# Solution Set 3


 UC Davis CHE 2A: General Chemistry  Instructor: Hayashi Chem 2A        Chem 2B        Chem 2C Reading Assignments Exercises Worksheets Homework Unit I: Atomic Theory            Unit II: Chemical Reactions           Unit III: Gases          Unit IV: Electronic Structure and Bonding

## 3.1

1. Write a balanced molecular equation for the dissociation of $$NaCl$$ in water.
2. Based on the equation in part a, is $$NaCl$$ a strong electrolyte or a weak electrolyte?
3. Hydrofluoric acid is a weak acid, and does not completely dissociate in water. Is it a strong electrolyte or a weak electrolyte?

A4.15S

1. $$NaCl \rightarrow Na^+ + Cl^-$$
2. $$NaCl$$ is a strong electrolyte
3. Hydrofluoric acid is a weak electrolyte.

Hint:

How do strong electrolytes dissociate in water vs. weak electrolytes.

Solution

1. $$NaCl$$ dissociates completely in water, therefore NaCl breaks up into one positively charged sodium ion ($$Na^+$$) and one negatively charged chlorine ion ($$Cl_-$$).
2. When a compound in water dissociates completely into ions it is a strong electrolyte. This is because the resulting solution is able to more easily conduct electricity.
3. Hydrofluoric acid does not dissociate completely (weak acid), therefore a solution of $$HF$$ and $$H_2O$$ would not conduct electricity easily. Therefore it is a weak electrolyte.

## 3.2

Write the balanced, complete ionic, and net ionic equations for the following reactions :

1. $$CoCl_2 + NaCl \rightarrow$$
2. $$KSO_4 + Ba(NO_3)_2 \rightarrow$$
3. $$NaOH + CaCl_2 \rightarrow$$

1. Balanced: $$CoCl_{2\;(aq)} + NaCl_{(aq)} \rightarrow CoCl_{2\;(aq)} + NaCl_{(aq)}$$
• Complete ionic: $$Co^{2+}_{(aq)} + 2Cl^-_{(aq)} + Na^+_{(aq)} + Cl^-_{(aq)} \rightarrow Co^{2+}_{(aq)} + 2Cl^-_{(aq)} + Na^+_{(aq)} + Cl^-_{(aq)}$$
• Net ionic: No reaction
1. Balanced: $$K_2SO_{4\;(aq)} + Ba(NO_3)_{2\;(aq)} \rightarrow BaSO_{4\;(s)} + 2KNO_{3\;(aq)}$$
• Complete ionic equation: $$2K^+_{(aq)} + SO^{2-}_{4\;(aq)} + Ba^{2+}_{(aq)} +2NO_{3\;(aq)}+ \rightarrow BaSO_{4\;(s)} + 2K^+_{(aq)} +2NO^+_{3\;(aq)}$$
• Net ionic equation: $$SO^{2-}_{4\;(aq)} + Ba^{2+}_{(aq)} \rightarrow BaSO_{4\;(s)}$$
1. Balanced: $$2NaOH_{(aq)} + CaCl_{2\;(aq)} \rightarrow 2NaCl_{(aq)} + Ca(OH)_{2\;(s)}$$
• Complete ionic equation: $$2Na^+_{(aq)} + 2OH^-_{(aq)} + Ca^{2+}_{(aq)} + 2Cl^-_{(aq)} \rightarrow 2Na^+_{(aq)} + 2Cl^-_{(aq)} + Ca(OH)_{2\;(s)}$$
• Net ionic equation: $$Ca^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Ca(OH)_{2\;(s)}$$

Hint:

Ions not used in the creation of a solid are spectator ions

Solution

1. The equation reads the same forwards as backwards, and all compounds in the balanced equation are soluble. Because the products and reactants are the same, the Net result is that no reaction occurs.
2. All reactants and products are soluble except for BaSO4(s), because although most sulfates are soluble, Barium is an exception to this rule. Therefore all other ions are spectator ions except for the ions involved in the synthesis of BaSO4(s)
3. The product NaCl is a soluble salt whereas Ca(OH)2(s) is an insoluble hydroxide. Ca(OH)2(s) is the only precipitate that forms during this reaction therefore all ions are spectator ions except for thos involved in the synthesis of Ca(OH)2(s).

## 3.3

What amount of 0.100 M $$KOH$$ is needed to neutralize 10.00 mL of 0.210 M $$HF$$?

21.0 mL

Hint:

Set up the balanced equation for this reaction.

