Chapter 9

Basics of Solution Stoichiometry Part D Fe2+(aq) + 2OH−(aq) → Fe(OH)2(s) If you had a 0.400L solution containing 0.0160M of Fe2+(aq), and you wished to add enough 1.27M NaOH(aq) to precipitate all of the metal, what is the minimum amount of the NaOH(aq) solution you would need to add? Assume that the NaOH(aq) solution is the only source of OH−(aq) for the precipitation.

0.0101 L ((0.400L*0.0160M)*(2mol OH))/1.27M NaOH = 0.01007

Basics of Solution Stoichiometry Part B For the described reaction, you have a 0.360L solution of 0.820M Ba3(PO4)2(aq). Using the solution map from part A as a guide, complete the following dimensional analysis that allows you to calculate the moles of BaSO4(s) that can be formed by placing the values of each conversion factor according to whether they should appear in the numerator or denominator.

0.360 L solution (0.820 mol Ba3(PO4)2(aq)/1L solution) x (3 mol BaSO4(s)/ 1 mol Ba3(PO4)2(aq)) = 0.8856 mol BaSO4(s)

Problem 8.15 Part A 1mol K+

1 Eq 1=1

Introduction to Solubility and Solution Formation Part B Steps in the solution formation process

1. NaCl crystal is added to water 2. The Na ions are attracted to the water molecules AND The Cl ions are attracted to the water molecules 3. NaCl solution

Part B What is the final volume V2 in milliliters when 0.793L of a 30.6% (m/v) solution is diluted to 22.5% (m/v)?

1080mL (0.793*30.6)/22.5=1.08L

Part C A 595mL NaClNaCl solution is diluted to a volume of 1.30L and a concentration of 8.00M . What was the initial concentration C1?

17.5 M (8.00*1300)/595=17.47

± Calculating Percent Concentration Part A Calculate the mass percent of a solution that is prepared by adding 58.9g of NaOH to 251g of H2O.

19.0%(m/m) 58.9g/(58.9g+251g) x100=19

Problem 8.34 Part C 48.5 g of Na2CO3 in 250.0 g of Na2CO3 solution

19.4% (48.5/250)x100

Problem 8.34 Part A Calculate the mass percent (m/m) for the solute in each of the following solutions. 74g of NaOH in 375g of NaOH solution

20% (74/375)x100=19.7

± Calculating Percent Concentration Part B Calculate the mass/volume percent of a NaCl solution in which 183g of NaCl is dissolved in enough water to give a total volume of 4.78L .

3.83%(m/v) 4.78L=4780g (183/4780)x100=3.83%

Problem 8.15 Part D 2mol CO₃²⁻

4 Eq 2x2=4

Problem 8.15 Part B 4mol OH⁻

4 Eq 4x1 = 4

Problem 8.41 A mouthwash contains 22.5{\rm \\%} (v/v) alcohol If the bottle of mouthwash contains 335mL , what is the volume, in milliliters, of the alcohol?

75.4 mL (0.225*335)=75.375

Problem 8.15 Part C 4mol Ca²⁺

8 Eq 4x2=8

Problem 8.34 Part B 2.0g of KOH in 23.0g of H2O

8.0% (2/23+2)x100= 8.0

Basics of Solution Stoichiometry Part C The following dimensional analysis setup could be used to determine the theoretical mass of AlBr3(s) (molecular mass = 266.69g/mol ) produced based on reacting 86.4g of a 0.083mol/L solution of Br2(l) (density = 1032g/L ) with excess Al(s) as described in the following equation: 3Br2(l) + 2Al(s) → 2AlBr3(s) Complete the dimensional analysis for calculating the mass of the product by placing the values of each conversion factor according to whether they should appear in the numerator or denominator when calculating the mass of AlBr3(s) produced from a sample of Br2(l).

86.4g Br2(l) x 1 L solution/1032g Br2(l) x 0.083mol Br2(l)/1L solution x 2mol AlBr3(s)/3mol Br2(l) x 266.69g AlBr3(s)/1mol AlBr3(s)

Animation - Evaluating Electrolytes and Nonelectrolytes Part C The student is told that one of the four solids tested is actually aluminum chloride,AlCl3. Which of the solids could be AlCl3? A. Bright B. Dim C. None D. Bright

AlCl3 is a strong electrolyte therefore A and D is the answer

Animation - Evaluating Electrolytes and Nonelectrolytes Part C Which of the solids is a weak electrolyte when dissolved in water? A. Bright light B. Dim C. None D. Bright

B. Dim

Solubility of Gases and Solids in Water Part B Imagine that you mix 50 g of a solute with 100 g of water at 40 ∘C. Classify each solute by whether it would completely or partially dissolve under these conditions.

