2.1 What water-soluble and water-insoluble substances do you know? Water-soluble substances include salt (NaCl), sugar, ethanol, road salt (CaCl2), ammonia, hydrochloric acid. Water-insoluble substances include oil, gasoline, wax, sand, plastics, many rocks. Water-soluble substances are generally very polar or ionic while many water-insoluble substances are low polarity or are substances that contain ‘networks’ of covalent bonds that are hard to break (e.g., sand or plastics).

2.2-2.4 What is observed when substances dissolve? These will be discussed in lecture during Week 4.

2.5-2.6 Which compounds are soluble in water? These will be discussed in lecture during Week 4.

2.8 Energy diagram for hexane dissolving in water (hexane (l) ® hexane (aq)). The net energy change for dissolving hexane in water is zero. So, although it takes E to separate the pure hexane molecules from one another, the same amount of E is released as the hexane molecules become ‘hydrated’ with water molecules and form a solution.

2.9 Predict relative solubilities

(a) 1-butanol can hydrogen bond to other 1-butanol molecules. There is a H directly bound to an O in this molecule, so the O-H bond is polar enough to make it possible for the H to H-bond through the lone pair on another molecule’s O.

(b) 1-butanol can H-bond with water in a way similar to that described in (a). So can diethyl ether – the H-bonds are all between H atoms of water and the lone pairs on the diethyl ether’s O. None of the H atoms on diethyl ether is involved in a polar enough bond (C-H bonds are ~nonpolar) for H-bonds to form involving them.

(c) 1-butanol is most soluble in water because it can form the most H-bonds with water. Pentane is the least soluble since it is non-polar and cannot set up significant IMFs with water (no H-bonds are possible).

(d) Pentane should be most soluble in nonpolar solvents like mineral oil because the two nonpolar materials can interact with each other (through London forces). 1-butanol has the very polar alcohol group, so it should be least soluble in mineral oii.

2.10 What are the interactions of glucose with water? There is a maximum of 3 H-bonds per methanol molecule between methanol and water (see Figure 2.3). In glucose, there are three possible H-bonds with water at each of the five O-H groups (15 at the O-H groups). There is also one ‘ring’ O atom that can make two H-bonds to water through its lone pairs. So, a maximum of 17 H-bonds can be made between one molecule of glucose and water molecules. All these H-bonds to water mean that glucose is very water-soluble.

2.11 Predict Relative Solubilties These are similar size compounds, so they’ll have similar London forces. Both have at least one O-H group which will become involved in 3 H-bonds with water molecules (similar to methanol-water). The PEG 200 has many more O atoms whose lone pairs can H-bond to the H atoms in water, so it should be more soluble.

2.12-2.13 Which solutions conduct an electric current? Distilled water and aqueous glucose solutions do not conduct current (the meter light bulb does not light when the electrodes are placed in solution). Aqueous NaCl (table salt) conducts electricity (causes the light bulb to light). See Figure 2.8 for a molecular-level picture of the solutions.

2.19 Writing a balanced chemical reaction equation.

Mg + O -> Mg²⁺ + O^2-

Note that the problem asks for the reaction between Mg and O atoms. Oxygen is present under ordinary conditions as O₂ molecules, not atoms.

2.20 Are attractions or repulsions stronger in a crystal? Attractive interactions are between ions that are very closely spaced. So, d in eqn 2.2 is very small, and this makes the (+)E of the attractive interaction very large. The distance between same charge ions is greater, so the (-)E of the repulsive interaction small. (Just the fact that the crystalline lattice exists tells you that the attractive forces must outweigh the repulsive forces.)

2.22 What temperature changes occur when solids dissolve? In a test experiment with all materials initially at ~22ºC, the following results were observed:

Room temp H₂O = 22ºC, NH₄Cl solution = 16ºC, CaCl₂solution = 28ºC, NaCl solution = 22ºC; all three solutions conduct electricity.

