chapter 4 hw solutions - chem 1, g-chem...

CHAPTER 4 HW – SOLUTIONS

AQUEOUS SOLUBILITY

1.) Use a solubility table to determine whether each salt would be soluble (aq) or insoluble (s) in water.

a. Na2S aq (rule 1) c. CaCrO4 s (rule 8) e. SrSO4 s (rule 6)

b. PbI2 s (rule 5) d. (NH4)2SO4 aq (rule 2) f. Li3PO4 aq (rule 1)

2.) Give an example of a soluble and insoluble salt that contains the following ions:

Soluble salt Insoluble salt

Contains a bromide ion NaBr, NH4Br, CaBr2 … AgBr, PbBr2, CuBr

Contains a phosphate ion Na3PO4, Rb3PO4, Cs3PO4 … Ag3PO4, FePO4, Zn3(PO4)2 …

3.) Select the microscopic picture (A-D) that represents each compound in an aqueous solution.

ELECTROLYTES

4.) One of these acids is strong, while the other is weak. Draw a beaker representation of the major solute component(s) in each solution, then explain the difference between a weak and strong acid.

HF (aq) HCl (aq)

A strong acid (like HCl) completely separates into ions in solution. A weak acid (like HF) forms very few ions in solution and most of the molecules are intact.

A B C D

+

+

+ +2-

2-

2-2+

2+ 2+

2+

-

- - -

--+

+ 2-

F–H+HFHF

Cl–H+Cl–Cl–

H+H+

a. Barium nitrite D Ba2+ NO2

− NO2−

b. Sodium perchlorate A

Na+ ClO4 –

c. Magnesium sulfate C

Mg2+ SO42−

d. Potassium carbonate B

K+ K+ CO32−

5.) Draw a beaker representation of the major solute component(s) in each aqueous solution. If the reagent is insoluble, draw it settled to the bottom of the beaker. For acids, show several solute particles so as to distinguish between weak and strong acids.

NaOH HC2H3O2 CH3OH PbS

HClO3 Na2CO3 HBr HCN

6.) List the major solute component(s) present in each aqueous solution.

a. NH4Cl NH4+ Cl– (ionic) d. HNO3 H+ NO3

– (strong acid)

b. C2H6O C2H6O (molecular) e. H3PO4 H3PO4 (weak acid)

c. KClO4 K+ ClO4– (ionic) f. C5H14ON3 C5H14ON3 (molecular)

7.) Which is a stronger electrolyte in each set? Briefly explain each.

a. AgNO3 vs. H2CO3

AgNO3 is a stronger electrolyte: it’s a soluble ionic compound (so produces many ions in solution, which makes something conductive). H2CO3 is a weak acid so makes few ions in solution.

b. AlCl3 vs. KCl

Both are soluble ionic compounds, but AlCl3 is a stronger electrolyte because it makes more ions: 4 ions (Al3+ Cl– Cl– Cl–) compared to only 2 (K+ Cl–).

OH–Na+

HC2H3O2HC2H3O2

C2H3O2–H+

CH3OH

PbS (s)

ClO3–H+

ClO3–ClO3–

H+H+

CO32–Na+

Na+

Br–H+Br–Br–

H+H+ HCN

HCN

CN–H+

8.) Which is a stronger electrolyte in each set? Briefly explain each.

a. HCl vs. C6H12O6

HCl is a strong acid so produces many ions in solution, making it a strong electrolyte. C6H12O6 is a molecular compound so produces no ions in solution and is a non-electrolyte (HCl is stronger).

b. MgCrO4 vs. Na2CrO4

Na2CrO4 is a stronger electrolyte because it is soluble so produces ions in solution. MgCrO4 is insoluble and so is a non-electrolyte.

9.) Which statements are true? For any false statements, briefly explain why they are false.

a. A concentrated aqueous solution will always contain a strong or weak electrolyte. False

A concentrated solution could contain a molecular compound, which is a non-electrolyte.

b. A strong electrolyte breaks up into ions when dissolved in water. True

c. An acid is a strong electrolyte. False

Acids can be weak or strong electrolytes (weak or strong acids).

d. All ionic compounds are strong electrolytes in water. False

Many are, but if an ionic compound is insoluble it won’t be an electrolyte.

PRECIPITATION REACTIONS

10.) When aqueous solutions of silver nitrate and sodium carbonate are combined, a white precipitate forms (see picture to the right). Write the balanced formula equation for this reaction, including phase designations.

