Chapter X | Chapter Title



CHAPTER 4 | Solution Chemistry and the Hydrosphere

4.10. Collect and Organize

The molarity of a solution is the moles of solute in one liter of solution. For each of these calculations, we are given the moles of solute and the volume in either liters or milliliters.

Analyze

To find molarity we need only divide the moles of solute by the volume of solution in liters. If the volume is expressed in milliliters, we can move the decimal over three places to obtain volume in liters (e.g., 100 mL = 0.100 L).

Solve

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Think about It

You cannot decide how concentrated a solution is unless you know both the moles of solute and the volume of the solution. Notice that in part d with only 0.045 mol of sucrose the solution is actually fairly concentrated because the volume of the solution is small.

4.13. Collect and Organize

We are asked to calculate the grams of a solute needed to prepare a solution of a specific concentration.

Analyze

First we can find the moles of solute needed for the solution by multiplying the molarity by the volume (in liters!). Once we have moles, we can then use the molar mass of the solute to calculate the mass of solute needed.

Solve

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Think about It

This is a practical calculation for preparing solutions when we know that we want to have a solution with a particular concentration.

4.16. Collect and Organize

To express the concentration of heavy metal toxins in milligrams per liter (mg/L) we have to convert to millimoles to milligrams.

Analyze

From the molar mass of an element it will be true that g/mol = mg/mmol because

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We can use this and molar mass values to convert mmol to mg for each of the toxins.

Solve

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Think about It

We often see contaminants expressed as mass per volume, so knowing how to convert from moles per volume helps in comparisons between different sets of data.

4.22. Collect and Organize

The concentration of Cu2+ in mol/L is to be calculated knowing the mass percent of CuSO4, the sample size, and the volume of the solution.

Analyze

We can find the mass of CuSO4 in the sample by multiplying the percent (0.07%) by the sample size. Next, we can find the moles of CuSO4 in the sample using the molar mass of CuSO4 (159.61 g/mol). Because 1 mole of Cu2+ is in 1 mole of CuSO4, this value is also the mole of Cu2+. The concentration of Cu2+, then, is the moles Cu2+ divided by the volume of the solution (2.0 L).

4.30. Collect and Organize

From a concentrated solution of SrCl2, we are asked to determine how much of that solution is required to prepare a less concentrated solution. We use the dilution equation to solve.

Analyze

Here, we have to be careful to use consistent units in the equation

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Minitial is given in M (mol/L) while Mfinal is given in mM (1 ( 10–3 mol/L). If we convert Mfinal to mol/L

(5.0 mM = 5.0 ( 10–3 mol/L) we have consistent units and can apply the dilution equation to solve for Vinitial.

Solve

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Think about It

Because the original concentration is so high compared to the final desired concentration, we need only a small volume of the concentrated solution for this dilution.

4.40. Collect and Organize

The molar concentration of the ions in solution must take into account not only the original salt concentration, but also the number of ions into which the salt dissociates. All solutions are 0.025 M in each of the salts.

Analyze

A salt that dissociates into one cation and one anion has a concentration of ions twice that of the salt solution concentration. Likewise, a salt that dissociates into one cation and two anions has a concentration of ions three times that of the salt solution concentration (because three ions are formed upon dissolution).

Solve

a) KCl is 0.025 M K+; 0.025 M Cl–

b) CuSO4 is 0.025 M Cu2+; 0.025 M SO42–

c) CaCl2 is 0.025 M Ca2+; 0.050 M Cl–

Think about It

The charge on the ion does not necessarily matter. As we see here, 0.025 M KCl and 0.025 M CuSO4, both of which dissociate in solution into two ions, have the same concentration of total dissolved ions.

4.44. Collect and Organize

We are to differentiate between a strong acid and a weak acid.

Analyze

Acids can be either weak or strong depending on the degree to which they dissociate in aqueous solution.

Solve

A strong acid is one that produces 100% H+ + A– in solution. A weak acid only partially dissociates and gives a mixture of H+, A–, and HA in solution.

Think about It

Likewise, a strong base completely forms OH– in aqueous solution.

4.48. Collect and Organize / Analyze

Bases can be either weak or strong depending on the degree to which they dissociate in aqueous solution.

Solve

A strong base is one that produces 100% OH– in solution. A weak base only partially dissociates and gives a mixture of B+, OH–, and BOH in solution.

Think about It

Likewise, a strong acid completely forms H+ in aqueous solution.

4.51. Collect and Organize

In each part of this problem, we identify the acid (proton donor) and base (proton acceptor). To write the net ionic equations we have to identify the spectator ions and remove them from the ionic equation.

Analyze

For each reaction, write the species present in aqueous solution (showing dissociation). From these species you can identify the acid and base. Then eliminate any spectator ions in the ionic equation to give the net ionic equation.

Solve

(a) Ionic and net ionic equation:

2 H+(aq) + SO42–(aq) + Ca2+(aq) + 2 OH–(aq) ( CaSO4(s) + 2 H2O[pic]

The acid is H2SO4; the base is Ca(OH)2.

(b) Ionic and net ionic equation:

PbCO3(s) + 2 H+(aq) + SO42–(aq) ( PbSO4(s) + CO2(g) + H2O[pic]

PbCO3 is the base; sulfuric acid is the acid.