Solution

This reaction is one to one:

$HF + KOH \rightarrow H_2O + KF$

Take the given information and let the equation guide the conversion:

$10.00 \; mL \; HF \times \dfrac{0.210\; mM \;HF}{1 \;mL \;HF} \times \dfrac{1\; mM\; KOH}{1\;mM \;HF} \times \dfrac{1\; mL\; KOH}{ 0.100\; mM\; KOH} = 21.0\; mL$

By the way: Like other sources of fluoride ions, $$KF$$ is poisonous, although lethal doses approach gram levels for humans. It is harmful by inhalation and ingestion. It is highly corrosive, and skin contact may cause severe burns.

## 3.4

A students has 97 mL of 0.80 M HCl which they would like to neutralize with 0.18M NaOH. What volume of NaOH should be used?

0.43 L

Hint:

Find the amount of moles of HCl present to find the amount of moles of NaOH are needed to neutralize the HCl.

Solution

First find how many moles of HCl are present in the initial solution, do this through the equation $$(0.80\; mol/L) \times (0.097\;L)$$ to find that 0.0776 moles of HCl are present. This indicates how many moles of NaOH are needed to neutralize this solution. To find the volume of NaOH needed to neutralize the moles of HCl use the equation

(0.0776 moles of HCl) = (0.18 moles of NaOH/ L) * (X amount L)

to find that the amount of Liters of NaOH needed is 0.43 L.

## 3.5

Below are a list of reactions. State whether each is an oxidation-reduction reaction and if it is an oxidation-reduction reaction, identify the reducing and oxidizing agents.

1. $$CO_2(g) + H_2(g) \rightarrow CO_{(g)} + H_2O_{(g)}$$
2. $$K_2SO_{4\, (aq)} + BaI2(aq) \rightarrow BaSO_{4\, (s)} + 2KI_{(aq)}$$
3. $$H_3PO_{4\,(aq)} + Se^{2-}_{(aq)} \rightarrow HPO_4^{2-}(aq) + H_2Se_{(aq)}$$
4. $$Zn (s) + CuSO_4 \rightarrow ZnSO_4(aq) + Cu(s)$$

1. redox; the reducing agent being oxidized is hydrogen and the oxidizing agent being reduced is carbon.
2. not redox
3. not redox
4. redox; zinc is the reducing agent that is oxidized and copper is the oxidizing agent that is reduced.

Hint:

Take note of the oxidation numbers of each element.

Solution

1. Carbon has an oxidation number of 4+ on the left side of the equation and 2+ on the right side. Oxygen’s oxidation number is 2- on both sides of the equation and hydrogen has an oxidation number of 0 on the left and 1+ on the right. This means that carbon is the oxidizing agent that steals electrons from hydrogen and hydrogen is the reducing agent that gives electrons to carbon.
2. The oxidation numbers of each element remain the same on either side of the equation therefore this is not a redox equation.
3. Same as the solution to part b.
4. The oxidation number of zinc is 0 on the left and 2+ on the right, the opposite is true of copper. This means that zinc is the reducing agent and copper is the oxidizing agent.

## 3.6

Indicate the oxidation state of the element assigned in the compounds below:

1. $$C$$ in $$CO_2$$
2. $$S$$ in $$CaSO_4$$
3. $$Sn$$ in $$SnO_2$$
4. $$Cd$$ in $$Cd$$
5. $$Co$$ in $$CoCl_2$$

1. Oxidation state of C: +4
2. Oxidation state of S: +6
3. Oxidation state of Sn: +4
4. Oxidation state of Cd: 0
5. Oxidation state of Co: +2

Hint:

Solution

1. (Oxidation state of C) + (2* Oxidation state of O) = 0

(Oxidation state of C) + (2*-2) = 0

(Oxidation state of C) = +4

1. (Oxidation state of Ca)+(Oxidation state of S)+(4*Oxidation state of O) = 0

(+2) + (Oxidation state of S) + (4*-2) = 0

(Oxidation state of S) = +6

1. (Oxidation state of Sn) + (2* Oxidation state of O) = 0

(Oxidation state of Sn) + (2* -2) = 0

(Oxidation state of Sn) = +4

1. (Oxidation State of Cd) =0
2. (Oxidation state of Co) + (2* Oxidation state of Cl) = 0

(Oxidation state of Co) + (2*-1) = 0

(Oxidation state of Co) = +2

## 3.7

When titrating a 4.00 mL volume of the diprotic acid, $$H_2SO_4$$ of 0.645 M concentration, what volume of 0.493 M $$NaOH$$ is necessary?

10.5 mL

Hint:

How do a strong base and weak polyprotic acid react together?

