30oC and no supersaturation occurs? All is dissolved at 70oC since thesolubility at that temperature is >60 g/100 H2O. At 30oC, only 24 g dissolves in 100 g H2O.Therefore, 30 g - 24 g = 6 grams KNO3 must precipitate when the substance is cooled.

----------------------------------------------------------------------------------------------------------------------------------------------A. a) What is the solubility of KNO3 at 10oC?

b) What is the lowest T at which 45g of KNO3 dissolves in 100g of water?

----------------------------------------------------------------------------------------------------------------------------------------------B. a) Which salt is least soluble at 60oC?

b) What happens if 1 gram of KNO3 is added to a solution at 55oC whose concentration is 25 g/100ml ofwater?

----------------------------------------------------------------------------------------------------------------------------------------------C. a) What is the lowest T at which 40 grams of KNO3 will dissolve in 100 g of water?

b) Describe what happens if a solution prepared by adding 25 g of NaCl and 25 g of KNO3 to 100 ml of

water at 40oC is cooled to 5oC?

----------------------------------------------------------------------------------------------------------------------------------------------



■Temperature Dependence of Solubility (continued)9. D. a) What is the solubility of KNO3 at 40oC?

b) How many grams of KNO3 must be added to 100 grams of water at 50oC to obtain a saturatedsolution?

----------------------------------------------------------------------------------------------------------------------------------------------E. a) What is the lowest T at which 55 grams of KNO3 will dissolve in 100 g of water?

b) How many grams of NaCl must be added to 100 grams of water at 65oC to obtain a saturatedsolution?

----------------------------------------------------------------------------------------------------------------------------------------------

✔✔✔■More Detail about Pressure Dependence of Solubility (Henry's Law)10a) State Henry's Law indicating the relationship between the pressure of a gas and its solubility. (p. 521)

----------------------------------------------------------------------------------------------------------------------------------------------b) S. If the solubility of carbon dioxide in water at 25oC and 402 torr of carbon dioxide is 5.69 M, what is the

solubility of carbon dioxide in water at 25oC if the pressure of carbon dioxide is doubled to 804 torr?Since the change in pressure is an easy multiple of the original pressure, you can find the answereasily by remembering that the solubility of a gas in a liquid is directly proportional to the pressure ofthat gas above the liquid. Therefore, if the pressure of carbon dioxide above water doubles, thesolubility of carbon dioxide in water will also double. That is, solubility = 2*5.69 M = 11.38 M whichrounds to 11.4 M.

----------------------------------------------------------------------------------------------------------------------------------------------A. If the solubility of carbon dioxide in water at 25oC and 402 torr of carbon dioxide is 5.69 M, what is the

solubility of carbon dioxide in water at 25oC if the pressure of carbon dioxide is halved to 201 torr?

----------------------------------------------------------------------------------------------------------------------------------------------B. If the solubility of carbon dioxide in water at 25oC and 402 torr of carbon dioxide is 5.69 M, what pressure of

carbon dioxide would be required to cause the solubility of carbon dioxide to triple?

---------------------------------------------------------------------------------------------------------------------------------------------- C. If the solubility of nitrogen in water at 25oC and 1.5 atm nitrogen is 1.1*10-2 M, what pressure of nitrogen is

required to cause the solubility of nitrogen to decrease to 1.1*10-3 M?

----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■More Detail about Pressure Dependence of Solubility (Henry's Law, continued)10b) D. If the solubility of nitrogen in water at 25oC and 1.5 atm nitrogen is 1.1*10-2 M, what is the solubility of

nitrogen in water at 25oC if the pressure of nitrogen is tripled to 4.5 atm?

----------------------------------------------------------------------------------------------------------------------------------------------E. If the solubility of nitrogen in water at 25oC and 1.5 atm nitrogen is 1.1*10-2 M, what is the solubility of

nitrogen in water at 25oC if the pressure of nitrogen is quadrupled to to 6.0 atm?

----------------------------------------------------------------------------------------------------------------------------------------------c) The solubility of carbon dioxide in water at 25oC and 402 torr of carbon dioxide is 5.69 M while that of

nitrogen at the same temperature and 1.5 atm nitrogen is 1.1*10-2 M. Use this information to answer thefollowing. Explain all answers in detail.S. What is the solubility of CO2 at 25oC when the pressure of CO2 is 567 torr?

Henry’s Law (Cg = k*Pg ) tells us that the solubility of a gas in a liquid is directly proportional to thepressure of the gas above the liquid. Therefore, the problem can be worked using ratio-and-proportion as follows:

5.69 M402 torr =

x567 torr

x = (5.69M

402 torr) * 567 torr = 8.025 M which rounds to 8.03 M

----------------------------------------------------------------------------------------------------------------------------------------------A. What is the solubility of N2 at 25oC when the pressure of N2 is 2.9 atm?