2.23 What energy changes occur when solids dissolve?

(a) All three materials dissolved in water and conducted electricity in solution. These are properties of ionic compounds.

(b) When NH₄Cl solution dissolved, the vial became cool which means that the solid absorbed energy from its surroundings (dissolved endothermically). The opposite is true for the CaCl₂solution, which made the vial warmer and so must have been dissolving exothermically.

(c) No surprises, but it’s interesting that some release/absorb a lot of energy while another (NaCl) dissolves without a big energy change. These solids all dissolve to produce ions in solution (conduct electricity).

2.24 How does Figure 2.15 correlate with Investigate This 2.22? The net energy change for dissolving of NaCl (s) is very small. This is consistent with my observation that the temperature of the NaCl solution was essentially the same as the initial temperature of the water.

2.27 What reactions of ions in solution can you observe? Answers are in Table 2.4 on p. 95.

2.28-2.29 How do you explain the reactions of ions in solution? All the solutions conduct electricity (contain ions) Only the CaCl₂ + Na₂SO₄ reaction mixture produced a solid. For that reaction, apparently either the Ca²⁺ and SO4^2- ions OR the Cl^- + Na⁺ combined to make a precipitate. But, since we know from experience that NaCl is very soluble in water, it must be the Ca²⁺ and SO₄^2- that combined to produce a solid. The ‘leftover’ Na⁺ and Cl^- ions are responsible for the observed conductivity. For the CaCl₂ + NaNO₃ mixture, there was no apparent reaction. The observed conductivity makes sense if we assume that all the reactant ions were simply left unreacted in solution.

2.32-2.33, 2.35 Which mixtures of ionic compounds yield precipitates? These will be discussed in lecture during Week 5.

2.36 Predicting solubilities

(a) Salts containing alkali metal cations or a halide ion tend to be soluble, so both Na₃PO₄ and CoCl₂ are likely to be soluble.

(b) The new combinations would be Na⁺ + Cl^- and Co²⁺+ PO₄^3-. A precipitate would not be expected for the first combination (see (a)). The second combination should yield a precipitate since compounds with both a multiply charged cation and anion are usually insoluble. See the rules on the bottom of p. 100.

2.37 What might cause some exceptions to the solubility rules?

(a) The potassium salts of Cl-, Br-, and I- are all soluble but those of silver ion are not. The problem says that potassium and silver ions are about the same size. Since ions of similar size and charge should have about the same attraction for water molecules, the degree of ‘hydration’ should be the same. SO, for this set of salts, it’s not the degree of hydration of the dissolved ion (the DE_hydration) but the energy required to separate the cations-anions in the solid lattice (DE_lattice) that matters.

(b) The data in Table 2.3 show that (as expected) all the potassium and silver halides have + DE_lattice values. The DE_hydration values are all -. The sum of DE_lattice and DE_hydrationis equal to DE_dissolving. The sign of DE_dissolving tells you whether the salt will dissolve enothermically or exothermically. The more endothermic the dissolving process, the less likely it is that the salt will dissolve. DE_dissolving is positive for the potassium halides but much more positive for the silver halide salts. So, the silver halide salts dissolve more endothermically and this makes them much less soluble than their potassium salt counterparts.

2.38 (and parts of 2.39, 2.40, 2.55, 2.59, 2.60, 2.62, 2.63, 2.66) Do Co²⁺(aq) and PO₄^3-(aq) reqct to form a precipitate? These will be discussed in lecture during Week 5.

2.67, 2.68 What are the properties of aqueous solutions of gases? These will be discussed in lecture during Week 6.

2.74 Which aqueous solutions of gas are acid? Which are basic?

The acidic solutions (pH <7) are CO2(aq), HCl(aq). The basic solution (pH >7) is NH3(aq). The neutral solution (pH = 7) is glucose(aq). Water also has a pH =7. The electrical conductivities of water and glucose are much lower than those of the acidic and basic solutions (see results of 2.67). The presence of hydronium/hydroxide ion from these acids/bases is probably responsible (more ions = higher conductivity).