2 AgNO3 (aq) + Na2CO3 (aq) → Ag2CO3 (s) + 2 NaNO3 (aq)

11.) Write balanced formula equations for each reaction, including phase designations.

a. CaI2 (aq) + Na2SO4 (aq) → CaSO4 (s) + 2 NaI (aq)

Ca2+ I– Na+ SO42–

b. CrCl3 (aq) + 3 NaOH (aq) → Cr(OH)3 (s) + 3 NaCl (aq)

Cr3+ Cl– Na+ OH–

12.) Write balanced formula equations for each reaction, including phase designations.

a. (NH4)2S (aq) + CdCl2 (aq) → CdS (s) + 2 NH4Cl (aq)

NH4+ S2– Cd2+ Cl–

b. 2 AgClO4 (aq) + CaBr2 (aq) → 2 AgBr (s) + Ca(ClO4)2 (aq)

Ag+ ClO4– Ca2+ Br–

c. 2 Fe(C2H3O2)3 (aq) + 3 K2CO3 (aq) → 6 KC2H3O2 (aq) + Fe2(CO3)3 (s)

Fe3+ C2H3O2– K+ CO3

2–

13.) When the following aqueous solutions are mixed, what precipitate (if any) will form?

a. MgI2 (aq) + K2CO3 (aq) MgCO3 (s)

Mg2+ I– K+ CO32–

b. Ni(NO3)2 (aq) + CaCl2 (aq) No precipitate (both new products are soluble)

Ni2+ NO3– Ca2+ Cl–

c. AlBr3 (aq) + Na2CrO4 (aq) Al2(CrO4)3 (s)

Al3+ Br– Na+ CrO42–

14.) Write the following equations for the reaction represented below, including phase designations.

Formula equation: K3PO4 (aq) + AlBr3

(aq) → AlPO4 (s) + 3 KBr (aq)

Complete Ionic equation:

3 K+ (aq) + PO43− (aq) + Al3+ (aq) + 3 Br– (aq) → AlPO4 (s) + 3 K+ (aq) + 3 Br– (aq)

Net Ionic equation: PO43− (aq) + Al3+ (aq) → AlPO4 (s)

+PO43– K+

K+K+ AlPO4 (s)

Al3+

Br–

Br–

Br–

Br–Br– Br–

K+

K+

K+

15.) Write the balanced complete ionic and net ionic equation for this reaction. Include phase designations.

2 CuNO2 (aq) + Li2CO3

(aq) → Cu2CO3 (s) + 2 LiNO2 (aq)


2 Cu+ (aq) + 2 NO2− (aq) + 2 Li+ (aq) + CO3

2− (aq) → Cu2CO3 (s) + 2 Li+ (aq) + 2 NO2− (aq)

Net Ionic equation: 2 Cu+ (aq) + CO32− (aq) → Cu2CO3 (s)

16.) What are the spectator ions in the previous reaction? Li+ and NO2–

ACID-BASE REACTIONS

17.) Select the microscopic picture (A-D) that represents each acid or base in aqueous solution.

18.) Write the balanced formula equations for the following acid-base reactions, include phases.

a. HC2H3O2 (aq) + NaOH (aq) → H2O (l) + NaC2H3O2 (aq)

H+ C2H3O2– Na+ OH–

b. 2 HF (aq) + Ca(OH)2 (aq) → 2 H2O (l) + CaF2 (aq)

H+ F– Ca2+ OH–

c. HCl (aq) + NaHCO3 (aq) → NaCl (aq) + H2O (l) + CO2 (g)

H+ Cl– Na+ HCO3– H2CO3 → H2O (l) + CO2 (g)

d. Na2CO3 (aq) + 2 HC2H3O2 (aq) → 2 NaC2H3O2 (aq) + H2O (l) + CO2 (g)

Na+ CO32– H+ C2H3O2

–

e. CaCO3 (s) + 2 HBr (aq) → CaBr2 (aq) + H2O (l) + CO2 (g)

Ca2+ CO32– H+ Br–

A B C D

+

+

+ +2-

2-

2-2+

2+ 2+

2+

-

- - -

--+

+ 2-

a. Nitric acid A H+ NO3

−

b. Sodium hydroxide A Na+ OH−

c. Calcium hydroxide D Ca2+ OH− OH−

19.) Write the balanced complete ionic and net ionic equations for each of the following acid-base reactions. Include phase designations.

a. HClO4 (aq) + NaOH (aq) → H2O (l) + NaClO4 (aq)


H+ (aq) + ClO4– (aq) + Na+ (aq) + OH– (aq) → H2O (l) + Na+ (aq) + ClO4

– (aq)

Net Ionic equation: H+ (aq) + OH– (aq) → H2O (l)

b. HCN (aq) + NaOH (aq) → H2O (l) + NaCN (aq)


HCN (aq) + Na+ (aq) + OH– (aq) → H2O (l) + Na+ (aq) + CN– (aq)

Note HCN is a weak acid, so remains mostly intact, not ionized.