----------------------------------------------------------------------------------------------------------------------------------------------B. What pressure of CO2 must be maintained at 25oC to raise the concentration of CO2 to 8.21 M?

---------------------------------------------------------------------------------------------------------------------------------------------- C. What is the solubility of carbon dioxide at 25oC if the pressure is 872 torr?

----------------------------------------------------------------------------------------------------------------------------------------------D. What pressure of N2 must be maintained at 25oC to obtain a solubility of N2 of 8.5*10-3 M?

----------------------------------------------------------------------------------------------------------------------------------------------



✔✔✔■More Detail about Pressure Dependence of Solubility (Henry's Law, continued)10c) E. What pressure of N2 must be maintained at 25oC to obtain a solubility of N2 of 5.3*10-4 M?

----------------------------------------------------------------------------------------------------------------------------------------------F. What is the solubility of carbon dioxide at 25oC under a carbon dioxide pressure of 4.3 atm?

----------------------------------------------------------------------------------------------------------------------------------------------■Bonding/Model Act iv i ty to Improve Abi l i ty to Visual ize in 3-D

Algorithm to write the electron-dot (Lewis)structure of covalent species (from Mod 8A,formal charged reviewed on pages just beforethis module in Handbook.)

Example: HNO2

❶ Draw the “skeleton” of the chemical species showinga single bond between each atom. If the species is apolyatomic ion, enclose it in brackets [] and put the chargeon it.

H - O - N - ONote: In most oxyacids, the acidic “H” is attached to an“O” atom in the same manner show in this skeleton.

❷ Add up the valence electrons, then add or subtractelectrons to compensate for the charge on the species (ifany).

Valence electrons on H = 1Valence electrons on N = 5Valence electrons on 2 O’s = 12

18❸ Subtract two electrons for each bond in the skeleton. There are two bonds in the skeleton and each bond

contains 2 electrons so subtract 6 electrons from the 18to get (18 - 6) = 12 .

❹ Add the other electrons so that each atom has a filledouter shell (8 electrons around all atoms except H). If thisrequires more electrons than are left in the valence e- pool,put in multiple bonds, one for each pair of e-'s you areshort. If you have electrons left over after you fill allouter shells, put them around the largest atom (which isusually the central one).

Now put the 12 electrons in to get: _ _

H - O - N - O|“N” atom does not have 8 electrons so shift to get _

H - O - N = O|All except “H” now have 8 electrons.

❺ Check your structure by adding up all of the electronsshown to make certain the total number electrons showingin the structure = total number of valence electrons (plusadjustment for charge if any).

There are 10 pairs of electrons, 20 total, in the structureabove. Since this is the same number of valenceelectrons as obtained in #2 above and everything has afilled shell, the structure is okay.

❻Write resonance structures if needed. No resonance possible for this structure since "X=O-H"structures are not stable.

❼Calculate the formula charge on each atom (# electronson a neutral atom - # electrons in lone pairs on atom -0.5*# electrons in bonds attached to atom). If not zero,write the formal charge above the atom enclosed in a smallcircle. Then, if resonance exists , STOP.

All formal charges are zero for this example so you have _

H - O - N = O|

❽If there is no resonance, look to see if the central atomcan have an expanded octet (i.e. if it is an element in the3rd or higher rows). If not , STOP.

NA

❾If an atom can have an expanded octet, see if thenumber and size of the formal charges can be reduced byexpanding the octet, if so shift them and recalculate formalcharges. IF NOT, STOP.

NA

❿If Lewis structure was changed in ❾ above, check againfor resonance structures.

NA



II. Formal Charges:■Using Formal Charges to Predict Most Stable Lewis Structures(Review from Module 8A)a) Define formal charge. (p. 385) The difference between the number of valence electrons on a free atom and the number

of electrons “assigned” to the atom in the molecule or ion.-----------------------------------------------------------------------------------------------------------------------------------------------

b) What two things does one need in order to calculate formal charges? (p. 386)• number of valence electrons on the free atom• number of electrons “belonging” to atom in a molecule or ion-----------------------------------------------------------------------------------------------------------------------------------------------

c) What are the two rules needed to determine which electrons “belong” to a given atom in a species? (p. 386)(Each “lone pair” of electrons contributes 2 electrons and each shared pair contributes 1 electron.Therefore, the number of electrons “belonging to” an atom for formal charges = 2*# lone pairs + # sharedpairs on the atom.)-----------------------------------------------------------------------------------------------------------------------------------------------

d) What is the formula used to calculate the formal charge on a given atom in a species? (p. 386)formal charge = number of valence electrons on free atom - 2*# lone pairs - # shared pairs on the atom)-----------------------------------------------------------------------------------------------------------------------------------------------

e) What is the relationship between formal charges and the charge on a chemical species? (p. 378) sum of formal charges of all atoms in a chemical species = overall charge on chemical species