Completely dissolved: KNO3, NaNO3, Pb(NO3)2 Partially dissolved: KClO3, K2Cr2O7, KCl

Introduction to Osmosis Part D A red blood cell is placed into each of the following solutions. Indicate whether crenation, hemolysis, or neither will occur. Solution A: 1.75% (m/v) NaCl Solution B: 0.57% (m/v) glucose Solution C: distilled H2O Solution D: 6.13% (m/v) glucose Solution E: 5.0% (m/v) glucose and 0.9% (m/v) NaCl

Crenation: A, D, E Hemolysis: C, B

Introduction to Solubility and Solution Formation Part C Factors affecting the dissolution process

Faster dissolution process: NaCl is added to water and solution is heated NaCl added to water and solution is stirred Finely powdered NaCl salt is added to water Slower dissolution process: A large piece of NaCl is added to water NaCl salt is added to water and solution is cooled

Introduction to Osmosis Part C A red blood cell has been placed into each of three different solutions. One solution is isotonic to the cell, one solution is hypotonic to the cell, and one solution is hypertonic to the cell. Sort each beaker into the appropriate bin based on the cell's reaction in each solution.

Hypertonic H20 out Isotonic balance Hypotonic h2o in

Solubility of Gases and Solids in Water Part A Determine whether each of the following solutes is more soluble in hot water or cold water.

More soluble in hot water: sodium chloride, fructose More soluble in cold water: (gasses are more soluble in cold water) oxygen, carbon dioxide, nitrogen

Strength of Electrolytes Part A Drag the appropriate chemical equations to their respective bins.

Nonelectrolyte: A(g)→A(g) AB(s) ← A(aq) + B(aq) Weak Electrolyte: AB(aq) ⇌ A(aq) + B(aq) AB(aq) + CD(l) ⇌ AD(aq) + BC(aq) Strong Electrolyte: AB(s) → A(aq) + B(aq)

Problem 8.57 Part D a 6.0{\rm \\%} (m/v) glucose solution prepared from 95mL of a 13{\rm \\%} (m/v) glucose solution

V= 210mL (95*13)/6=205.8333

Problem 8.57 Part A a 0.800M HCl solution prepared from 30.0mL of a 6.00M HCl solution

V= 230 mL (6.00*30.0)/0.800=225 mL V₂=M₁V₁/M₂

Problem 8.57 Part B a 2.0{\rm \\%} (m/v) LiCl solution prepared from 10.0mL of a 10.0{\rm \\%} (m/v) LiCl solution

V= 50mL (10*10)/2=50mL V₂=M₁V₁/M₂

Problem 8.57 Part C a 0.500M H3PO4 solution prepared from 60.0mL of a 6.00M H3PO4 solution

V= 720mL (60*6)/0.5 = 720

Strength of Electrolytes Part C Rank from weakest to strongest electrolyte Amount of acid that dissociates from 1.00mol H₃AsO₄ 6.82x10⁻² mol HIO₃ 1.53x10⁻¹ mol H₂S 3.60x10⁻⁴ mol H₂Se 3.54x10⁻² mol

Weakest H₂S 3.60x10⁻⁴ mol H₂Se 3.54x10⁻² mol H₃AsO₄ 6.82x10⁻² mol HIO₃ 1.53x10⁻¹ mol Strongest

Problem 8.2 Part C 1.0mL of Br2 and 150.0mL of methylene chloride a. Bromine is the solute, methylene chloride is the solvent. b. Methylene chloride is the solute, bromine is the solvent.

a. Bromine is the solute, methylene chloride is the solvent.

Dialysis Part A Each of the following is placed in a dialyzing bag and the bag is then immersed in distilled water. Which substances will be found outside the bag? a. NaCl solution b. albumin solution c. starch solution

a. NaCl solution

Dialysis Part B Each of the following is placed in a dialyzing bag and the bag is then immersed in distilled water. Which substances will be found inside the bag? a. NaCl solution b. albumin solution c. starch solution

a. NaCl solution b. albumin solution c. starch solution

Introduction to Osmosis Part A A semipermeable membrane is placed between the following solutions. Which solution will decrease in volume? a. Solution A: 1.6% (m/v) starch b. Solution B: 4.26% (m/v) starch

a. Solution A: 1.6% (m/v) starch

Introduction to Solubility and Solution Formation Part A The properties of a solute and its solubility in liquid solvents a. Table salt dissolves in water to form a solution. b. Zinc sulfate is soluble in water. c. Carbon dioxide gas will be more soluble in water when the partial pressure is high. d. Ethanol and hexane are miscible. e. Gases are readily soluble in water at higher temperature

a. Table salt dissolves in water to form a solution. b. Zinc sulfate is soluble in water. c. Carbon dioxide gas will be more soluble in water when the partial pressure is high.