Net Ionic equation: HCN (aq) + OH– (aq) → H2O (l) + CN– (aq)

20.) Write the balanced formula and net ionic equations for each acid-base reaction, including phases.

a. Formula equation: NaOH (aq) + HCl (aq) → H2O (l) + NaCl (aq)

Na+ OH– H+ Cl–

Net ionic equation: H+ (aq) + OH– (aq) → H2O (l)

b. Formula equation: HF (aq) + KOH (aq) → H2O (l) + KF (aq)

H+ F– K+ OH–

Net ionic equation: HF (aq) + OH– (aq) → H2O (l) + F– (aq)

c. Formula equation: H3C4HO6 (aq) + 3 LiOH (aq) → 3 H2O (l) + Li3C4HO6 (aq)

H+ C4HO6 – Li+ OH–

Net ionic equation: H3C4HO6 (aq) + 3 OH– (aq) → 3 H2O (l) + C4HO63– (aq)

CONCENTRATION (MOLARITY)

21.) A normal fasting blood glucose level should be between 3.9 and 7.1 mmol/L, meaning the blood glucose level for an average person is about 5.5 mmol/L (or 0.0055 mol/L). Imagine that you were to ingest 41 g of sugars (assume these are all glucose, C6H12O6), the amount in one can of regular Pepsi soda. Although blood glucose is tightly regulated by your body, imagine that all of the glucose dissolved in your blood. Considering an average person has 5.0 L of blood, what would be the concentration of glucose in your blood (in M) after drinking the soda?

41 g C6H12O6 × "#$%&'()*+'",-.",/

= 0.23 mol C6H12O6

MMC6H12O6 = 6 (12.01) + 12 (1.01) + 6 (16.00) = 180.18 g/mol

M = #$%0

= -.12#$%&'()*+'

3.-0 = 0.046 M C6H12O6

(For this theoretical situation, I’m ignoring the baseline glucose already in your blood.)

(Normal blood glucose is 0.0055 mol/L, so this is about 8 times as concentrated .)

22.) Calculate the molarity of a solution made by dissolving 0.1846 g of K2Cr2O7 (MM=294.20 g/mol) in enough water to make 500.0 mL of solution.

0.1846 g K2Cr2O7 × "#$%145.1-/

= 0.0006275 mol K2Cr2O7

500.0 mL × "0

"---#0 = 0.5000 L

M = 𝑚𝑜𝑙𝐿 = -.---:1;3#$%<*&=*+>

-.3---0 = 0.001255 M

23.) How many moles of NaCl are in 250.0 mL of a 0.072 M NaCl solution?

250.0 mL × "0

"---#0× -.-;2#$%?@&%

"0 = 0.018 mol NaCl

24.) What volume of 2.00 M CaCl2 contains 0.993 moles of CaCl2?

0.993 mol CaCl2 × "0

1.--#$%&@&%* = 0.497 L

25.) How many grams of NaOH are in 2.00 L of a 0.250 M NaOH solution?

2.00 L × -.13-#$%?@+(

"0 ×

5-.--/"#$%?@+(

= 20.0 g NaOH

MM = 22.99 + 1.01 + 16.00 = 40.00 g/mol

26.) Calculate the concentration of all ions in 100.0 mL of solution that contains 0.100 mol Ca(NO3)2.

Ca(NO3)2(aq) → Ca2+(aq) + 2 NO3-(aq)

0.100 mol Ca(NO3)2 × "#$%&@*A

"#$%&@(?+C)* = 0.100 mol Ca2+

0.100 mol Ca(NO3)2 × 1#$%?+CE

"#$%&@(?+C)* = 0.200 mol NO3

–

V = 100.0 mL × "0

"---#0 = 0.1000 L

MCa2+ = -."--#$%&@*A

-."---0 = 1.00 M Ca2+ MNO3– =

-.1--#$%?+CE

-."---0 = 2.00 M NO3

–

27.) Which has the larger amount of chloride ions, 125.0 mL of a 0.224 M AlCl3 solution or 100.0 mL of a 0.60 M MgCl2 solution ?

