Chapter 4 Reactions in Aqueous Solution

Example 2

Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH.

Given: identity of solute and volume and molarity of solution

Asked for: amount of solute in moles

Strategy:

Use either Equation 4.5 or Equation 4.6, depending on the units given in the problem.

Solution:

Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation 4.5 is more useful:

$$\text{moles NaOH }=V_{\text{L}}{\text{M}}_{\text{mol/L}}=\text{ (2}\text{.50 }\cancel{\text{L}}\text{)}\left(\frac{\text{0}\text{.100 mol}}{\cancel{\text{L}}}\right)=\text{ 0}\text{.250 mol NaOH}$$

Exercise

Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine.

Answer: 41.6 mmol

Example 3

The solution in Figure 4.6 "Preparation of a Solution of Known Concentration Using a Solid Solute" contains 10.0 g of cobalt(II) chloride dihydrate, CoCl₂·2H₂O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl₂·2H₂O?

Given: mass of solute and volume of solution

Asked for: concentration (M)

Strategy:

To find the number of moles of CoCl₂·2H₂O, divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters.

Solution:

The molar mass of CoCl₂·2H₂O is 165.87 g/mol. Therefore,

$${\text{moles CoCl}}_{\text{2}}{\text{•2H}}_{\text{2}}\text{O }=\left(\frac{\text{10}\text{.0 }\cancel{\text{g}}}{\text{165}\text{.87 }\cancel{\text{g}}\text{/mol}}\right)=\text{ 0}\text{.0603 mol}$$

The volume of the solution in liters is

$$\text{volume }=\text{ 500 mL}\left(\frac{\text{1 L}}{\text{1000 }\cancel{\text{mL}}}\right)=\text{ 0}\text{.500 L}$$

Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is

$$\text{molarity }=\frac{\text{0}\text{.0603 mol}}{\text{0}\text{.500 L}}=\text{ 0}\text{.121 M }={\text{CoCl}}_{\text{2}}{\text{•H}}_{\text{2}}\text{O}$$

Exercise

The solution shown in Figure 4.7 "Preparation of 250 mL of a Solution of (NH" contains 90.0 g of (NH₄)₂Cr₂O₇ in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate?

Answer: (NH₄)₂Cr₂O₇ = 1.43 M

Example 4

The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol.

Given: molarity, volume, and molar mass of solute

Asked for: mass of solute

Strategy:

A Calculate the number of moles of glucose contained in the specified volume of solution by multiplying the volume of the solution by its molarity.

B Obtain the mass of glucose needed by multiplying the number of moles of the compound by its molar mass.

Solution:

A We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution:

$$\begin{array}{c}{V}_{\text{L}}{\text{M}}_{\text{mol/L}}=\text{ moles}\\ \text{500 }\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}\text{.310 mol glucose}}{\text{1}\cancel{\text{L}}}\right)=\text{ 0}\text{.155 mol glucose}\end{array}$$

B We then convert the number of moles of glucose to the required mass of glucose:

$$\text{mass of glucose }=\text{ 0}\text{.155 }\cancel{\text{mol glucose}}\left(\frac{\text{180}\text{.16 g glucose}}{\text{1 }\cancel{\text{mol glucose}}}\right)=\text{ 27}\text{.9 g glucose}$$

Exercise

Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution.

Answer: 2.3 g NaCl

Example 5

What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example 4?

Given: volume and molarity of dilute solution

Asked for: volume of stock solution

Strategy:

A Calculate the number of moles of glucose contained in the indicated volume of dilute solution by multiplying the volume of the solution by its molarity.

B To determine the volume of stock solution needed, divide the number of moles of glucose by the molarity of the stock solution.

Solution:

A The D5W solution in Example 4 was 0.310 M glucose. We begin by using Equation 4.7 to calculate the number of moles of glucose contained in 2500 mL of the solution:

$$\text{moles glucose }=\text{ 2500}\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}\text{.310 mol glucose}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 0}\text{.775 mol glucose}$$

B We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose:

$$\text{volume of stock soln }=\text{ 0}\text{.775 }\cancel{\text{mol glucose}}\left(\frac{\text{1 L}}{\text{3}\text{.00 }\cancel{\text{mol glucose}}}\right)=\text{ 0}\text{.258 L or 258 mL}$$

In determining the volume of stock solution that was needed, we had to divide the desired number of moles of glucose by the concentration of the stock solution to obtain the appropriate units. Also, the number of moles of solute in 258 mL of the stock solution is the same as the number of moles in 2500 mL of the more dilute solution; only the amount of solvent has changed. The answer we obtained makes sense: diluting the stock solution about tenfold increases its volume by about a factor of 10 (258 mL → 2500 mL). Consequently, the concentration of the solute must decrease by about a factor of 10, as it does (3.00 M → 0.310 M).

We could also have solved this problem in a single step by solving Equation 4.7 for V_s and substituting the appropriate values:

$$V_{\text{s}}=\frac{\text{(}{V}_{\text{d}}{\text{)(M}}_{\text{d}}\text{)}}{{\text{M}}_{\text{s}}}=\frac{\text{(2}\text{.500 L)(0}\text{.310 }\cancel{\text{M}}\text{)}}{\text{3}\text{.00 }\cancel{\text{M}}}=\text{ 0}\text{.258 L}$$

As we have noted, there is often more than one correct way to solve a problem.

Exercise

What volume of a 5.0 M NaCl stock solution is necessary to prepare 500 mL of normal saline solution (0.16 M NaCl)?

Answer: 16 mL

Example 6

What are the concentrations of all species derived from the solutes in these aqueous solutions?

1. 0.21 M NaOH
2. 3.7 M (CH₃)CHOH
3. 0.032 M In(NO₃)₃

Given: molarity

Asked for: concentrations

Strategy:

A Classify each compound as either a strong electrolyte or a nonelectrolyte.

B If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution.

Solution:

1. Sodium hydroxide is an ionic compound that is a strong electrolyte (and a strong base) in aqueous solution:

   $$\text{NaOH(s)}\stackrel{{\text{H}}_{\text{2}}\text{O(l)}}{\to }{\text{Na}}^{+}\text{(aq)}+{\text{OH}}^{–}\text{(aq)}$$

   B Because each formula unit of NaOH produces one Na⁺ ion and one OH⁻ ion, the concentration of each ion is the same as the concentration of NaOH: [Na⁺] = 0.21 M and [OH⁻] = 0.21 M.

2. A The formula (CH₃)₂CHOH represents 2-propanol (isopropyl alcohol) and contains the –OH group, so it is an alcohol. Recall from Section 4.1 "Aqueous Solutions" that alcohols are covalent compounds that dissolve in water to give solutions of neutral molecules. Thus alcohols are nonelectrolytes.

   B The only solute species in solution is therefore (CH₃)₂CHOH molecules, so [(CH₃)₂CHOH] = 3.7 M.