Problem 8.2 Part A Identify the solute and the solvent in each solution composed of the following. 5mL of acetic acid and 200mL of water a. Water is the solvent, acetic acid is the solute. b. Acetic acid is the solvent, water is the solute

a. Water is the solvent, acetic acid is the solute.

Problem 8.23 Part C Adding 471g sugar to 134g H2O a. saturated solution b. unsaturated solution

a. saturated solution

Problem 8.75 Part C NaCl solution and starch (colloid) a. sodium and chloride ions b. sodium ions and starch molecules c. chloride ions and starch molecules d. distilled water only

a. sodium and chloride ions

Problem 8.75 Part D urea solution a. urea molecules b. distilled water only

a. urea molecules

± Dilution Part A a. C2=V2/C1V1 b. C2=C1V1/V2 c. C2=V1V2/C1 d. C2=C1V1V2

b. C2=C1V1/V2

Reaction Types and Stoichiometry Part B Which reaction is a decomposition reaction? a. C2H4(g) + O2(g) → CO2(g) + H2O(g) b. NH4NO3(s) → N2O(g) + H2O(g) c. CO(g) + 12O2(g) → CO2(g) d. Li(s) + N2(g) → Li3N(s)

b. NH4NO3(s) → N2O(g) + H2O(g) C → A +B

Introduction to Osmosis Part B A semipermeable membrane is placed between the following solutions. Which solution will increase in volume? a. Solution C: 7.02% NaCl b. Solution D: 15.0% NaCl

b. Solution D: 15.0% NaCl

Part D You have a large amount of 9.00M stock solution. You need 1.40L of 3.00M solution for an experiment. How would you prepare the desired solution without wasting any stock solution? a. Start with 1.40L of the stock solution. Add water until you reach a total volume of 4.20L . b. Start with 0.467L of the stock solution. Add water until you reach a total volume of 1.40L . c. Start with 1.40L of water. Add stock solution until you reach a total volume of 2.10L . d. Mix 0.700L each of stock solution and water.

b. Start with 0.467L of the stock solution. Add water until you reach a total volume of 1.40L . Concentrated stock solutions are commonly used in laboratories for the initial solution in a dilution process. For instance, a laboratory might have a 12.0 M HCl solution that can be used to make less concentrated HCl solutions.

Problem 8.2 Part B 200.0g of water and 5.0g of sugar a. Sugar is the solvent, water is the solute. b. Water is the solvent, sugar is the solute.

b. Water is the solvent, sugar is the solute.

Problem 8.75 Part B starch (colloid) and alanine (amino acid) solution a. starch molecules b. alanine molecules c. starch and alanine molecules d. distilled water only

b. alanine molecules

Problem 8.23 Part A Adding 23.0g KCl to 100g H2O a. saturated solution b. unsaturated solution

b. unsaturated

Problem 8.23 Part B Adding 12.0g NaNO3 to 36.1g H2O a. saturated solution b. unsaturated solution

b. unsaturated (100/36.1g)(12.0g)<88.0g

Animation - Evaluating Electrolytes and Nonelectrolytes Part A Watch the animation, then check off the samples that will conduct electricity. a. Sugar solution b. Solid Sugar c. NaCl solution d. NaCl solid

c. NaCl solution

Dialysis Part C Consider a mixture of NaCl, albumin solution, and starch solutions in a dialyzing bag. The dialyzing bag is then immersed in distilled water. Which of the following actions should be taken to remove the most salt from the mixture? a. Leave the dialyzing bag in the distilled water for an extended period of time. b. Replace the dialyzing bag periodically. c. Replace the water with fresh distilled water periodically.

c. Replace the water with fresh distilled water periodically.

Problem 8.75 Part A NaCl solution a. sodium ions b. chloride ions c. sodium and chloride ions d. distilled water only

c. sodium and chloride ions

Reaction Types and Stoichiometry Part A Which of the following general reactions appropriately models a combination reaction? a. A + B → A + B b. C → A + B c. A + B → C + D d. A + B → C

d. A + B → C

Strength of Electrolytes Part C light bulb produced a

dim/flickered glow = weak electrolyte bright glow = strong electrolyte light bulb remained off = nonelectrolyte

Basics of Solution Stoichiometry Part A Into a 0.25 M solution of Ba3(PO4)2(aq), excess Na2SO4(aq) was added to form BaSO4(s). Ba3(PO4)2(aq) + 3Na2SO4(aq) → 3BaSO4(s) + 2Na3PO4(aq) If you knew the volume of the solution containing Ba3(PO4)2(aq), determine how you would predict the mass of BaSO4(s) formed by completing the following solution map.

volume Ba3(PO4)2(aq) → moles Ba3(PO4)2(aq) → moles BaSO4(s) → mass BaSO4(s)