Moles are related to quantity- no need to use Avogadro’s number and count ions directly. The solution with the greater moles of Cl– has more Cl–.

125.0 mL × "0

"---#0× -.115#$%F%&%C

"0 ×

2#$%&%–

"#$%F%&%C = 0.0840 mol Cl–

100.0 mL × "0

"---#0× -.:-#$%H/&%*

"0 ×

1#$%&%–

"#$%H/&%* = 0.12 mol Cl–

There are more Cl– ions in the MgCl2 solution.

DILUTIONS

28.) When water is added to a solution of acetic acid, what does not change?

a. The molarity of acetic acid c. The volume of solution

b. The moles of acetic acid d. None of the above change

29.) What is the concentration of a solution made by diluting 50.00 mL of 0.725 M NaBr to 0.375 L?

M1V1 = M2V2

M2 = H)I)I*

= (-.;13H)(3-.--#0)

(2;3#0) = 0.0967 M

V = 0.375 L × "---#0"0

= 375 mL

30.) 10.00 mL of a 2.900 M nitric acid solution is mixed with 350.0 mL of water. What is the molarity of the new solution?

M1V1 = M2V2

M2 = H)I)I*

= (1.4--H)("-.--#0)

(2:-.-#0) = 0.08056 M HNO3

V2 = 10.00 mL + 350.0 mL = 360.0 mL

31.) What volume of 1.00 M NaOH should be diluted to 2.00 L in order to make a 0.250 M NaOH solution?

M1V1 = M2V2

V1 = H*I*H)

= (-.13-H)(1.--0)

(".--H) = 0.500 L

32.) 230. mL of a 0.275 M CaCl2 aqueous solution is left on a hotplate overnight. The next morning the solution is now 1.10 M. What volume of water evaporated overnight?

M1V1 = M2V2

V2 = H)I)H*

= (-.1;3H)(12-.#0)

("."-H) = 57.5 mL

The amount of water that evaporated is 230. mL – 57.5 mL = 173 mL

TITRATION CALCULATIONS

33.) An Alka-Seltzer tablet contains aspirin, sodium bicarbonate, and citric acid. The sodium bicarbonate and citric acid react to form bubbles when dissolved in water through the following reaction.

H3C6H5O7 (aq) + 3 NaHCO3 (aq) → Na3C6H5O7 (aq) + 3 CO2 (g) + 3 H2O (l)

If it takes 13.49 mL of a 0.56 M H3C6H5O7 solution to neutralize the sodium bicarbonate in one Alka-Seltzer tablet, how many grams of NaHCO3 are present in the tablet?

13.49 mL H3C6H5O7 × "0

"---#0 × -.3:#$%(C&'(J+>

"0 = 0.0076 mol H3C6H5O7

0.0076 mol H3C6H5O7 × 2#$%?@(&+C"#$%(C&'(J+>

× ,5.-"/

"#$%?@(&+C = 1.9 g NaHCO3

MMNaHCO3 = 22.99 + 1.01 + 12.01 + 3(16.00) = 84.01 g/mol

34.) 50.00 mL of 0.100 M KHP (HKC8H5O4) is titrated with 0.250 M KOH in a buret through the following equation. How many mL of KOH is needed to neutralize the KHP solution?

HKC8H5O4 (aq) + KOH (aq) → H2O (l) + K2C8H5O4 (aq)

50.00 mL KHP × "0

"---#0 × -."--#$%<(K

"0 = 0.00500 mol KHP

0.00500 mol KHP × "#$%<+("#$%<(K

× "0

-.13-#$%<+( × "---#0"0

= 20.0 mL KOH

Note: Don’t use M1V1 = M2V2 (dilution); this doesn’t take into account any mole ratio! Dilution problems will only involve adding water, and will typically not be used when an equation is given. If an equation is shown, the problem will use mole ratio at some point (“go to moles”).

35.) Some over-the-counter antacids contain Mg(OH)2 (for example Milk of Magnesia). Others instead contain CaCO3 (tradename Tums), which enables the tablets to also be used as a source of dietary calcium. If titration of one Tums tablet, which contains 0.500 g of CaCO3, takes 20.19 mL of an HCl solution to neutralize it, what is the concentration of the HCl solution?

CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + H2O (l) + CO2 (g)

0.500 g CaCO3 × "#$%&@&+C"--.-4/

× 1#$%(&%"#$%&@&+C

= 0.00999 mol HCl

MMCaCO3 = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol

20.19 mL × "0

"---#0 = 0.02019 L

M = #$%0

= -.--444#$%(&%

-.-1-"40 = 0.495 M HCl

36.) Carminic acid, a naturally occurring red pigment extracted from the female cochineal insect was commonly used as a dye in the 1700’s and early 1800’s. A titration of 0.3602 g of carminic acid (a monoprotic acid) requires 18.02 mL of a 0.0406 M NaOH solution to neutralize it. Calculate the molar mass of carminic acid.

HA (aq) + NaOH (aq) → H2O (l) + NaA (aq) – generic monoprotic equation

18.02 mL NaOH × "0

"---#0 × -.-5-:#$%?@+(

"0 ×

"#$%(F"#$%?@+(

= 0.000732 mol carminic acid

MM = /#$%

= -.2:-1/

-.---;21#$% = 492 g/mol

OTHER SOLUTION CALCULATIONS

37.) Some plants are toxic to humans as their stems and leaves contain high levels of oxalates, either in the form of oxalic acid (H2C2O4) or sodium oxalate (Na2C2O4). When ingested, these substances cause swelling of the respiratory tract and in high enough quantities suffocation. A standard analysis for determining the quantity of oxalates is to precipitate them via reaction with CaCl2.

Na2C2O4 (aq) + CaCl2 (aq) → CaC2O4 (s) + 2 NaCl (aq)

What is the concentration of oxalate (C2O42–) in 35.0 mL of a Na2C2O4 solution if reaction with

excess CaCl2 forms 1.77 g of CaC2O4 (MM=128.10) precipitate?

1.77 g CaC2O4× "#$%&@&*+L"1,."-/

× "#$%?@*&*+L"#$%&@&*+L

× "#$%&*+L*E

"#$%?@*&*+L = 0.0138 mol C2O4

2–

35.0 mL × "0

"---#0 = 0.0350 L

M = #$%0

= -.-"2,#$%&*+L*E

-.-23-0 = 0.395 M C2O4

2–

38.) What volume (in L) of 0.100 M K3PO4 is required to precipitate all the lead(II) ions from 150.0 mL of 0.250 M Pb(NO3)2, via this balanced equation?

2 K3PO4 (aq) + 3 Pb(NO3)2 (aq) → Pb3(PO4)2 (s) + 6 KNO3 (aq)

150.0 mL Pb(NO3)2 × "0

"---#0 × -.13-#$%KM(?+C)*

"0 = 0.0375 mol Pb(NO3)2

0.0375 mol Pb(NO3)2 × 1#$%<CK+L

2#$%KM(?+C)* ×

"0-."--#$%<CK+L

= 0.250 L K3PO4

39.) What mass of barium sulfate can be produced when 100.0 mL of a 0.100 M solution of barium chloride is mixed with 100.0 mL of a 0.100 M solution of iron(III) sulfate and react via this unbalanced equation?

3 BaCl2(aq) + Fe2(SO4)3(aq) → 3 BaSO4(s) + FeCl3(aq)

100.0 mL BaCl2 × "0

"---#0 × -."--#$%N@&%*

"0 = 0.0100 mol BaCl2

Same calculation with Fe2(SO4)3 so 0.0100 mol Fe2(SO4)3

Since equal moles are present, but 3 moles of BaCl2 are needed per mol of Fe2(SO4)3, the BaCl2will run out first and is the LR. To prove this,

0.0100 mol BaCl2 × "#$%OP*(Q+L)C2#$%N@&%*

= 0.00333 mol Fe2(SO4)3 needed (have excess)

0.0100 mol BaCl2 × 2#$%N@Q+L2#$%N@&%*

× 122.24/

"#$%N@Q+L = 2.33 g BaSO4

MMBaSO4 = 137.33 + 32.06 + 4(16.00) = 233.39 g/mol

40.) What mass of iron(III) hydroxide can be made by reacting 75.0 mL of a 0.105 M iron(III) nitrate solution with 125.0 mL of 0.150 M sodium hydroxide?