3. A Indium nitrate is an ionic compound that contains In³⁺ ions and NO₃⁻ ions, so we expect it to behave like a strong electrolyte in aqueous solution:

   $${\text{In(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(s)}\stackrel{{\text{H}}_{\text{2}}\text{O(l)}}{\to }{\text{In}}^{\text{3+}}\text{(aq)}+{\text{3NO}}_{\text{3}}{}^{–}\text{(aq)}$$

   B One formula unit of In(NO₃)₃ produces one In³⁺ ion and three NO₃⁻ ions, so a 0.032 M In(NO₃)₃ solution contains 0.032 M In³⁺ and 3 × 0.032 M = 0.096 M NO₃^–—that is, [In³⁺] = 0.032 M and [NO₃⁻] = 0.096 M.

Exercise

What are the concentrations of all species derived from the solutes in these aqueous solutions?

1. 0.0012 M Ba(OH)₂
2. 0.17 M Na₂SO₄
3. 0.50 M (CH₃)₂CO, commonly known as acetone

Answer:

1. [Ba²⁺] = 0.0012 M; [OH⁻] = 0.0024 M
2. [Na⁺] = 0.34 M; [SO₄²⁻] = 0.17 M
3. [(CH₃)₂CO] = 0.50 M

Example 7

Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN)₂]⁻ ion. Gold is then recovered by reduction with metallic zinc according to the following equation:

Zn(s) + 2[Au(CN)₂]⁻(aq) → [Zn(CN)₄]²⁻(aq) + 2Au(s)

What mass of gold would you expect to recover from 400.0 L of a 3.30 × 10⁻⁴ M solution of [Au(CN)₂]⁻?

Given: chemical equation and molarity and volume of reactant

Asked for: mass of product

Strategy:

A Check the chemical equation to make sure it is balanced as written; balance if necessary. Then calculate the number of moles of [Au(CN)₂]⁻ present by multiplying the volume of the solution by its concentration.

B From the balanced chemical equation, use a mole ratio to calculate the number of moles of gold that can be obtained from the reaction. To calculate the mass of gold recovered, multiply the number of moles of gold by its molar mass.

Solution:

A The equation is balanced as written, so we can proceed to the stoichiometric calculation. We can adapt Figure 4.10 "An Expanded Flowchart for Stoichiometric Calculations" for this particular problem as follows:

As indicated in the strategy, we start by calculating the number of moles of [Au(CN)₂]⁻ present in the solution from the volume and concentration of the [Au(CN)₂]⁻ solution:

$$\begin{array}{ll}{\text{moles [Au(CN)}}_{\text{2}}{\text{]}}^{–}\hfill & =V_{\text{L}}{\text{M}}_{\text{mol/L}}\hfill \\ \hfill & =\text{ 400}\text{.0 }\cancel{\text{L}}\left(\frac{\text{3}\text{.30}\times {\text{10}}^{\text{4–}}{\text{ mol [Au(CN)}}_{\text{2}}{\text{]}}^{–}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 0}{\text{.132 mol [Au(CN)}}_{\text{2}}{\text{]}}^{–}\hfill \end{array}$$

B Because the coefficients of gold and the [Au(CN)₂]⁻ ion are the same in the balanced chemical equation, if we assume that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN)₂]⁻ we started with (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so we need to convert the number of moles of gold to the corresponding mass using the molar mass of gold:

$$\begin{array}{ll}\text{mass of Au }\hfill & =\text{ (moles Au)(molar mass Au)}\hfill \\ \hfill & =\text{ 0}\text{.132 }\cancel{\text{mol Au}}\left(\frac{\text{196}\text{.97 g Au}}{\text{1 }\cancel{\text{mol Au}}}\right)=\text{ 26}\text{.0 g Au}\hfill \end{array}$$

At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth $1170.

$$\text{26}\text{.0 }\cancel{\text{g Au}}\times \frac{\text{1 }\cancel{\text{troy oz}}}{\text{31}\text{.10 }\cancel{\text{g}}}\times \frac{\text{\$1400}}{\text{1 }\cancel{\text{troy oz Au}}}=\text{ \$1170}$$

Exercise

What mass of solid lanthanum(III) oxalate nonahydrate [La₂(C₂O₄)₃·9H₂O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl₃ by adding a stoichiometric amount of sodium oxalate?

Answer: 3.89 g

Example 8

Because the consumption of alcoholic beverages adversely affects the performance of tasks that require skill and judgment, in most countries it is illegal to drive while under the influence of alcohol. In almost all US states, a blood alcohol level of 0.08% by volume is considered legally drunk. Higher levels cause acute intoxication (0.20%), unconsciousness (about 0.30%), and even death (about 0.50%). The Breathalyzer is a portable device that measures the ethanol concentration in a person’s breath, which is directly proportional to the blood alcohol level. The reaction used in the Breathalyzer is the oxidation of ethanol by the dichromate ion:

$${\text{3CH}}_{\text{3}}{\text{CH}}_{\text{2}}\text{OH(aq)}+\underset{\text{yellow-orange}}{{\text{2Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}–}}\text{(aq)}+{\text{16H}}^{\text{ + }}\text{(aq)}\underset{{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}}{\overset{{\text{Ag}}^{\text{ + }}}{\to }}\underset{\text{green}}{{\text{3CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H(aq)}+{\text{4Cr}}^{\text{3+}}\text{(aq)}+{\text{11H}}_{\text{2}}\text{O(l)}}$$

When a measured volume (52.5 mL) of a suspect’s breath is bubbled through a solution of excess potassium dichromate in dilute sulfuric acid, the ethanol is rapidly absorbed and oxidized to acetic acid by the dichromate ions. In the process, the chromium atoms in some of the Cr₂O₇²⁻ ions are reduced from Cr⁶⁺ to Cr³⁺. In the presence of Ag⁺ ions that act as a catalyst, the reaction is complete in less than a minute. Because the Cr₂O₇²⁻ ion (the reactant) is yellow-orange and the Cr³⁺ ion (the product) forms a green solution, the amount of ethanol in the person’s breath (the limiting reactant) can be determined quite accurately by comparing the color of the final solution with the colors of standard solutions prepared with known amounts of ethanol.

A typical Breathalyzer ampul contains 3.0 mL of a 0.25 mg/mL solution of K₂Cr₂O₇ in 50% H₂SO₄ as well as a fixed concentration of AgNO₃ (typically 0.25 mg/mL is used for this purpose). How many grams of ethanol must be present in 52.5 mL of a person’s breath to convert all the Cr⁶⁺ to Cr³⁺?

Given: volume and concentration of one reactant

Asked for: mass of other reactant needed for complete reaction

Strategy:

A Calculate the number of moles of Cr₂O₇²⁻ ion in 1 mL of the Breathalyzer solution by dividing the mass of K₂Cr₂O₇ by its molar mass.

B Find the total number of moles of Cr₂O₇²⁻ ion in the Breathalyzer ampul by multiplying the number of moles contained in 1 mL by the total volume of the Breathalyzer solution (3.0 mL).

C Use the mole ratios from the balanced chemical equation to calculate the number of moles of C₂H₅OH needed to react completely with the number of moles of Cr₂O₇²⁻ ions present. Then find the mass of C₂H₅OH needed by multiplying the number of moles of C₂H₅OH by its molar mass.

