Chem I Homework Page, Exam 2 Material

Homework Page with Visible Answers

This page has all of the required homework for the material covered in the second exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12^th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.

Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.

These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.

Aqueous Reactions (Chapter Four)

1. What ions will be present when each of the following substances is placed in water?
   • BaCl₂

   Answer

   Ba²⁺ and Cl^-

   • AgCl

   Answer

   AgCl is considered insoluble, so our answer would be that no ions are formed.

   • ZnSO₄

   Answer

   Zn²⁺ and SO₄^2-

   • HNO₃

   Answer

   H⁺ and NO₃^-

   • Na₂CO₃

   Answer

   Na⁺ and CO₃^2-

   • BaSO₄

   Answer

   BaSO₄ is considered insoluble, so we say that no ions are formed.

   • NaOH

   Answer

   Na⁺ and OH^-

2. What will be dissolved in the solution when each of the following weak acids are placed in water?
   • Hypochlorous acid, HClO

   Answer

   H⁺, ClO^- and HClO

   • Propionic acid, C₂H₅COOH or HC₃H₅O₂

   Answer

   H⁺, C₂H₅COO^- and C₂H₅COOH

3. Write the molecular, ionic and net ionic equations when the following substances are placed together in an aqueous solution.
   • CaBr₂ and K₂CO₃

   Answer

   CaBr₂(aq) + K₂CO₃(aq) → CaCO₃(s) + 2KBr(aq)
   
   Ca²⁺(aq) + 2Br^-(aq) + 2K⁺(aq) + CO₃^2-(aq) → CaCO₃(s) + 2K⁺(aq) + 2Br^-(aq)
   
   Ca²⁺(aq) + CO₃^2-(aq) → CaCO₃(s)

   • CdSO₄ and Ba(C₂H₃O₂)₂

   Answer

   CdSO₄(aq) + Ba(C₂H₃O₂)₂(aq) → BaSO₄(s) + Cd(C₂H₃O₂)₂(aq)
   
   Cd²⁺(aq) + SO₄^2-(aq) + Ba²⁺(aq) + 2C₂H₃O₂^-(aq)
       → BaSO₄(s) + Cd²⁺(aq) + 2C₂H₃O₂^-(aq)
   
   Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s)

   • HNO₃ and PbCO₃

   Answer

   2HNO₃(aq) + PbCO₃(s) → CO₂(g) + H₂O(l) + Pb(NO₃)₂(aq)
   
   2H⁺(aq) + 2NO₃^-(aq) + PbCO₃(s) → CO₂(g) + H₂O(l) + Pb²⁺(aq) 2NO₃^-(aq)
   
   2H⁺(aq) + PbCO₃(s) → CO₂(g) + H₂O(l) + Pb²⁺(aq)

   • (NH₄)₃PO₄ and CuCl₂

   Answer

   2(NH₄)₃PO₄(aq) + 3CuCl₂(aq) → Cu₃(PO₄)₂(s) + 6NH₄Cl(aq)
   
   6NH₄⁺(aq) + 2PO₄^3-(aq) + 3Cu²⁺(aq) + 6Cl^-(aq)
       → Cu₃(PO₄)₂(s) + 6NH₄⁺(aq) + 6Cl^-(aq)
   
   3Cu²⁺(aq) + 2PO₄^3-(aq) → Cu₃(PO₄)₂(s)

   • MgBr₂ and CaSO₄

   Answer

   MgBr₂(aq) + CaSO₄(aq) → CaBr₂(aq) + MgSO₄(aq)
   
   Mg²⁺(aq) + 2Br^-(aq) + Ca²⁺(aq) + SO₄^2-(aq)
       → Mg²⁺(aq) + 2Br^-(aq) + Ca²⁺(aq) + SO₄^2-(aq)
   
   No Net Ionic Equation.

4. A solution may contain any or all of the following ions: Ba²⁺, Ni²⁺, or Pb²⁺. From the following information, which ions are present? Show reasoning.

   When HCl is added to the solution a precipitate is formed. The precipitate is filtered from the solution. When K₂SO₄ is added to the filtered solution no precipitate is formed. When NaOH is then added to the filtered solution a precipitate is formed.