Fe(NO3)3 (aq) + 3 NaOH (aq) → Fe(OH)3 (s) + 3 NaNO3 (aq)

75.0 mL Fe(NO3)3 × "0

"---#0 × -."-3#$%OP(?+C)C

"0 = 0.00788 mol Fe(NO3)3

125.0 mL NaOH × "0

"---#0 × -."3-#$%?@+(

"0 = 0.0188 mol NaOH

Check if Fe(NO3)3 is LR

0.00788 mol Fe(NO3)3 × 2#$%<+(

"#$%OP(?+C)C = 0.0236 mol NaOH used (can’t, NaOH = LR)

LR to product

0.0188 mol NaOH × "#$%OP(+()C2#$%?@+(

× "-:.,;/OP(+()C"#$%OP(+()C

= 0.670 g Fe(OH)3

MMFe(OH)3 = 55.85 + 3(16.00) + 3(1.01) = 106.87 g (0.668 g with no rounding)

41.) Solid AgCl is produced when 100.0 mL of 0.20 M silver nitrate is reacted with 100.0 mL of 0.15 M calcium chloride.

a. What mass of AgCl can be produced?

2 AgNO3 (aq) + CaCl2 (aq) → 2 AgCl (s) + Ca(NO3)2 (aq)

100.0 mL AgNO3 × "0

"---#0 × -.1-#$%F/?+C

"0 = 0.020 mol AgNO3

100.0 mL CaCl2 × "0

"---#0 × -."3#$%&@&%*

"0 = 0.015 mol CaCl2

Find LR

0.015 mol CaCl2 × 1#$%F/?+C"#$%&@&%*

= 0.030 mol AgNO3 used (can’t, AgNO3 = LR)

LR to product

0.020 mol AgNO3 × 1#$%F/&%1#$%F/?+C

× "52.5/

"#$%F/&% = 2.9 g AgCl

MMAgCl = 107.9 + 35.45 = 143.4 g

b. Calculate the concentration of NO3– (aq) in solution after the precipitation is complete.

Combined volume = 2(100.0 mL) = 200.0 mL × "0

"---#0 = 0.2000 L

0.020 mol AgNO3 × "#$%?+CE"#$%F/?+C

= 0.020 mol NO3–. M =

-.-1-#$%?+C–

-.1---0 = 0.10 M NO3

–

c. Calculate the concentration of Ca2+ (aq) in solution after the precipitation is complete.

0.015 mol CaCl2 × "#$%&@A"#$%&@&%*

= 0.015 mol Ca2+

M = -.-"3#$%&@*A

-.1---0 = 0.075 M Ca2+

d. Calculate the concentration of Cl– (aq) in solution after the precipitation is complete.

Initial Cl– ions: 0.015 mol CaCl2 × 1#$%&%E"#$%&@&%*

= 0.030 mol Cl–

0.020 mol AgNO3 (L.R.) × 1#$%F/&%1#$%F/?+C

× "#$%&%–

"#$%F/&% = 0.020 mol Cl– reacted

Excess Cl– ions = Initial – Used: 0.030 mol – 0.020 mol = 0.010 mol Cl– left over

M = -.-"-#$%&%–

-.1---0 = 0.050 M Cl–

REDOX REACTIONS

42.) Assign oxidation numbers for all elements in each species.

a. CO O = –2, C = +2 f. P4O6 O = –2, P = +3

4x +6(–2) = 0

b. NiO2 O = –2, Ni = +4 g. C6H12O6 H = +1, O = –2, C = 0

x +2(–2) = 0 6x +12(1) +6(–2) = 0

c. O2 O = 0 (elemental) h. SF4 F = –1, S = +4

x + 4(–1) = 0

d. S2O32– S = +2, O = –2 i. HPO4

2– H = +1, O = –2, P = +5

2x +3(–2) = –2 1 + x + 4(–2) = –2

e. NH4+ H = +1, N = –3 j. KMnO4 K = +1, O = –2, Mn = +7

x + 4(1) = +1 K+ MnO4– x + 4(–2) = –1

43.) Give the number of protons and electrons in each species. Then identify whether each process corresponds to oxidation or reduction.

a. Ba → Ba2+ Oxidation or reduction: oxidation (loss of e–)

p+ 56 e– 56 p+ 56 e– 54

b. N2 (each N) → N3– Oxidation or reduction: reduction (gain of e–)

p+ 7 e– 7 p+ 7 e– 10

c. Cu2+ → Cu+ Oxidation or reduction: reduction (gain of e–)

p+ 29 e– 27 p+ 29 e– 28

44.) Identify which of the following are redox reactions.

a. Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g) Yes redox

b. NaOH (aq) + HCl (aq) → H2O (l) + NaCl (aq) No, Acid-base

c. AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3 (aq) No, Precipitation

d. 2 Zn (s) + O2 (g) → 2 ZnO (s) Yes Redox

e. MnO (aq) + BiO3– (aq) → MnO4

– (aq) + Bi3+ (aq) Yes Redox

45.) Corrosion is the deterioration of a metal through oxidation. The economic consequences of the corrosion of iron (rusting) are enormous. It’s estimated that up to 25% of the iron produced in the United States is used to replace bridges, building, and other structures that have been destroyed by corrosion. The following is the process of iron forming rust (Fe2O3).

4 Fe (s) + 3 O2 (g) → 2 Fe2O3 (s)

Fe → Fe3+ is a loss of electrons (oxidation)

O2 → O2– is a gain of electrons (reduction)

Answer the questions about this redox reaction.

What is oxidized? Fe What is reduced? O or O2

What is the oxidizing agent? O2 What is the reducing agent? Fe

46.) A rod of copper plunged into an aqueous solution of AgNO3 will over time grow silver crystals on it through the following redox reaction.

Cu (s) + 2 AgNO3 (aq) → Cu(NO3)2 (aq) + 2 Ag (s)

Cu → Cu2+ is a loss of electrons (oxidation)

Ag+→ Ag is a gain of electrons (reduction)

What is oxidized? Cu What is reduced? Ag or Ag+

What is the oxidizing agent? AgNO3 What is the reducing agent? Cu

47.) Inexpensive alkaline batteries used in flashlights and other household items use a zinc powder and paste of KOH as one reactant, separated from a paste of powdered carbon and manganese(IV) oxide (MnO2) as the other reactant. A graphite rod with metal caps sticks into the MnO2 paste to provide electrical contact. When the rod is connected by a wire, the following redox reaction occurs.

Zn (s) + 2 MnO2 (s) → ZnO (s) + Mn2O3 (s)

Zn → Zn2+ is a loss of electrons (oxidation)

Mn4+→ Mn3+ is a gain of electrons (reduction)

What is oxidized? Zn What is reduced? Mn in MnO2

Which element gains electrons in the process? Mn

What (if any) elements are spectators in this reaction? Oxygen is a spectator

In the balanced reaction, a total of 2 e– are transferred from Zn to Mn4+.

48.) The thermite reaction is a classic demonstration as it produces a tremendous amount of energy in an impressive display of fire and smoke. The reaction involves aluminum with rust (Fe2O3) and produces so much heat that it generates molten iron that often melts through the ceramic container and drains out the bottom. Historically the thermite reaction has been used to weld railroad ties together.

2 Al (s) + Fe2O3 (s) → 2 Fe (s) + Al2O3 (s)

Al → Al3+ is a loss of electrons (oxidation)

Fe3+→ Fe is a gain of electrons (reduction)

What is oxidized? Al What is reduced? Fe in Fe2O3 What is the oxidizing agent? Fe2O3

In the balanced reaction, a total of 6 e– are transferred from Al to Fe3+.

49.) In each redox reaction, identify which element is oxidized and which is reduced.

Reaction Oxidized Reduced

a. CH4 (g) + H2O (g) → CO (g) + 3 H2 (g) –4 +4 +1 –2 +2 –2 0 C H

b. 3 Cu (s) + 8 HNO3 (aq) → 0 +1 +5 –2 3 Cu(NO3)2 (aq) + 2 NO (g) + 4 H2O (l) +2 +5 –2 +2 –2 +1 –2

Cu N

50.) Use the half-reaction method to balance the following redox reaction under acidic conditions. Add water and H+ where needed.

I– (aq) + ClO– (aq) → I3– (aq) + Cl– (aq)

Half Rxns I– → I3– ClO– → Cl–

Elements 3 I– → I3– “

Oxygen (H2O) “ ClO– → Cl– + H2O

Hydrogen (H+) “ ClO– + 2 H+ → Cl– + H2O

Charge (e–) 3 I– → I3

– + 2 e– ClO– + 2 H+ + 2 e–→ Cl– + H2O

Electrons “ “

3 I– + ClO– + 2 H+ + 2 e– → Cl– + H2O + I3– + 2 e–

3 I– + ClO– + 2 H+ → Cl– + H2O + I3–

51.) The dichromate ion is capable of reacting with the chloride ion through a redox reaction. During this process the chromium changes from an orange color (of Cr2O7

2–) to a green color (of Cr3+). Use the half-reaction method to balance this redox reaction under acidic conditions. Add water and H+ where needed.