Solution:

A In any stoichiometry problem, the first step is always to calculate the number of moles of each reactant present. In this case, we are given the mass of K₂Cr₂O₇ in 1 mL of solution, which we can use to calculate the number of moles of K₂Cr₂O₇ contained in 1 mL:

$$\frac{{\text{moles K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{1 mL}}=\frac{\text{(0}\text{.25 }\cancel{\text{mg}}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{)}}{\text{mL}}\left(\frac{\text{1 }\cancel{\text{g}}}{\text{1000 }\cancel{\text{mg}}}\right)\left(\frac{\text{1 mol}}{\text{294}\text{.18 }\cancel{\text{g}}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}\right)=8.5\times {10}^{–7}\text{ moles}$$

B Because 1 mol of K₂Cr₂O₇ produces 1 mol of Cr₂O₇²⁻ when it dissolves, each milliliter of solution contains 8.5 × 10⁻⁷ mol of Cr₂O₇²⁻. The total number of moles of Cr₂O₇²⁻ in a 3.0 mL Breathalyzer ampul is thus

$${\text{moles Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}–}=\left(\frac{\text{8}\text{.5}\times {\text{10}}^{–\text{7}}\text{ mol}}{\text{1 }\cancel{\text{mL}}}\right)(\text{3}\text{.0 }\cancel{\text{mL}})=\text{ 2}\text{.6}\times {\text{10}}^{–\text{6}}{\text{ mol Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{2–}$$

C The balanced chemical equation tells us that 3 mol of C₂H₅OH is needed to consume 2 mol of Cr₂O₇²⁻ ion, so the total number of moles of C₂H₅OH required for complete reaction is

$${\text{moles of C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH }=(\text{2}\text{.6}\times {\text{10}}^{–\text{6}}\cancel{{\text{mol Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}–}})\left(\frac{{\text{3 mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{\text{2 }\cancel{{\text{mol Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}–}}}\right)=\text{ 3}\text{.9}\times {\text{10}}^{–\text{6}}{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$$

As indicated in the strategy, this number can be converted to the mass of C₂H₅OH using its molar mass:

$${\text{mass C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH }=(\text{3}\text{.9}\times {\text{10}}^{–\text{6}}\cancel{{\text{mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}})\left(\frac{\text{46}\text{.07 g}}{\cancel{{\text{mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}}\right)=\text{ 1}\text{.8}\times {\text{10}}^{–\text{4}}{\text{ g C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$$

Thus 1.8 × 10⁻⁴ g or 0.18 mg of C₂H₅OH must be present. Experimentally, it is found that this value corresponds to a blood alcohol level of 0.7%, which is usually fatal.

Exercise

The compound para-nitrophenol (molar mass = 139 g/mol) reacts with sodium hydroxide in aqueous solution to generate a yellow anion via the reaction

Because the amount of para-nitrophenol is easily estimated from the intensity of the yellow color that results when excess NaOH is added, reactions that produce para-nitrophenol are commonly used to measure the activity of enzymes, the catalysts in biological systems. What volume of 0.105 M NaOH must be added to 50.0 mL of a solution containing 7.20 × 10⁻⁴ g of para-nitrophenol to ensure that formation of the yellow anion is complete?

Answer: 4.93 × 10⁻⁵ L or 49.3 μL

Example 9

When aqueous solutions of silver nitrate and potassium dichromate are mixed, an exchange reaction occurs, and silver dichromate is obtained as a red solid. The overall chemical equation for the reaction is as follows:

2AgNO₃(aq) + K₂Cr₂O₇(aq) → Ag₂Cr₂O₇(s) + 2KNO₃(aq)

What mass of Ag₂Cr₂O₇ is formed when 500 mL of 0.17 M K₂Cr₂O₇ are mixed with 250 mL of 0.57 M AgNO₃?

Given: balanced chemical equation and volume and concentration of each reactant

Asked for: mass of product

Strategy:

A Calculate the number of moles of each reactant by multiplying the volume of each solution by its molarity.

B Determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient in the balanced chemical equation.

C Use mole ratios to calculate the number of moles of product that can be formed from the limiting reactant. Multiply the number of moles of the product by its molar mass to obtain the corresponding mass of product.

Solution:

A The balanced chemical equation tells us that 2 mol of AgNO₃(aq) reacts with 1 mol of K₂Cr₂O₇(aq) to form 1 mol of Ag₂Cr₂O₇(s) (Figure 4.11 "What Happens at the Molecular Level When Solutions of AgNO"). The first step is to calculate the number of moles of each reactant in the specified volumes:

$$\begin{array}{l}{\text{moles K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}=\text{ 500 }\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}{\text{.17 mol K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 0}{\text{.085 mol K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\\ {\text{moles AgNO}}_{\text{3}}=\text{ 250 }\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}{\text{.57 mol AgNO}}_{\text{3}}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 0}{\text{.14 mol AgNO}}_{\text{3}}\end{array}$$

B Now we can determine which reactant is limiting by dividing the number of moles of each reactant by its stoichiometric coefficient:

$$\begin{array}{l}{\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{: }\frac{\text{0}\text{.085 mol}}{\text{1 mol}}=\text{ 0}\text{.085}\\ {\text{AgNO}}_{\text{3}}\text{: }\frac{\text{0}\text{.14 mol}}{\text{2 mol}}=\text{ 0}\text{.070}\end{array}$$

Because 0.070 < 0.085, we know that AgNO₃ is the limiting reactant.

C Each mole of Ag₂Cr₂O₇ formed requires 2 mol of the limiting reactant (AgNO₃), so we can obtain only 0.14/2 = 0.070 mol of Ag₂Cr₂O₇. Finally, we convert the number of moles of Ag₂Cr₂O₇ to the corresponding mass:

$${\text{mass of Ag}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}=\text{ 0}\text{.070 }\cancel{\text{mol}}\left(\frac{\text{431}\text{.72 g}}{\text{1 }\cancel{\text{mol}}}\right)={\text{30 g Ag}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}$$

Exercise

Aqueous solutions of sodium bicarbonate and sulfuric acid react to produce carbon dioxide according to the following equation:

2NaHCO₃(aq) + H₂SO₄(aq) → 2CO₂(g) + Na₂SO₄(aq) + 2H₂O(l)

If 13.0 mL of 3.0 M H₂SO₄ are added to 732 mL of 0.112 M NaHCO₃, what mass of CO₂ is produced?

Answer: 3.4 g

Example 11

Using the information in Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.

1. Aqueous solutions of barium chloride and lithium sulfate are mixed.
2. Aqueous solutions of rubidium hydroxide and cobalt(II) chloride are mixed.
3. Aqueous solutions of strontium bromide and aluminum nitrate are mixed.
4. Solid lead(II) acetate is added to an aqueous solution of ammonium iodide.

Given: reactants

Asked for: reaction and net ionic equation

Strategy:

A Identify the ions present in solution and write the products of each possible exchange reaction.

B Refer to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water" to determine which, if any, of the products is insoluble and will therefore form a precipitate. If a precipitate forms, write the net ionic equation for the reaction.

Solution:

1. A Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions initially gives an aqueous solution that contains Ba²⁺, Cl⁻, Li⁺, and SO₄²⁻ ions. The only possible exchange reaction is to form LiCl and BaSO₄:

   

    

   B We now need to decide whether either of these products is insoluble. Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water" shows that LiCl is soluble in water (rules 1 and 4), but BaSO₄ is not soluble in water (rule 5). Thus BaSO₄ will precipitate according to the net ionic equation

   Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

   Although soluble barium salts are toxic, BaSO₄ is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine BaSO₄ particles in water.