Solution

Write the equation out the neutralization reaction:

$H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$

So then,

$4.00\;mL\; H_2SO_4 \times \dfrac{0.645\; mmol\; H_2SO_4}{1.00\; mL\; H_2SO_4}$

$= 2.58\; mmol\; H_2SO_4 \times \dfrac{2\; mmol\; NaOH}{1\; mmol\; H_2SO_4} \times {1\; mL\; NaOH}{0.493\; mmol\; NaOH} = 10.5\; mL$

## 3.8

Write down a balanced formula equation for an acid-base reaction that forms potassium nitrate.

$KOH_{(aq)} + HNO_{3\, (aq)} \rightarrow 2KNO_{3\, (aq)} + H_2O_{(l)}$

Hint:

• $$H^+$$ from the acid and $$OH^-$$ from the base produces water
• $$K^+$$ and $$NO_3^-$$ comes from either the acid or the base.

Solution

In an acid- base reaction, the products are always water and a salt (which may or may not be soluble). In this case, the salt is $$KNO_3$$. Water is produced by the $$H^+$$ and $$OH^-$$ from the acid and base. The acid and the base in the reaction must also contains $$K^+$$ and $$NO_3^-$$ to form $$KNO_3$$. So the acid is $$HNO_3$$ and the base is $$KOH$$.

## 3.9

Balance the following equations in both acidic and basic environments:

1. $$H_2(g) + O_2(g) \rightarrow H_2O(l)$$
2. $$Cr_2O_7^{2-}(aq) + C_2H_5OH (l) \rightarrow Cr^{3+}(aq) + CO_2(g)$$
3. $$Fe^{2+}(aq) + MnO_4^-(aq) \rightarrow Fe^{3+}(aq) + Mn^{2+}(aq)$$
4. $$Zn (s) + NO_3^-(aq) \rightarrow Zn^{2+}(aq) + NO(g)$$
5. $$Al (s) + H_2O (l) + O_2 (g) \rightarrow [Al(OH)_4]^-(aq)$$
6. Balance in only acidic Solutions
1. $$Mn^{2+} + BiO_3^- \rightarrow MnO4^- + Bi^{3+}$$
2. $$ClO_3^- + Cl - \rightarrow Cl_2 + ClO_2$$
7. Balance in only Basic Solution
1. $$Zn + NO_3^ - \rightarrow Zn(OH)_4^{2-} + NH_3$$
2. $$Al + OH^- \rightarrow AlO_2^- + H_2$$

a) $$H_2(g) + O_2(g) \rightarrow H_2O(l)$$

• Acidic Answer: $$2H_{2\;(g)} + O_{2\;(g)} \rightarrow 2H_2O_{(l)}$$
• Basic Answer: $$2H_{2\;(g)} + O_{2\;(g)} \rightarrow 2H_2O_{(l)}$$

b) $$Cr_2O_7^{2-}(aq) + C_2H_5OH (l) \rightarrow Cr^{3+}(aq) + CO_2(g)$$

• Acidic Answer: $$2Cr_2O_7^{2-}(aq) + C_2H_5OH_{(l)} + 16H^+_{(aq)} \rightarrow 4Cr^{3+}_{(aq)} + 2CO_{2\;(g)} + 11H_2O_{(l)}$$
• Basic Answer: $$2Cr_2O_7^{2-}(aq) + C_2H_5OH(l) + 5 H_2O_{(l)} \rightarrow 4Cr^{3+}_{(aq)} + 2CO_{2\;(g)} + 16OH^-(aq)$$

c) $$Fe^{2+}(aq) + MnO_4^-(aq) \rightarrow Fe^{3+}(aq) + Mn^{2+}(aq)$$

• Acidic Answer: $$MnO_4^-(aq) + 5Fe^{2+}_{(aq)} + 8H^+_{(aq)} \rightarrow Mn^{2+}_{(aq)} + 5Fe^{3+}_{(aq)} + 4H_2O_{(l)}$$
• Basic Answer: $$MnO_4^-(aq) + 5Fe^{2+}_{(aq)} + 4H_2O_{(l)} \rightarrow Mn^{2+}_{(aq)} + 5Fe^{3+}_{(aq)} + 8OH^-_{(aq)}$$

d) $$Zn (s) + NO_3^-(aq) \rightarrow Zn^{2+}(aq) + NO(g)$$

• Acidic Answer: $$3Zn(s) + 2NO_3^-(aq) + 8H^+_{(aq)} \rightarrow 3Zn^{2+}_{(aq)} + 2NO_{(g)} + 4H_2O_{(l)}$$
• Basic Answer: $$3Zn(s) + 2NO_3^-(aq) + 4H_2O_{(l)} \rightarrow 3Zn^{2+}_{(aq)} + 2NO_{(g)} + 8OH^-_{(aq)}$$