   Answer

   A reaction table showing what precipitates are possible would look like the following:
   

   	Ba2+	Ni2+	Pb2+
   HCl	BaCl2(aq)	NiCl2(aq)	PbCl2(s)
   K2SO4	BaSO4(s)	NiSO4(aq)	--
   NaOH	--	Ni(OH)2(s)	--

   A precipitate with HCl means that Pb²⁺ must be present. No precipitate with K₂SO₄ means that Ba²⁺ can not be present. A precipitate with NaOH means that Ni²⁺ must be present.

   Answer: Pb²⁺ and Ni²⁺ are in the solution.

5. A solution may contain any or all of the following ions: Sr²⁺, Fe²⁺, or Ca²⁺. From the following information, which ions are present? Show reasoning.

   When KOH is added to the solution a precipitate is formed. The precipitate is filtered from the solution. When Na₂SO₄ is added to the filtered solution a precipitate is again formed. This new precipitate is then filtered from the solution. When (NH₄)₂CO₃ is then added to the filtered solution a precipitate is not formed.

   Answer

   A reaction table showing what precipitates are possible would look like the following:
   

   	Sr2+	Fe2+	Ca2+
   KOH	Sr(OH)2(aq)	Fe(OH)2(s)	Ca(OH)2(aq)
   Na2SO4	SrSO4(s)	--	CaSO4(aq)
   (NH4)2CO3	--	--	CaCO3(s)

   A precipitate with KOH means that Fe²⁺ must be present. A precipitate with Na₂SO₄ means that Sr²⁺ is present. No precipitate with NaOH means that Ca²⁺ is not present.

   Answer: Sr²⁺ and Fe²⁺ are in the solution.

6. Answer the following oxidation/reduction related questions:
   • What element has the strongest attraction for electrons within a bond?

   Answer

   Flourine, F

   • What is the attraction for electrons within a bond called?

   Answer

   Electronegativity

   • Can oxidation occur without reduction?

   Answer

   No

   • What do we know about the number of electrons lost compared to the number of electrons gained?

   Answer

   They must be equal.

   • What is the trend for electronegativity across and down the periodic table?

   Answer

   The electronegativity increases going from left to right and decreases going down the periodic table.

   • True or False: The assignment of oxidation numbers assumes that the most electronegative element will get the electrons, or in other words that the bonds will be like ionic bonds.

   Answer

   True.

   • What is the oxidation number of the alkali metals?

   Answer

   1+

   • What is the oxidation number of flourine?

   Answer

   1-

   • What is the oxidation number of a single element ion?

   Answer

   The charge on the ion.

   • What is the oxidation number of elements in their standard state?

   Answer

   Zero.

   • What is the oxidation number rule for polyatomic ions?

   Answer

   All of the oxidation numbers must add up to the charge on the ion.

   • What is the oxidation number rule for molecules that don't have any charge?

   Answer

   The oxidation numbers must add up to zero.

   • What is the normal oxidation number of oxygen?

   Answer

   2-, but there are some exceptions.

7. What is the oxidation number of each element in each of the following substances?
   • HClO

   Answer

   H: +1, O: -2, Cl: +1

   • C₂H₅OH

   Answer

   H: +1, O: -2, C: -2

   • HSO₄^-

   Answer

   H: +1, O: -2, S: +6

   • PO₄^3-

   Answer

   O: -2, P: +5

   • Cl₂(g)

   Answer

   Cl: 0 (zero)

   • COCl₂

   Answer

   Cl: -1, O: -2, C: +4

   • KMnO₄

   Answer

   K: +1, O: -2, Mn: +7

   • Al(NO₃)₃

   Answer

   O: -2, N: +5, Al: +3

8. For each of the following reactions: (a) Use the oxidation number method to balance the reaction (show work, i.e., oxidation numbers, electrons transferred, etc.). (b) Identify the element that is reduced and the element that is oxidized. (c) Identify the oxidizing agent and the reducing agent.
   • Cd(s) + Ag⁺(aq) → Cd²⁺(aq) + Ag(s)

   Answer

   

   • Fe(NO₃)₂(aq) + Al(s) → Fe(s) + Al(NO₃)₃(aq)

   Answer

   

   • Cr²⁺(aq) + MnO₄^-(aq) + H⁺(aq) → Cr³⁺(aq) + Mn²⁺(aq) + H₂O(l)

   Answer

   

   • P₄(s) + HClO(aq) + H₂O → H₃PO₄(aq) + HCl(aq)

   Answer

   

   • C₂H₅OH(l) + O₂(g) → H₂O(l) + CO₂(g)

   Answer

   