   

   An x-ray of the digestive organs of a patient who has swallowed a “barium milkshake.” A barium milkshake is a suspension of very fine BaSO₄ particles in water; the high atomic mass of barium makes it opaque to x-rays.

2. A Rubidium hydroxide and cobalt(II) chloride are strong electrolytes, so when aqueous solutions of these compounds are mixed, the resulting solution initially contains Rb⁺, OH⁻, Co²⁺, and Cl⁻ ions. The possible products of an exchange reaction are rubidium chloride and cobalt(II) hydroxide):

   

    

   B According to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", RbCl is soluble (rules 1 and 4), but Co(OH)₂ is not soluble (rule 5). Hence Co(OH)₂ will precipitate according to the following net ionic equation:

   Co²⁺(aq) + 2OH⁻(aq) → Co(OH)₂(s)

3. A When aqueous solutions of strontium bromide and aluminum nitrate are mixed, we initially obtain a solution that contains Sr²⁺, Br⁻, Al³⁺, and NO₃⁻ ions. The two possible products from an exchange reaction are aluminum bromide and strontium nitrate:

   

    

   B According to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", both AlBr₃ (rule 4) and Sr(NO₃)₂ (rule 2) are soluble. Thus no net reaction will occur.

4. A According to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", lead acetate is soluble (rule 3). Thus solid lead acetate dissolves in water to give Pb²⁺ and CH₃CO₂⁻ ions. Because the solution also contains NH₄⁺ and I⁻ ions, the possible products of an exchange reaction are ammonium acetate and lead(II) iodide:

   

    

   B According to Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", ammonium acetate is soluble (rules 1 and 3), but PbI₂ is insoluble (rule 4). Thus Pb(C₂H₃O₂)₂ will dissolve, and PbI₂ will precipitate. The net ionic equation is as follows:

   Pb²⁺ (aq) + 2I⁻(aq) → PbI₂(s)

Exercise

Using the information in Table 4.2 "Guidelines for Predicting the Solubility of Ionic Compounds in Water", predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs.

1. An aqueous solution of strontium hydroxide is added to an aqueous solution of iron(II) chloride.
2. Solid potassium phosphate is added to an aqueous solution of mercury(II) perchlorate.
3. Solid sodium fluoride is added to an aqueous solution of ammonium formate.
4. Aqueous solutions of calcium bromide and cesium carbonate are mixed.

Answer:

1. Fe²⁺(aq) + 2OH⁻(aq) → Fe(OH)₂(s)
2. 2PO₄³⁻(aq) + 3Hg²⁺(aq) → Hg₃(PO₄)₂(s)
3. NaF(s) dissolves; no net reaction
4. Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s)

Example 12

A silver recovery unit can process 1500 L of photographic silver waste solution per day. Adding excess solid sodium chloride to a 500 mL sample of the waste (after removing the thiosulfate as described previously) gives a white precipitate that, after filtration and drying, consists of 3.73 g of AgCl. What mass of NaCl must be added to the 1500 L of silver waste to ensure that all the Ag⁺ ions precipitate?

Given: volume of solution of one reactant and mass of product from a sample of reactant solution

Asked for: mass of second reactant needed for complete reaction

Strategy:

A Write the net ionic equation for the reaction. Calculate the number of moles of AgCl obtained from the 500 mL sample and then determine the concentration of Ag⁺ in the sample by dividing the number of moles of AgCl formed by the volume of solution.

B Determine the total number of moles of Ag⁺ in the 1500 L solution by multiplying the Ag⁺ concentration by the total volume.

C Use mole ratios to calculate the number of moles of chloride needed to react with Ag⁺. Obtain the mass of NaCl by multiplying the number of moles of NaCl needed by its molar mass.

Solution:

We can use the data provided to determine the concentration of Ag⁺ ions in the waste, from which the number of moles of Ag⁺ in the entire waste solution can be calculated. From the net ionic equation, we can determine how many moles of Cl⁻ are needed, which in turn will give us the mass of NaCl necessary.

A The first step is to write the net ionic equation for the reaction:

Cl⁻(aq) + Ag⁺(aq) → AgCl(s)

We know that 500 mL of solution produced 3.73 g of AgCl. We can convert this value to the number of moles of AgCl as follows:

$$\text{moles AgCl }=\frac{\text{grams AgCl}}{\text{molar mass AgCl}}=\text{ 3}\text{.73 }\cancel{\text{g AgCl}}\left(\frac{\text{1 mol AgCl}}{\text{143}\text{.32 }\cancel{\text{g AgCl}}}\right)=\text{ 0}\text{.0260 mol AgCl}$$

Therefore, the 500 mL sample of the solution contained 0.0260 mol of Ag⁺. The Ag⁺ concentration is determined as follows:

$${\text{[Ag}}^{\text{ + }}\text{] }=\frac{{\text{moles Ag}}^{+}}{\text{liters soln}}=\frac{\text{0}\text{.0260 mol AgCl}}{\text{0}\text{.500 L}}=\text{ 0}\text{.0520 M}$$

B The total number of moles of Ag⁺ present in 1500 L of solution is as follows:

$${\text{moles Ag}}^{+}=\text{ 1500 }\cancel{\text{L}}\left(\frac{\text{0}\text{.520 mol}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 78}{\text{.1 mol Ag}}^{+}$$

C According to the net ionic equation, one Cl⁻ ion is required for each Ag⁺ ion. Thus 78.1 mol of NaCl are needed to precipitate the silver. The corresponding mass of NaCl is

$$\text{mass NaCl }=\text{ 78}\text{.1 }\cancel{\text{mol NaCl}}\left(\frac{\text{58}\text{.44 g NaCl}}{\text{1 }\cancel{\text{mol NaCl}}}\right)=\text{ 4560 g NaCl }=\text{ 4}\text{.56 kg NaCl}$$

Note that 78.1 mol of AgCl correspond to 8.43 kg of metallic silver, which is worth about $7983 at 2011 prices ($32.84 per troy ounce). Silver recovery may be economically attractive as well as ecologically sound, although the procedure outlined is becoming nearly obsolete for all but artistic purposes with the growth of digital photography.

Exercise

Because of its toxicity, arsenic is the active ingredient in many pesticides. The arsenic content of a pesticide can be measured by oxidizing arsenic compounds to the arsenate ion (AsO₄³⁻), which forms an insoluble silver salt (Ag₃AsO₄). Suppose you are asked to assess the purity of technical grade sodium arsenite (NaAsO₂), the active ingredient in a pesticide used against termites. You dissolve a 10.00 g sample in water, oxidize it to arsenate, and dilute it with water to a final volume of 500 mL. You then add excess AgNO₃ solution to a 50.0 mL sample of the arsenate solution. The resulting precipitate of Ag₃AsO₄ has a mass of 3.24 g after drying. What is the percentage by mass of NaAsO₂ in the original sample?