-----------------------------------------------------------------------------------------------------------------------------------------------g) What is the overall rule for using formal charges? (p. 381, The most stable Lewis structures are those with a) the

formal charges closest to zero and b) negative formal charges on the the more electronegative atoms.)-----------------------------------------------------------------------------------------------------------------------------------------------

h) Calculate the formal charge on each atom in the following Lewis structure.•f.c. H = 1 val e-

- 0 lp e- ‘s - ( 1shared pair) = 0•f.c. C = 4 val e-

- 0 lp e- ‘s - (4shared pairs) = 0•f.c. O = 6 val e-

- 2*2 lp - (2shared pairs ) = 0

C OH H

H

H 0 0

0

0 0

0

√ Sum of formal charges = 0,the charge on the species

i) When writing Lewis structures, it is best to go through the following scheme:• Draw the Lewis structure so that every atom has an octet (except H).• If there is resonance at this point, draw the resonance structures and stop.• If there is no resonance, see if any atom can have an expanded octet. If not, stop .• If an atom can have an expanded octet, see if the number and size of the formal charges can be reduced by expanding

the octet. If not, stop .• If the answer to the above is yes, do it and then check again for resonance structures.

Write the BEST Lewis structure for the following species using formal charges to evaluate various possibilities. Hint:Start by writing it like we would have done above and then try expanded shells. (pp. 378-81)

S. ClO2-1Cl = 1*7 = 71O = 1*6 = 61O = 1*6 = 6charge = 1 20 -4 16

This is the Lewis structure you would get without considering expanded shells.

It probably isn't as good as the following which uses the "d" orbitals on Cl because the following has smaller formal charges. Both the following and that above have negative formal charges on the more electronegative element "O".

ClO O[ ]- -1 -1 +1

ClO O[ ]- 0 0 -1 0 0 -1

ClO O[ ]-The second of the structures (the one with expanded shells and resonance) is more likely because it hassmaller charges overall.



■Bonding/Model Activi ty to Improve Abil i ty to Visualize in 3-D (continued)A. a) Draw the Lewis structure of nitric acid.

----------------------------------------------------------------------------------------------------------------------------------------------B. a) Draw the Lewis structure of chloric acid.

----------------------------------------------------------------------------------------------------------------------------------------------C. a) Draw the Lewis structure of sulfuric acid.

----------------------------------------------------------------------------------------------------------------------------------------------D. a) Draw the Lewis structure of sulfurous acid.

----------------------------------------------------------------------------------------------------------------------------------------------E. a) Draw the Lewis structure of acetic acid.

----------------------------------------------------------------------------------------------------------------------------------------------



■Bonding/Model Activi ty to Improve Abil i ty to Visualize in 3-D (continued)F. a) Draw the Lewis structure of phosphoric acid.

----------------------------------------------------------------------------------------------------------------------------------------------A-F b) Then use your model set to assemble a model of the species.

■Challenge QuestionsA. A saturated solution of potassium nitrate at 50oC is cooled to 30oC. How much KNO3 precipitates?

Refer to the diagram below.|

60_ |S (grams |o 50_ |l of |u 40_ |b solute |i 30_ |l per |i 20_ |t 100 g |y 10_ |

of water) |0_ | . . . .

|0 20 40 60 80temp (oC) -->

B. Thirty one grams of KNO3 is placed in 100.0 ml of water at 10oC and the mixture is stirred until solute stopsdissolving. How many grams of solute remain undissolved? Use the diagram above.

C. 42 grams of KNO3 are placed in 100.0 grams of water at 20oC and allowed to come to equilibrium. What percent ofthe KNO3 dissolves? Use the diagram above.

D. The solubility of CuS in water is 9.3*10-23 M. Calculate the solubility of CuS in grams CuS/100 grams of water.Make any reasonable assumptions.

E. The solubility of PbSO4 is 1.0*10-4 M. Calculate the solubility of PbSO4/100 grams of water.Make any reasonable assumptions.

F. 30.0 grams of KNO3 is added to 100 grams of water and the mixture is heated to 60oC so that all solute dissolves.

The solution is then cooled to 10oC. How much KNO3 precipitates from the solution as it iscooled? Refer to the diagram above.

Note: All parts of this module must befilled out when you arrive at drill. Inaddition, you should also complete thefollowing skills module (logarithms) andbe ready to take a quiz on both theregular and skills modules.Revised by MA, 5/7/2002, 5/5/2003


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