9. Use the half reaction method to complete and balance the following reactions.
   • SO₂ + NO₃^- → SO₄^2- + NO (acidic solution)

   Answer

   

   • ClO₃^- + Br^- → Cl₂ + Br₂ (acidic solution)

   Answer

   

   • Cr²⁺ + MnO₄^- → Cr³⁺ + Mn²⁺ (acidic solution)

   Answer

   

   • Sn(OH)₄^2- + BrO^- → SnO₂ + Br^- (basic solution)

   Answer

   

   • S^- + NO₃^- → S + NO (basic solution)

   Answer

   

   • I^- + OCl^- → IO₃^- + Cl^-

   Answer

   

   • NO₂^- + Ga → NH₄⁺ + GaO₂^-

   Answer

   

10. Answer each of these molarity problems.
    • What is the definition of molarity?

    Answer

    molarity is defined as moles of solute divided by liters of solution.

    • What is the difference between 0.4 mol and 0.4 M?

    Answer

    0.4 mol is the number of moles while 0.4 M is the number of moles per liter of solution.

    • What is the molarity when 1.2 moles of NaCl is placed in a solution where the total volume is 2.4 L?

    Answer

    molarity =

    1.2 mole NaCl   2.4 L soln

    =

    0.5 mole NaCl   1 L soln

    = 0.5 M

    • What is the molarity when 5.0 g of KMnO₄ are placed in water to make a total solution having a volume of 500 mL?

    Answer

    (

    5 g KMnO4

    )

    (

    1 mole KMnO4   158.04 g KMnO4

    )

    = 0.0316 mole KMnO4

    molarity =

    0.0316 mole KMnO4   0.5 L soln

    =

    0.0633 mole KMnO4   1 L soln

    = 0.0633 M KMnO4

    • How many moles of KCl are in 250 mL of a 1.2 M KCl solution?

    Answer

    (

    0.25 L soln

    )

    (

    1.2 mole KCl   1 L soln

    )

    = 0.3 moles KCl

    • What volume of 2.2 M LiBr is needed to have 1.1 mole of LiBr?

    Answer

    (

    1.1 mole LiBr

    )

    (

    1 L soln   2.2 mole LiBr

    )

    = 0.5 L soln

    • How many grams of NaOH are needed to make 500 mL of a 1.5 M NaOH solution?

    Answer

    (

    0.5 L soln

    )

    (

    1.5 mole NaOH   1 L soln

    )

    (

    39.99 g NaOH   1 mole NaOH

    )

    = 29.99 g NaOH

    • What concentration should a stock solution be in order to use 10 mL of it to prepare 250 mL of a 0.2 M solution?

    Answer

    (

    0.250 L soln

    )

    (

    0.2 mole solute   1 L soln

    )

    = 0.05 mole solute

    0.05 mole solute   0.010 L soln

    = 5 M

    • How much of a 1.5 M stock solution should you use to prepare 100 mL of a 0.5 M solution?

    Answer

    (

    0.1 L soln

    )

    (

    0.5 mole solute   1 L soln

    )

    (

    1 L stock soln   1.5 mole solute

    )

    = 0.033 L stock soln

    0.033 L or 33 mL of the stock soln are needed

    • A condition called hyponatremia occurs when the sodium concentration in the blood falls below 0.130 M. A typical adult has 4.7 L of blood. (a) How many grams of sodium are in the blood of a typical adult? (b) How much sodium would a person need to get into their blood if the sodium concentration was 0.12 M?

    Answer

    (

    4.7 L blood

    )

    (

    0.130 mole Na   1 L blood

    )

    (

    22.99 g Na   1 mole Na

    )

    = 14.05 g Na (a)

    (

    4.7 L blood

    )

    (

    0.120 mole Na   1 L blood

    )

    (

    22.99 g Na   1 mole Na

    )

    = 12.97 g Na

    Amount needed = 14.05 g - 12.97 g = 1.08 g Na needed (b)

    • Many places have laws that say a person is legally drunk when they have a concentration of more than 0.013 M in ethanol, C₂H₅OH. A typical adult has 4.7 L of blood. (a) How many grams of ethanol are in the blood of a typical adult when legally drunk? (b) A can of beer typically has about 20 g of ethanol. How many cans of beer will provide the legal limit?