Answer: 91.0%

Example 14

Calcium propionate is used to inhibit the growth of molds in foods, tobacco, and some medicines. Write a balanced chemical equation for the reaction of aqueous propionic acid (CH₃CH₂CO₂H) with aqueous calcium hydroxide [Ca(OH)₂] to give calcium propionate. Do you expect this reaction to go to completion, making it a feasible method for the preparation of calcium propionate?

Given: reactants and product

Asked for: balanced chemical equation and whether the reaction will go to completion

Strategy:

Write the balanced chemical equation for the reaction of propionic acid with calcium hydroxide. Based on their acid and base strengths, predict whether the reaction will go to completion.

Solution:

Propionic acid is an organic compound that is a weak acid, and calcium hydroxide is an inorganic compound that is a strong base. The balanced chemical equation is as follows:

2CH₃CH₂CO₂H(aq) + Ca(OH)₂(aq) → (CH₃CH₂CO₂)₂Ca(aq) + 2H₂O(l)

The reaction of a weak acid and a strong base will go to completion, so it is reasonable to prepare calcium propionate by mixing solutions of propionic acid and calcium hydroxide in a 2:1 mole ratio.

Exercise

Write a balanced chemical equation for the reaction of solid sodium acetate with dilute sulfuric acid to give sodium sulfate.

Answer: 2CH₃CO₂Na(s) + H₂SO₄(aq) → Na₂SO₄(aq) + 2CH₃CO₂H(aq)

Example 15

Assume that the stomach of someone suffering from acid indigestion contains 75 mL of 0.20 M HCl. How many Tums tablets are required to neutralize 90% of the stomach acid, if each tablet contains 500 mg of CaCO₃? (Neutralizing all of the stomach acid is not desirable because that would completely shut down digestion.)

Given: volume and molarity of acid and mass of base in an antacid tablet

Asked for: number of tablets required for 90% neutralization

Strategy:

A Write the balanced chemical equation for the reaction and then decide whether the reaction will go to completion.

B Calculate the number of moles of acid present. Multiply the number of moles by the percentage to obtain the quantity of acid that must be neutralized. Using mole ratios, calculate the number of moles of base required to neutralize the acid.

C Calculate the number of moles of base contained in one tablet by dividing the mass of base by the corresponding molar mass. Calculate the number of tablets required by dividing the moles of base by the moles contained in one tablet.

Solution:

A We first write the balanced chemical equation for the reaction:

2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + H₂CO₃(aq)

Each carbonate ion can react with 2 mol of H⁺ to produce H₂CO₃, which rapidly decomposes to H₂O and CO₂. Because HCl is a strong acid and CO₃²⁻ is a weak base, the reaction will go to completion.

B Next we need to determine the number of moles of HCl present:

$$\text{75 }\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}\text{.20 mol HCl}}{\cancel{\text{L}}}\right)=0.\text{015 mol HCl}$$

Because we want to neutralize only 90% of the acid present, we multiply the number of moles of HCl by 0.90:

(0.015 mol HCl)(0.90) = 0.014 mol HCl

We know from the stoichiometry of the reaction that each mole of CaCO₃ reacts with 2 mol of HCl, so we need

$${\text{moles CaCO}}_{\text{3}}=\text{ 0}\text{.014 }\cancel{\text{mol HCl}}\left(\frac{{\text{1 mol CaCO}}_{\text{3}}}{\text{2 }\cancel{\text{mol HCl}}}\right)=\text{ 0}{\text{.0070 mol CaCO}}_{\text{3}}$$

C Each Tums tablet contains

$$\left(\frac{\text{500 }\cancel{{\text{mg CaCO}}_{\text{3}}}}{\text{1 Tums tablet}}\right)\left(\frac{\text{1 }\cancel{\text{g}}}{\text{1000 }\cancel{{\text{mg CaCO}}_{\text{3}}}}\right)\left(\frac{{\text{1 mol CaCO}}_{\text{3}}}{\text{100}\text{.1 }\cancel{\text{g}}}\right)=\text{ 0}{\text{.00500 mol CaCO}}_{\text{3}}$$

Thus we need $\frac{\text{0}\text{.0070 }\cancel{{\text{mol CaCO}}_{\text{3}}}}{\text{0}\text{.00500 }\cancel{{\text{mol CaCO}}_{\text{3}}}}=\text{ 1}\text{.4}$ Tums tablets.

Exercise

Assume that as a result of overeating, a person’s stomach contains 300 mL of 0.25 M HCl. How many Rolaids tablets must be consumed to neutralize 95% of the acid, if each tablet contains 400 mg of NaAl(OH)₂CO₃? The neutralization reaction can be written as follows:

NaAl(OH)₂CO₃(s) + 4HCl(aq) → AlCl₃(aq) + NaCl(aq) + CO₂(g) + 3H₂O(l)

Answer: 6.4 tablets

Example 16

1. What is the pH of a 2.1 × 10⁻² M aqueous solution of HClO₄?
2. The pH of a vinegar sample is 3.80. What is its hydrogen ion concentration?

Given: molarity of acid or pH

Asked for: pH or [H⁺]

Strategy:

Using the balanced chemical equation for the acid dissociation reaction and Equation 4.41 or 4.40, determine [H⁺] and convert it to pH or vice versa.

Solution:

1. HClO₄ (perchloric acid) is a strong acid, so it dissociates completely into H⁺ ions and ClO₄⁻ ions:

   HClO₄(l) → H⁺(aq) + ClO₄⁻(aq)

   The H⁺ ion concentration is therefore the same as the perchloric acid concentration. The pH of the perchloric acid solution is thus

   pH = −log[H⁺] = −log(2.1 × 10⁻²) = 1.68

   The result makes sense: the H⁺ ion concentration is between 10⁻¹ M and 10⁻² M, so the pH must be between 1 and 2.

   Note: The assumption that [H⁺] is the same as the concentration of the acid is valid for only strong acids. Because weak acids do not dissociate completely in aqueous solution, a more complex procedure is needed to calculate the pH of their solutions, which we will describe in Chapter 16 "Aqueous Acid–Base Equilibriums".

2. We are given the pH and asked to calculate the hydrogen ion concentration. From Equation 4.40,

   10^−pH = [H⁺]

Thus [H⁺] = 10^−3.80 = 1.6 × 10⁻⁴ M.

Exercise

1. What is the pH of a 3.0 × 10⁻⁵ M aqueous solution of HNO₃?
2. What is the hydrogen ion concentration of turnip juice, which has a pH of 5.41?

Answer:

1. pH = 4.52
2. [H⁺] = 3.9 × 10⁻⁶ M

Example 17

Arsenic acid (H₃AsO₄) is a highly poisonous substance that was once used as a pesticide. The reaction of elemental zinc with arsenic acid in acidic solution yields arsine (AsH₃, a highly toxic and unstable gas) and Zn²⁺(aq). Balance the equation for this reaction using oxidation states:

H₃AsO₄(aq) + Zn(s) → AsH₃(g) + Zn²⁺(aq)

Given: reactants and products in acidic solution

Asked for: balanced chemical equation using oxidation states

Strategy:

Follow the procedure given in Table 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method" for balancing a redox equation using oxidation states. When you are done, be certain to check that the equation is balanced.

Solution:

1. Write a chemical equation showing the reactants and the products. Because we are given this information, we can skip this step.