e) $$Al (s) + H_2O (l) + O_2 (g) \rightarrow [Al(OH)_4]^-(aq)$$

• Acidic Answer: $$4Al(s) + 3O_{2\;(g)} + 10 H_2O_{(l)} \rightarrow 4[Al(OH)_4]^-(aq) + 4H^+_{(aq)}$$
• Basic Answer: $$4Al(s) + 3O_{2\;(g)} + 6H_2O_{(l)} + 4OH^-_{(aq)} \rightarrow 4[Al(OH)_4]^-_{(aq)}$$

f) Balance in only acidic Solutions

1. $$14 H^+ + 2 Mn^{2+} + 5 BiO_3^- \rightarrow 2 MnO_4 - + 5 Bi ^{3+} + 7 H_2O$$
2. $$4 H^+ + 2 ClO_3^- + 2 Cl^- \rightarrow 2 ClO_2 + 2 H_2O + Cl_2$$

g) Balance in in only Basic Solution

1. $$6 H_2O + NO_3^- + 7 OH^- + 4 Zn \rightarrow NH_3 + 4 Zn(OH)_4^{2-}$$
2. $$2 Al + 2 H_2O + 2 OH^- \rightarrow 2 AlO_2^- + 3 H_2$$

## 3.10

Write the products of these acid-base reactions. Balance the complete reaction.

1. $$HClO_{4\; (aq)} + Mg(OH)_{2\; (aq)} \rightarrow$$
2. $$H_2S_{(aq)} + NaOH_{(aq)} \rightarrow$$

1. $$2HClO_4 (aq) + Mg(OH)_2 (aq) \rightarrow Mg(ClO_4)_2 (aq) + 2H_2O (l)$$
2. $$H_2S(aq) + 2NaOH (aq) \rightarrow Na_2S (aq) + 2H_2O (l)$$

Hint

Solution

Acid + Base -> Salt + Water

1. $$Mg^{2+}$$ is soluble in $$ClO_4^-$$. Balance the equation.
2. $$Na^+$$ is soluble in $$S^{2-}$$. Balance the equation.

## 3.11

Write chemical reactions for the dissociation of the following compounds in water and determine if they are strong electrolytes or weak electrolytes: $$HCl$$, $$NaOH$$, $$MgCl_2$$, $$Al_2S_3$$

A.4.25S -

1. strong electrolyte: $$HCl \rightarrow H^+ + Cl^-$$
2. strong electrolyte: $$NaOH \rightarrow Na^+ + OH^-$$
3. strong electrolyte: $$MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$$
4. weak electrolyte: $$Al_2S_3 \rightarrow Al_2S_3$$

Hint

Solution

1. $$HCl$$ is a strong acid and dissociates in water completely. It dissociates into the ions $$H^+$$ and $$Cl^-$$ therefore creating an electrolytic solution and being a strong electrolyte.
2. $$NaOH$$ is a strong electrolyte and dissociates in water completely. It dissociates into the ions $$Na^+$$ and $$OH^-$$, therefore creating an electrolytic solution and being a strong electrolyte.
3. $$MgCl_2$$ is a salt that dissociates in water completely. It dissociates into $$Mg^{2+}$$ ions and $$Cl^-$$ ions, therefore creating an electrolytic solution and being a strong electrolyte.
4. $$Al_2S_3$$ is a sulfide compound, Solubility rules indicate that sulfide compounds are only soluble in water when paired with an element from either group 1 or group 2. Aluminum is a group 3 element, therefore $$Al_2S_3$$ is insoluble and a weak electrolyte.

## 3.12

Dichloromethane ($$CH_2Cl_2$$) is a useful solvent for many chemical processes in the organic chemistry lab. In the food industry, dichloromethane can be used to decaffeinated coffee and tea as well as to prepare extracts of hops and other flavorings.

Figure: Sample of dichloromethane. from Wikipedia.

Dichloromethane's vapor pressure at 25 °C is 57.3 kPa. Convert this pressure to the following pressure units.