    Answer

    (

    4.7 L blood

    )

    (

    0.013 mole C2H5OH   1 L blood

    )

    (

    62 g C2H5OH   1 mole C2H5OH

    )

    = 3.79 g C2H5OH (a)

    (

    3.79 g C2H5OH

    )

    (

    1 can   20 g C2H5OH

    )

    = 0.19 can Only one fifth of a can! (b)

11. What mass of Na₃PO₄ is needed to completely precipitate the Ni²⁺ ions that are in 15 mL of a solution that is 0.8 M NiCl₂?

    Answer

    2Na₃PO₄(aq) + 3NiCl₂(aq) → Ni₃(PO₄)₂(s) + 6NaCl(aq)

    (

    0.015 L NiCl2 solution

    )

    (

    0.8 mole NiCl2   1 L NiCl2 solution

    )

    = 0.012 mole NiCl2

    (

    0.012 mole NiCl2

    )

    (

    2 mole Na3PO4   3 mole NiCl2

    )

    (

    164 g Na3PO4   1 mole Na3PO4

    )

    = 1.3 g Na3PO4

12. How many mL of 0.5 M NaOH is needed to completely neutralize 100 mL of 0.2 M HCl?

    Answer

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

    (

    0.1 L HCl soln

    )

    (

    0.2 mole HCl   1 L HCl soln

    )

    = 0.02 mole HCl

    (

    0.02 mole HCl

    )

    (

    1 mole NaOH   1 mole HCl

    )

    (

    1 L NaOH soln   0.5 mole NaOH

    )

    = 0.04 L or 40 mL NaOH soln

13. 15 mL of 0.1 M NaOH is needed to completely neutralize 20 mL of an H₂SO₄ solution. What was the concentration of the original H₂SO₄ solution?

    Answer

    2NaOH(aq) + H₂SO₄(aq) → 2H₂O(l) + Na₂SO₄(aq)

    (

    0.015 L NaOH soln

    )

    (

    0.1 mole NaOH   1 L NaOH soln

    )

    = 0.0015 mole NaOH

    (

    0.0015 mole NaOH

    )

    (

    1 mole H2SO4   2 mole NaOH

    )

    = 0.00075 mole H2SO4

    [H2SO4] =

    0.00075 mole H2SO4   0.02 L soln

    =

    = 0.0375 M H2SO4

14. Assume a silver acetate solution that is 1.4 M AgC₂H₃O₂ when doing the following problems.
    • Write the balanced chemical equation for the reaction of silver acetate with sodium chloride in an aqueous solution.

    Answer

    AgC₂H₃O₂(aq) + NaCl(aq) → AgCl(s) + NaC₂H₃O₂(aq)

    • How many grams of NaCl are needed to completely react with 200 mL of the AgC₂H₃O₂ solution?

    Answer

    (

    0.2 L AgC2H3O2 solution

    )

    (

    1.4 mole AgC2H3O2   1 L AgC2H3O2 solution

    )

    = 0.28 mole AgC2H3O2

    (

    0.28 mole AgC2H3O2

    )

    (

    1 mole NaCl   1 mole AgC2H3O2

    )

    (

    58.5 g NaCl   1 mole NaCl

    )

    = 16.4 g NaCl

    • How much silver chloride is created when 100 mL of the AgC₂H₃O₂ solution is combined with 200 mL of 0.6 M NaCl?

    Answer

    Determine the moles of each reactant. We have:

    (

    0.1 L AgC2H3O2 soln

    )

    (

    1.4 mole AgC2H3O2   1 L AgC2H3O2 soln

    )

    = 0.14 mole AgC2H3O2

    (

    0.2 L NaCl soln

    )

    (

    0.6 mole NaCl   1 L NaCl soln

    )

    = 0.12 mole NaCl

    Determine limitiing reactant by choosing one reactant and determining the amount of the other one that is needed to completely react. I have chosen NaCl below.

    (

    0.12 mole NaCl

    )

    (

    1 mole AgC2H3O2   1 mole NaCl

    )

    = 0.12 mole AgC2H3O2 needed

    We need 0.12 mole AgC₂H₃O₂, but have 0.14 mole. We have extra AgC₂H₃O₂ and so NaCl is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.

    (

    0.12 mole NaCl

    )

    (

    1 mole AgCl   1 mole NaCl

    )

    (

    143.4 g AgCl   1 mole AgCl

    )

    = 17.2 g AgCl

    • What are the concentrations of each of the ions in the final solution when 100 mL of the AgC₃H₃O₂ solution is combined with 200 mL of 0.6 M NaCl? Use information calculated in the last problem and assume the volumes are additive (the total volume is the sum of the volumes that are added together).