2. Assign oxidation states using the procedure described in Section 3.5 "Classifying Chemical Reactions" and determine which atoms change oxidation state. The oxidation state of arsenic in arsenic acid is +6, and the oxidation state of arsenic in arsine is −3. Conversely, the oxidation state of zinc in elemental zinc is 0, and the oxidation state of zinc in Zn²⁺(aq) is +2:

   $${\text{H}}_{\text{3}}\stackrel{\text{+5}}{\text{As}}{\text{O}}_{\text{4}}\text{(aq)}+\stackrel{\text{0}}{\text{Zn}}\text{(s)}\to \underset{\text{–3}}{\text{As}}{\text{H}}_{\text{3}}\text{(g)}+\stackrel{\text{+2}}{{\text{Zn}}^{\text{2+}}}\text{(aq)}$$

3. Write separate equations for oxidation and reduction. The arsenic atom in H₃AsO₄ is reduced from the +5 to the −3 oxidation state, which requires the addition of eight electrons:

   $${\text{Reduction: As}}^{\text{5+}}+{\text{8e}}^{-}\to \underset{\text{–3}}{{\text{As}}^{\text{3–}}}$$

   Each zinc atom in elemental zinc is oxidized from 0 to +2, which requires the loss of two electrons per zinc atom:

   $$\text{Oxidation: Zn}\to {\text{Zn}}^{\text{2+}}+{\text{2e}}^{-}$$

4. Multiply the oxidation and reduction equations by appropriate coefficients so that both contain the same number of electrons. The reduction equation has eight electrons, and the oxidation equation has two electrons, so we need to multiply the oxidation equation by 4 to obtain

   Reduction (× 1): As⁵⁺ + 8e^– → As^3– Oxidation (× 4): 4Zn⁰ → 4Zn²⁺ + 8e^–

5. Write the oxidation and reduction equations showing the actual chemical forms of the reactants and the products, adjusting coefficients as necessary to give the numbers of atoms shown in step 4. Inserting the actual chemical forms of arsenic and zinc and adjusting the coefficients gives

   Reduction: H₃AsO₄(aq) + 8e^– → AsH₃(g) Oxidation: 4Zn(s) → 4Zn²⁺(aq) + 8e^–

6. Add the two equations and cancel the electrons. The sum of the two equations in step 5 is

   $${\text{H}}_{\text{3}}{\text{AsO}}_{\text{4}}\text{(aq) }+\text{ 4Zn(s)}+\cancel{{\text{8e}}^{–}}\to {\text{AsH}}_{\text{3}}\text{(g)}+{\text{4Zn}}^{\text{2}+}\text{(aq)}+\cancel{{\text{8e}}^{–}}$$

   which then yields

   H₃AsO₄(aq) + 4Zn(s) → AsH₃(g) + 4Zn²⁺(aq)

7. Balance the charge by adding H⁺or OH⁻ions as necessary for reactions in acidic or basic solution, respectively. Because the reaction is carried out in acidic solution, we can add H⁺ ions to whichever side of the equation requires them to balance the charge. The overall charge on the left side is zero, and the total charge on the right side is 4 × (+2) = +8. Adding eight H⁺ ions to the left side gives a charge of +8 on both sides of the equation:

   H₃AsO₄(aq) + 4Zn(s) + 8H⁺(aq) → AsH₃(g) + 4Zn²⁺(aq)

8. Balance the oxygen atoms by adding H₂O molecules to one side of the equation. There are 4 O atoms on the left side of the equation. Adding 4 H₂O molecules to the right side balances the O atoms:

   H₃AsO₄(aq) + 4Zn(s) + 8H⁺(aq) → AsH₃(g) + 4Zn²⁺(aq) + 4H₂O(l)

   Although we have not explicitly balanced H atoms, each side of the equation has 11 H atoms.

9. Check to make sure that the equation is balanced in both atoms and total charges. To guard against careless errors, it is important to check that both the total number of atoms of each element and the total charges are the same on both sides of the equation:

Atoms: 1As + 4Zn + 4O + 11H = 1As + 4Zn + 4O + 11H Total charge: 8(+1) = 4(+2) = +8

The balanced chemical equation for the reaction is therefore:

H₃AsO₄(aq) + 4Zn(s) + 8H⁺(aq) → AsH₃(g) + 4Zn²⁺(aq) + 4H₂O(l)

Exercise

Copper commonly occurs as the sulfide mineral CuS. The first step in extracting copper from CuS is to dissolve the mineral in nitric acid, which oxidizes the sulfide to sulfate and reduces nitric acid to NO. Balance the equation for this reaction using oxidation states:

CuS(s) + H⁺(aq) + NO₃⁻(aq) → Cu²⁺(aq) + NO(g) + SO₄²⁻(aq)

Answer: 3CuS(s) + 8H⁺(aq) + 8NO₃⁻(aq) → 3Cu²⁺(aq) + 8NO(g) + 3SO₄²⁻(aq) + 4H₂O(l)

Example 18

The commercial solid drain cleaner, Drano, contains a mixture of sodium hydroxide and powdered aluminum. The sodium hydroxide dissolves in standing water to form a strongly basic solution, capable of slowly dissolving organic substances, such as hair, that may be clogging the drain. The aluminum dissolves in the strongly basic solution to produce bubbles of hydrogen gas that agitate the solution to help break up the clogs. The reaction is as follows:

Al(s) + H₂O(aq) → [Al(OH)₄]⁻(aq) + H₂(g)

Balance this equation using oxidation states.

Given: reactants and products in a basic solution

Asked for: balanced chemical equation

Strategy:

Follow the procedure given in Table 4.4 "Procedure for Balancing Oxidation–Reduction Reactions by the Oxidation State Method" for balancing a redox reaction using oxidation states. When you are done, be certain to check that the equation is balanced.

Solution:

We will apply the same procedure used in Example 17 but in a more abbreviated form.

1. The equation for the reaction is given, so we can skip this step.

2. The oxidation state of aluminum changes from 0 in metallic Al to +3 in [Al(OH)₄]⁻. The oxidation state of hydrogen changes from +1 in H₂O to 0 in H₂. Aluminum is oxidized, while hydrogen is reduced:

   $$\stackrel{\text{0}}{\text{Al}}\text{(s)}+{\stackrel{\text{+1}}{\text{H}}}_{\text{2}}\text{O(aq)}\to {\text{[}\stackrel{\text{+3}}{\text{Al}}{\text{(OH)}}_{\text{4}}\text{]}}^{\text{–}}\text{(aq)}+\stackrel{\text{0}}{{\text{H}}_{\text{2}}}\text{(g)}$$

3.  

   $$\begin{array}{l}{\text{Reduction: Al}}^{\text{0}}\to {\text{Al}}^{\text{3+}}+{\text{3e}}^{-}\\ {\text{Oxidation: H}}^{\text{ + }}+{\text{e}}^{-}\to {\text{H}}^{\text{0}}{\text{ (in H}}_{\text{2}}\text{)}\end{array}$$

4. Multiply the reduction equation by 3 to obtain an equation with the same number of electrons as the oxidation equation:

   Reduction: 3H⁺ + 3e^– → 3H⁰ (in H₂) Oxidation: Al⁰ → Al³⁺ + 3e^–

5. Insert the actual chemical forms of the reactants and products, adjusting the coefficients as necessary to obtain the correct numbers of atoms as in step 4. Because a molecule of H₂O contains two protons, in this case, 3H⁺ corresponds to 3/2H₂O. Similarly, each molecule of hydrogen gas contains two H atoms, so 3H corresponds to 3/2H₂.

   $$\begin{array}{l}\text{Reduction: }\frac{3}{2}{\text{H}}_{\text{2}}\text{O}+{\text{3e}}^{-}\to \frac{3}{2}{\text{H}}_{\text{2}}\\ \text{Oxidation: Al}\to {\left[\text{Al}{\left(\text{OH}\right)}_{\text{4}}\right]}^{–}+{\text{3e}}^{–}\end{array}$$

6. Adding the equations and canceling the electrons gives

   $$\begin{array}{c}\text{Al}+\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}\text{O}+\cancel{{\text{3e}}^{-}}\to {{\text{[Al(OH)}}_{\text{4}}\text{]}}^{-}+\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}+\cancel{{\text{3e}}^{-}}\\ \text{Al}+\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}\text{O}\to {{\text{[Al(OH)}}_{\text{4}}\text{]}}^{-}+\frac{\text{3}}{\text{2}}{\text{H}}_{\text{2}}\end{array}$$

   To remove the fractional coefficients, multiply both sides of the equation by 2:

   2Al + 3H₂O → 2[Al(OH)₄]^– + 3H₂

7. The right side of the equation has a total charge of −2, whereas the left side has a total charge of 0. Because the reaction is carried out in basic solution, we can balance the charge by adding two OH⁻ ions to the left side:

   2Al + 2OH^– + 3H₂O → 2[Al(OH)₄]^– + 3H₂

8. The left side of the equation contains five O atoms, and the right side contains eight O atoms. We can balance the O atoms by adding three H₂O molecules to the left side:

   2Al + 2OH^– + 6H₂O → 2[Al(OH)₄]^– + 3H₂

9. Be sure the equation is balanced:

   Atoms: 2Al + 8O + 14H = 2Al + 8O + 14H Total charge: (2)(0) + (2)(–1) + (6)(0) = (2)(–1) + (3)(0) −2 = −2

   The balanced chemical equation is therefore

   2Al(s) + 2OH^–(aq) + 6H₂O(l) → 2[Al(OH)₄]^–(aq) + 3H₂(g)

   Thus 3 mol of H₂ gas are produced for every 2 mol of Al.

Exercise

The permanganate ion reacts with nitrite ion in basic solution to produce manganese(IV) oxide and nitrate ion. Write a balanced chemical equation for the reaction.

Answer: 2MnO₄⁻(aq) + 3NO₂⁻(aq) + H₂O(l) → 2MnO₂(s) + 3NO₃⁻(aq) + 2OH⁻(aq)

Example 20

The calcium salt of oxalic acid [Ca(O₂CCO₂)] is found in the sap and leaves of some vegetables, including spinach and rhubarb, and in many ornamental plants. Because oxalic acid and its salts are toxic, when a food such as rhubarb is processed commercially, the leaves must be removed, and the oxalate content carefully monitored.

The reaction of MnO₄⁻ with oxalic acid (HO₂CCO₂H) in acidic aqueous solution produces Mn²⁺ and CO₂:

$$\underset{\text{purple}}{{\text{MnO}}_{\text{4}}{}^{-}\text{(aq)}}+{\text{HO}}_{\text{2}}{\text{CCO}}_2\text{H(aq)}+{\text{H}}^{\text{ + }}\text{(aq)}\to \underset{\text{colorless}}{{\text{Mn}}_{}^{2+}\text{(aq)}}+{\text{CO}}_{\text{2}}\text{(g)}+{\text{H}}_{\text{2}}\text{O(l)}$$

Because this reaction is rapid and goes to completion, potassium permanganate (KMnO₄) is widely used as a reactant for determining the concentration of oxalic acid.

Suppose you stirred a 10.0 g sample of canned rhubarb with enough dilute H₂SO₄(aq) to obtain 127 mL of colorless solution. Because the added permanganate is rapidly consumed, adding small volumes of a 0.0247 M KMnO₄ solution, which has a deep purple color, to the rhubarb extract does not initially change the color of the extract. When 15.4 mL of the permanganate solution have been added, however, the solution becomes a faint purple due to the presence of a slight excess of permanganate (Figure 4.23 "The Titration of Oxalic Acid with Permanganate"). If we assume that oxalic acid is the only species in solution that reacts with permanganate, what percentage of the mass of the original sample was calcium oxalate?

Given: equation, mass of sample, volume of solution, and molarity and volume of titrant

Asked for: mass percentage of unknown in sample

Strategy:

A Balance the chemical equation for the reaction using oxidation states.

B Calculate the number of moles of permanganate consumed by multiplying the volume of the titrant by its molarity. Then calculate the number of moles of oxalate in the solution by multiplying by the ratio of the coefficients in the balanced chemical equation. Because calcium oxalate contains a 1:1 ratio of Ca²⁺:⁻O₂CCO₂⁻, the number of moles of oxalate in the solution is the same as the number of moles of calcium oxalate in the original sample.

C Find the mass of calcium oxalate by multiplying the number of moles of calcium oxalate in the sample by its molar mass. Divide the mass of calcium oxalate by the mass of the sample and convert to a percentage to calculate the percentage by mass of calcium oxalate in the original sample.

Solution:

A As in all other problems of this type, the first requirement is a balanced chemical equation for the reaction. Using oxidation states gives

2MnO₄⁻(aq) + 5HO₂CCO₂H(aq) + 6H⁺(aq) → 2Mn²⁺(aq) + 10CO₂(g) + 8H₂O(l)

Thus each mole of MnO₄⁻ added consumes 2.5 mol of oxalic acid.

B Because we know the concentration of permanganate (0.0247 M) and the volume of permanganate solution that was needed to consume all the oxalic acid (15.4 mL), we can calculate the number of moles of MnO₄⁻ consumed:

$$\text{15}\text{.4 }\cancel{\text{mL}}\left(\frac{\text{1 }\cancel{\text{L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}{\text{.0247 mol MnO}}_4{}^{-}}{\text{1 }\cancel{\text{L}}}\right)=\text{ 3}\text{.80}\times {\text{10}}^{-\text{4}}{\text{ mol MnO}}_{\text{4}}{}^{-}$$

The number of moles of oxalic acid, and thus oxalate, present can be calculated from the mole ratio of the reactants in the balanced chemical equation:

$$\begin{array}{ll}{\text{moles HO}}_{\text{2}}{\text{CCO}}_{\text{2}}\text{H }\hfill & =\text{ 3}\text{.80}\times {\text{10}}^{-\text{4}}\cancel{{\text{mol MnO}}_{\text{4}}{}^{-}}\left(\frac{{\text{5 mol HO}}_{\text{2}}{\text{CCO}}_{\text{2}}\text{H}}{\text{2 }\cancel{{\text{mol MnO}}_{\text{4}}{}^{-}}}\right)\hfill \\ \hfill & =\text{ 9}\text{.50}\times {\text{10}}^{-\text{4}}{\text{ mol HO}}_{\text{2}}{\text{CCO}}_{\text{2}}\text{H}\hfill \end{array}$$