(c) Ionic equation:

Ca2+(aq) + 2 OH–(aq) + 2 CH3COOH(aq) ( Ca2+(aq) + 2 CH3COO–(aq) + 2 H2O[pic]

Calcium is a spectator ion. Ca(OH)2 is the base; CH3COOH is the acid.

Net ionic equation:

OH–(aq) + CH3COOH(aq) ( CH3COO–(aq) + H2O[pic]

Think about It

Reactions a and c are neutralization reactions whereas reaction b is an acid–base reaction that also forms a precipitate (PbSO4) and a gas (CO2).

4.52. Collect and Organize

For each neutralization reaction shown, in which an acid reacts with a base, we are asked to write the balanced chemical reactions, name the products, and write the net ionic equation.

Analyze

Neutralization reactions always give water and salt as the products. We can name the salts using the rules for naming (Chapter 2). The net ionic equation does not include any spectator ions.

Solve

(a) HBr(aq) + KOH(aq) ( H2O[pic]+ KBr(aq)

Potassium bromide and water are produced.

Ionic equation:

H+(aq) + Br–(aq) + K+(aq) + OH–(aq) ( H2O[pic]+ K+(aq) + Br–(aq)

Net ionic equation:

H+(aq) + OH–(aq) ( H2O[pic]

(b) 2 H3PO4(aq) + 3 Ba(OH)2(aq) ( 6 H2O[pic]+ Ba3(PO4)2(s)

Barium phosphate and water are produced.

Ionic and net ionic equation:

2 H3PO4(aq) + 3 Ba2+(aq) + 6 OH–(aq) ( 6 H2O[pic]+ Ba3(PO4)2(s)

(c) Al(OH)3(s) + 3 HCl(aq) ( AlCl3(aq) + 3 H2O[pic]

Aluminum chloride and water are produced.

Ionic equation:

Al(OH)3(s) + 3 H+(aq) + 3 Cl–(aq) ( Al3+(aq) + 3 Cl–(aq) + 3 H2O[pic]

Net ionic equation:

Al(OH)3(s) + 3 H+(aq) ( Al3+(aq) + 3 H2O[pic]

(d) 2 CH3COOH(aq) + Sr(OH)2(aq) ( 2 H2O[pic] + Sr(CH3COO)2(aq)

Water and strontium acetate are produced.

2 CH3COOH(aq) + Sr2+(aq) + 2 OH–(aq) ( 2 H2O[pic] + Sr2+(aq) + 2 CH3COO–(aq)

2 CH3COOH(aq) + OH–(aq) ( H2O[pic] + CH3COO–(aq)

Think about It

For any neutralization reaction between soluble acids and bases in which the salt produced in the reaction is soluble, the net ionic equation always has the form shown in part a.

4.57. Collect and Organize

All of the titrations involve a neutralization reaction. The moles of base (OH–) required must be equal to the moles of acid (H+) in the sample.

Analyze

First, we need to calculate the number of moles of acid from the volume and concentration of acid in the samples. Because the stoichiometry of the neutralization reaction is 1 mol OH– : 1 mol H+, the moles of base required is equal to the moles of H+ in the sample. We can find the volume of base needed by dividing moles of OH– required by the concentration of base used.

Solve

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Think about It

Because sulfuric acid has two H+ ions (it is a diprotic acid), we need twice as many moles of OH– to neutralize it compared to the same concentration of a monoprotic acid such as HNO3.

4.59. Collect and Organize

Using the solubility of Ca(OH)2 we first have to calculate the moles of Ca(OH)2 in the solution; then we can find the volume of the HCl(aq) solution to neutralize the Ca(OH)2 solution.

Analyze

To find the moles of Ca(OH)2 in the saturated solution, we first have to multiply the volume of the solution by the solubility of Ca(OH)2. This gives the grams of Ca(OH)2 in the solution, which can then be converted into moles by dividing the grams of Ca(OH)2 by the molar mass of Ca(OH)2. Because there are two moles of OH– in Ca(OH)2, the moles of OH– to neutralize must be twice the moles of Ca(OH)2. We can then use the 1:1 molar ratio of OH– to H+ in the neutralization reaction and the concentration of the HCl solution to find the volume of HCl required to neutralize the Ca(OH)2 solution.

Solve

Moles of Ca(OH)2 in the saturated solution:

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Volume (mL) of HCl required to neutralize:

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Think about It

Calcium hydroxide is not very soluble in water, but this neutralization requires a large volume of HCl solution because the HCl solution is fairly dilute.

4.61. Collect and Organize

This problem asks us to calculate how much acid of a concentration of 0.10 M could one dose of antacid neutralize.

Analyze

First, we need to calculate the moles of Mg(OH)2 in the dose using the molar mass of Mg(OH)2. Next, we have to use the balanced chemical equation for the neutralization:

Mg(OH)2(s) + 2 HCl(aq) ( MgCl2(aq) + 2 H2O[pic]

which shows that 2 mol of HCl are required for 1 mol of Mg(OH)2 to determine the moles of HCl that would be neutralized. The volume of HCl can then be found from the moles of HCl neutralized and the concentration of the HCl in stomach acid.

Solve

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Think about It

One 10 mL dose can neutralize nearly 300 mL of stomach acid.

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