1. atm
2. torr
3. psi
4. mm H
5. bar

1. 0.566 atm
2. 430 torr
3. 8.32 psi
4. 430 mm Hg
5. 0.573 bar

Hint

Solution

1. 1 atm =101.325 kPa, so (1 atm/101.325 kPa)(57.3 kPa)=0.566atm
2. 1 atm=760 torr, so (760 torr/1 atm)(0.56 6atm)= 430 torr
3. 1 atm=14.7 psi, so (14.7 psi/1 atm)(0.566 atm)= 8.32 psi
4. 1 torr=1mm Hg, so (1mm Hg/1 torr)(430 torr)= 430 mm Hg
5. 1 atm=1.01325 bar, so (1.01325 bar/1atm)(0.566 atm)= 0.573 bar

## 3.13

An ideal gas is contained in a flask with a volume of 10 mL. If the volume of the container is increased to 50 mL, what is the final pressure, $$P_f$$ in the flask? Assume the original pressure is $$P_i$$ and the temperature is maintained constant.

• $$P_f=\dfrac{P_i}{5}$$

Hint

Solution

For a given mass, at constant temperature, the pressure times the volume is a constant. If the same substance under two different sets of condition the law can be expressed as follow.

$P_iV_i=P_fV_f=C$

So

$P_i \times 10\; mL = P_f \times 50\; mL$

$P_f=\dfrac{P_i}{5}$

## 3.14

A truck tire will blow out if its internal pressure increases to 2 atm. In a morning, a driver filled the tire will air to 1.50 atm at $$18 ^oC$$. Driving the truck at noon increases the temperature to $$32 ^oC$$. Assuming the volume of the tire is constant, what’s the final condition of the tire?

Figure: Overpressurizing a tire can result in catastrophic failure (i.e., "a blown tire").

• Since $$P_f= 1.57\; atm < 2\; atm$$, the tire will not blow out.

Hint

Solution

Step 1: Write down your given information:

• $$P_i=1.5\;atm$$
• $$P_f =?$$
• $$V_f=V_i$$
• $$n_f=n_i$$
• $$R = 0.0820574 \; L*atm*mol^{-1}K^{-1}$$
• $$T_i= 18^oC$$
• $$T_f=32 ^oC$$

Step 2: Convert as necessary:

• Temperature into Kelvin:

$T_i= 18^oC + 273\;K = 291 K$

$T_f=32 ^oC+273\; K =305\; K$

Step 3: Plug in the variables into the ideal gas law and solve for volume:

$V=\dfrac{nRT}{P}$

$P=\left(\dfrac{nR}{V}\right) T$

$\dfrac{P}{T}=\dfrac{nR}{V}$

$P_f=\dfrac{n_fRT_f}{V}$

$P_f=\dfrac{nR}{V} \times T_f$

$P_f=\dfrac{P_i}{T_i} \times T_f$

$P_f = \dfrac{1.50\;atm}{291\;K} \times 305\;K$

• Since $$P_f= 1.57\; atm < 2\; atm$$, the tire will not blow out.

## 3.15

Please calculate the density of nitrogen gas at STP.

• $$\rho =1.251g/L$$

Hint

Solution

• STP corresponds to 273 K (0° Celsius) and 1 atm

Step 1:Write down your given information:

• $$P=1\;atm$$
• $$V=1\;L$$
• $$n=?$$
• $$R=0.0820574 \;L*atm*mol^{-1}K^{-1}$$
• $$T=273\;K$$

Step 2: Convert as necessary:

Temperature is already given in Kelvin. The pressure is given in units to match those of $$R$$

Step 3: Plug in the variables into the appropriate equation:

$V=\dfrac{nRT}{P}$

$n=\dfrac{VP}{RT}$

$n=\dfrac{1\; L \times 1\; atm}{0.0820574 L*atm*mol^{-1}K^{-1} \times 273\; K}$

therefore, there are $$n=0.0446\; mol$$ of $$N_2$$, but we need mass, not moles.

$m=MM \times n$

$m=28.01\; g \times 0.0446\; mol$

$m=1.25\; g$

$\rho=\dfrac{m}{V}$

$\rho=\dfrac{1.251\; g}{1\; L}$

$\rho=1.251\; g/L$

## 3.16

A chemist notices that one of many organic solvents is leaking from the hood. The rate of effusion of this organic gas is 2.866 mL/sec while the effusion of oxygen gas is 3.640 mL/sec. What is the molar mass of this organic solvent?

• $$MM= 58.1\; g/mol$$

Hint

Solution

Graham's Law states that the rate of effusion of two different gases at the same conditions are inversely proportional to the square roots of their molar masses as given by the following equation:

$\dfrac{\text{Rate of effusion of A}}{\text{Rate of effusion of B}} =\sqrt{\dfrac{MM_B}{MM_A}}$

Assuming A is oxygen gas and B is the organic solvent

$\dfrac{3.640\; mL/sec}{2.866\; mL/sec}=\sqrt{\dfrac{MM_B}{36.0\;g/mol}}$

$MM_B= 58.1\; g/mol$

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