C The problem asks for the percentage of calcium oxalate by mass in the original 10.0 g sample of rhubarb, so we need to know the mass of calcium oxalate that produced 9.50 × 10⁻⁴ mol of oxalic acid. Because calcium oxalate is Ca(O₂CCO₂), 1 mol of calcium oxalate gave 1 mol of oxalic acid in the initial acid extraction:

Ca(O₂CCO₂)(s) + 2H⁺(aq) → Ca²⁺(aq) + HO₂CCO₂H(aq)

The mass of calcium oxalate originally present was thus

$$\begin{array}{ll}{\text{mass of CaC}}_{\text{2}}{\text{O}}_{\text{4}}\hfill & =\text{ 9}\text{.50}\times {\text{10}}^{-\text{4}}\cancel{{\text{mol HO}}_{\text{2}}{\text{CCO}}_{\text{2}}\text{H}}\left(\frac{\text{1 }\cancel{{\text{mol CaC}}_{\text{2}}{\text{O}}_{\text{4}}}}{\text{1 }\cancel{{\text{mol HO}}_{\text{2}}{\text{CCO}}_{\text{2}}\text{H}}}\right)\left(\frac{\text{128}{\text{.10 g CaC}}_{\text{2}}{\text{O}}_{\text{4}}}{\text{1 }\cancel{{\text{mol CaC}}_{\text{2}}{\text{O}}_{\text{4}}}}\right)\hfill \\ \hfill & =\text{ 0}{\text{.122 g CaC}}_{\text{2}}{\text{O}}_{\text{4}}\hfill \end{array}$$

The original sample contained 0.122 g of calcium oxalate per 10.0 g of rhubarb. The percentage of calcium oxalate by mass was thus

$${\text{\% CaC}}_{\text{2}}{\text{O}}_{\text{4}}=\frac{\text{0}\text{.122 g}}{\text{10}\text{.0 g}}\times \text{100 }=\text{ 1}\text{.22\%}$$

Because the problem asked for the percentage by mass of calcium oxalate in the original sample rather than for the concentration of oxalic acid in the extract, we do not need to know the volume of the oxalic acid solution for this calculation.

Exercise

Glutathione is a low-molecular-weight compound found in living cells that is produced naturally by the liver. Health-care providers give glutathione intravenously to prevent side effects of chemotherapy and to prevent kidney problems after heart bypass surgery. Its structure is as follows:

Glutathione is found in two forms: one abbreviated as GSH (indicating the presence of an –SH group) and the other as GSSG (the disulfide form, in which an S–S bond links two glutathione units). The GSH form is easily oxidized to GSSG by elemental iodine:

2GSH(aq) + I₂(aq) → GSSG(aq) + 2HI(aq)

A small amount of soluble starch is added as an indicator. Because starch reacts with excess I₂ to give an intense blue color, the appearance of a blue color indicates that the equivalence point of the reaction has been reached.

Adding small volumes of a 0.0031 M aqueous solution of I₂ to 194 mL of a solution that contains glutathione and a trace of soluble starch initially causes no change. After 16.3 mL of iodine solution have been added, however, a permanent pale blue color appears because of the formation of the starch-iodine complex. What is the concentration of glutathione in the original solution?

Answer: 5.2 × 10⁻⁴ M

Example 21

The structure of vitamin C (ascorbic acid, a monoprotic acid) is as follows:

An absence of vitamin C in the diet leads to the disease known as scurvy, a breakdown of connective tissue throughout the body and of dentin in the teeth. Because fresh fruits and vegetables rich in vitamin C are readily available in developed countries today, scurvy is not a major problem. In the days of slow voyages in wooden ships, however, scurvy was common. Ferdinand Magellan, the first person to sail around the world, lost more than 90% of his crew, many to scurvy. Although a diet rich in fruits and vegetables contains more than enough vitamin C to prevent scurvy, many people take supplemental doses of vitamin C, hoping that the extra amounts will help prevent colds and other illness.

Suppose a tablet advertised as containing 500 mg of vitamin C is dissolved in 100.0 mL of distilled water that contains a small amount of the acid–base indicator bromothymol blue, an indicator that is yellow in acid solution and blue in basic solution, to give a yellow solution. The addition of 53.5 mL of a 0.0520 M solution of NaOH results in a change to green at the endpoint, due to a mixture of the blue and yellow forms of the indicator (Figure 4.24 "The Titration of Ascorbic Acid with a Solution of NaOH"). What is the actual mass of vitamin C in the tablet? (The molar mass of ascorbic acid is 176.13 g/mol.)

Given: reactant, volume of sample solution, and volume and molarity of titrant

Asked for: mass of unknown

Strategy:

A Write the balanced chemical equation for the reaction and calculate the number of moles of base needed to neutralize the ascorbic acid.

B Using mole ratios, determine the amount of ascorbic acid consumed. Calculate the mass of vitamin C by multiplying the number of moles of ascorbic acid by its molar mass.

Solution:

A Because ascorbic acid acts as a monoprotic acid, we can write the balanced chemical equation for the reaction as

HAsc(aq) + OH⁻(aq) → Asc⁻(aq) + H₂O(l)

where HAsc is ascorbic acid and Asc⁻ is ascorbate. The number of moles of OH⁻ ions needed to neutralize the ascorbic acid is

$${\text{moles OH}}^{-}=\text{ 53}\text{.5 }\cancel{\text{mL}}\left(\frac{\cancel{\text{1 L}}}{\text{1000 }\cancel{\text{mL}}}\right)\left(\frac{\text{0}{\text{.0520 mol OH}}^{-}}{\cancel{\text{1 L}}}\right)=\text{ 2}\text{.78}\times {\text{10}}^{-\text{3}}{\text{ mol OH}}^{{}^{-}}$$

B The mole ratio of the base added to the acid consumed is 1:1, so the number of moles of OH⁻ added equals the number of moles of ascorbic acid present in the tablet:

$$\begin{array}{cc}\text{mass ascorbic acid}& =\text{ 2}\text{.78}\times {\text{10}}^{-\text{3}}\cancel{\text{mol HAsc}}\end{array}\left(\frac{\text{176}\text{.13 g HAsc}}{\text{1 }\cancel{\text{mol HAsc}}}\right)=\text{ 0}\text{.490 g HAsc}$$

Because 0.490 g equals 490 mg, the tablet contains about 2% less vitamin C than advertised.

Exercise

Vinegar is essentially a dilute solution of acetic acid in water. Vinegar is usually produced in a concentrated form and then diluted with water to give a final concentration of 4%–7% acetic acid; that is, a 4% m/v solution contains 4.00 g of acetic acid per 100 mL of solution. If a drop of bromothymol blue indicator is added to 50.0 mL of concentrated vinegar stock and 31.0 mL of 2.51 M NaOH are needed to turn the solution from yellow to green, what is the percentage of acetic acid in the vinegar stock? (Assume that the density of the vinegar solution is 1.00 g/mL.)

Answer: 9.35%