    Answer

    From the last problem we see that the NaCl was the limiting reactant. That means that all of the Cl will be in AgCl(s) and the concentration of the Cl^- is zero: [Cl^-] = 0.

    There was extra Ag⁺. The amount used was 0.12 moles, but we had 0.14 moles. That means that there was 0.02 moles extra that would remain in the solution after the reaction was complete. Those 0.02 moles would be in a total of 300 mL, so

    [Ag+] =

    0.02 mole Ag+   0.3 L soln

    =

    0.067 mole Ag+   1 L soln

    = 0.067 M Ag+

    Na⁺ doesn't make a precipitate, so all of the original amount would be in the final solution and

    [Na+] =

    0.12 mole Na+   0.3 L soln

    =

    0.4 mole Na+   1 L soln

    = 0.4 M Na+

    C₂H₃O₂^- doesn't make a precipitate, so all of the original amount would be in the final solution and

    [C2H3O2-] =

    0.14 mole C2H3O2-   0.3 L soln

    =

    0.467 mole C2H3O2-   1 L soln

    = 0.467 M C2H3O2-

15. A 100 mL solution of 1.2 M KOH is combined with 150 mL of 1.0 M ZnCl₂.
    • How many grams of Zn(OH)₂ can be formed?

      Hint

      • What is the balanced chemical equation?
      • How many moles of each reactant do you start with? (Use molarity.)
      • What is the limiting reactant? (Use mole ratios from balanced eqn.)
      • How many moles of product are formed? (Use mole ratios from balanced eqn.)
      • How many grams of product are formed? (Use periodic table.)
      • How many moles of each original ion is left over?
      • What is the total volume?
      • What is the final molarity of each ion? (Moles left over divided by total volume.)

    Answer

    2KOH(aq) + ZnCl₂(aq) → 2KCl(aq) + Zn(OH)₂(s)

    Determine the moles of each reactant. We have:

    (

    0.1 L KOH soln

    )

    (

    1.2 mole KOH   1 L KOH soln

    )

    = 0.12 mole KOH

    (

    0.15 L ZnCl2 soln

    )

    (

    1.0 mole ZnCl2   1 L ZnCl2 soln

    )

    = 0.15 mole ZnCl2

    Determine limitiing reactant by choosing one reactant and determining the amount of the other one that is needed to completely react. Using KOH:

    (

    0.12 mole KOH

    )

    (

    1 mole ZnCl2   2 mole KOH

    )

    = 0.06 mole ZnCl2 needed

    We need 0.06 mole ZnCl₂, but have 0.15 mole. We have extra ZnCl₂ and so KOH is the limiting reactant. Use the amount of limiting reactant that we have to determine the amount of product produced.

    (

    0.12 mole KOH

    )

    (

    1 mole Zn(OH)2   2 mole KOH

    )

    (

    99.4 g Zn(OH)2   1 mole Zn(OH)2

    )

    = 5.96 g Zn(OH)2

    • What are the concentrations of each of the ions in the final solution? Assume the volumes are additive (the total volume is the sum of the volumes that are added together).

    Answer

    From the last problem we see that the KOH was the limiting reactant. That means that all of the OH^- will be in Zn(OH)₂(s) and the concentration of the OH^- is zero: [OH^-] = 0.

    There was extra Zn²⁺. The amount used was 0.06 moles ZnCl₂, but we had 0.15 moles. That means that there was 0.15 - 0.06 = 0.09 moles extra Zn²⁺ that would remain in the solution after the reaction was complete. Those 0.09 moles would be in a total of 250 mL, so

    [Zn2+] =

    0.09 mole Zn2+   0.25 L soln

    =

    0.36 mole Zn2+   1 L soln

    = 0.36 M Zn2+

    K⁺ doesn't make a precipitate, so all of the original amount, 0.12 mole KOH, would be in the final solution and

    [K+] =

    0.12 mole K+   0.25 L soln

    =

    0.48 mole K+   1 L soln

    = 0.48 M K+

    Cl^- doesn't make a precipitate, so all of the original amount, (0.15 mole ZnCl₂)(2 mole Cl^-/mole ZnCl₂) = 0.3 mole Cl^-, would be in the final solution and

    [Cl-] =

    0.3 mole Cl-   0.25 L soln

    =

    1.2 mole Cl-   1 L soln

    = 1.2 M Cl-