Cr2O72– (aq) + 6 Cl– (aq) → 3 Cl2 (aq) + 2 Cr3+ (aq)

Half Rxns Cr2O72– → Cr3+ Cl– → Cl2

Elements Cr2O72– → 2 Cr3+ 2 Cl– → Cl2

Oxygen Cr2O72– → 2 Cr3+ + 7 H2O “

Hydrogen 14 H+ + Cr2O72– → 2 Cr3+ + 7 H2O “

Charge 6 e– + 14 H+ + Cr2O72– → 2 Cr3+ + 7 H2O 2 Cl– → Cl2 + 2 e–

Electrons “ 6 Cl– → 3 Cl2 + 6 e–

6 e– + 14 H+ + Cr2O72– + 6 Cl– → 2 Cr3+ + 7 H2O + 3 Cl2 + 6 e–

14 H+ + Cr2O72– + 6 Cl– → 2 Cr3+ + 7 H2O + 3 Cl2

52.) Calcium levels in the blood can be determined by adding oxalate (C2O42–) to precipitate CaC2O4,

followed by dissolving the precipitate in acid and titrating the resulting acid (H2C2O4) with permanganate ion. Balance the reaction of H2C2O4 with MnO4

– under acidic conditions.

H2C2O4 (aq) + MnO4– (aq) → CO2 (g) + Mn2+ (aq)

Half Rxns H2C2O4 → CO2 MnO4– → Mn2+

Elements H2C2O4 → 2 CO2 “

Oxygen “ MnO4– → Mn2+ + 4 H2O

Hydrogen H2C2O4 → 2 CO2 + 2 H+ 8 H+ + MnO4– → Mn2+ + 4 H2O

Charge H2C2O4 → 2 CO2 + 2 H+ + 2 e– 5 e– + 8 H+ + MnO4– → Mn2+ + 4 H2O

Electrons 5 H2C2O4 → 10 CO2 + 10 H+ + 10 e– 10 e– + 16 H+ + 2 MnO4– → 2 Mn2+ + 8 H2O

5 H2C2O4 + 10 e– + 16 H+ + 2 MnO4– → 10 CO2 + 10 H+ + 10 e– + 2 Mn2+ + 8 H2O

5 H2C2O4 + 6 H+ + 2 MnO4–→ 10 CO2 + 2 Mn2+ + 8 H2O

53.) Balance the following redox reaction under basic conditions. Add water, H+, or OH– where needed.

NO2– (aq) + Al (s) → NH3 (g) + AlO2

– (aq)

Half Rxns NO2– → NH3 Al → AlO2

–

Elements “ “

Oxygen NO2– → NH3 + 2 H2O Al + 2 H2O → AlO2

–

Hydrogen NO2– + 7 H+ → NH3 + 2 H2O Al + 2 H2O → AlO2

–+ 4 H+

Charge NO2– + 7 H+ + 6 e– → NH3 + 2 H2O Al + 2 H2O → AlO2

– + 4 H+ + 3 e–

Electrons “ 2 Al + 4 H2O → 2 AlO2– + 8 H+ + 6 e–

NO2– + 7 H+ + 6 e– + 2 Al + 4 H2O → NH3 + 2 H2O + 2 AlO2

– + 8 H+ + 6 e–

NO2– + 2 Al + 2 H2O → NH3 + 2 AlO2

– + 1 H+

Since this redox is in basic solution, OH– is added to neutralize the H+.

(1 OH–) + NO2– + 2 Al + 2 H2O → NH3 + 2 AlO2

– + 1 H+ + (1 OH–)

OH– + NO2– + 2 Al + 2 H2O → NH3 + 2 AlO2

– + H2O

OH– + NO2– + 2 Al + H2O → NH3 + 2 AlO2

–

54.) Balance the following half-reaction under acidic conditions. Add water and H+ where needed.

ClO3– (aq) → ClO2 (aq)

Half-reactions ClO3– → ClO2

Oxygen ClO3– → ClO2 + H2O

Hydrogen ClO3– + 2 H+→ ClO2 + H2O

Electrons ClO3– + 2 H+ + 1 e– → ClO2 + H2O

chapter 4 hw solutions - chem 1, g-chem...

Documents