2.75 How many moles of hydronium ion are present in water?

The pH of the solution is –log₁₀(10^-4) = 4. This is less than the pH of neutral solutions (pH = 7), so the compound dissolved in water is functioning as a Bronsted-Lowry acid. Acids react with water to bring the pH lower than 7.

2.76 Reactions of Bronsted-Lowry acids with water.

H₂Te(aq) + H₂O(l) = H₃O⁺(aq) + HTe^-(aq)

HF(aq) + H₂O(l) = H₃O⁺(aq) + F^-(aq)

2.77 Structures and reactions for Bronsted-Lowry oxyacids and oxyanions.

(a) Below is just one structure – carbonate. When the hydrogen ion is lost to form the corresponding anion, the electron pair previously involved in the bond between O-H is left behind on the oxygen of the anion.

(b) For nitric acid only,

Conventional: H₂O(l) + HNO₃(aq) = H₃0⁺(aq) NO₃^-(aq)

Structural: HOH + HONO₂ = H₃O⁺ + ONO₂^-

Lewis:

2.79 What are the reactions in an aqueous solution of carbon dioxide?

(a) When CO₂ dissolves in water, it produces (HO)₂CO (aq), a weak acid. So the addition of CO2 makes the water acidic. The ability of CO2 to interact with water increases its solubility.

(b) Since every mole of CO2 that dissolves ends up as (HO)₂CO(aq), there will be 0.035 mol of (HO)₂CO (aq) per Liter produced which, in turn, transfers protons to water. If this proton transfer ‘ goes to completion’ then the concentration of hydronium ion in solution will also be 0.035 mole/L. The pH of this solution would be –log₁₀(0.035) = ~2. The pH is actually more like 4-5, so there isn’t as much hydronium as predicted. We can conclude that the reaction 2.23 does not go to completion.

(c) The conductivity of carbon dioxide is lower than for HCl. Apparently, HCl transfers its protons more completely to water (makes more hydronium ions) than carbon dioxide does.

2.80 Proton transfer from acetic (ethanoic) acid to water

(a) The reaction HCl(aq) + H₂O(l) = Cl-(aq) + H₃O+(aq) must occur to a greater extent than the reaction CH3COOH + H₂O(l) = CH3COO-(aq) + H₃O+(aq). The reaction that produces the greater number of ions (which are responsible for conduction) will result in a solution with higher conductivity.

2.81-2.82 Are CO₂ solutions affected by added acids and bases? This will be covered in class during Week 6.

2.84 What is the stoichiometry of addition of base in Investigate This 2.81?

(a) moles of OH- = moles of NaOH = M of NaOH * V of NaOH = 6M * 0.006L = 0.036 mole

(b) When OH- is added, the balloon deflates which must mean the amount of CO2 gas is decreasing (more CO₂ is dissolving in solution). So, reaction 2.27 is shifting to the right.

(c) In 2.81, 75 mL of water saturated with CO₂ was in the flask. Since every mole of CO2 that dissolves ends up as (HO)₂CO (see 2.79), the concentration of (HO)2CO is 0.035M. The number of moles of (HO)₂CO is 0.035M * 0.075L = 0.0026 mole. From Example 2.83 (see the text), 0.007 mole MORE (HO)₂CO is produced by the reaction of HCl and HOCO2-. All this ends up in the water solution (the balloon deflated). From Example 2.83 the moles of HCl added is 0.018 mole. With 0.007 mole of this involved in the reaction with HOCO₂-, there are 0.018 – 0.007 - = 0.011 mole of HCl left unreacted. This can react with added hydroxide ion.

(d) There are 0.036 mole of OH-, 0.011 mole of HCl, and 0.0026 mole of (HO)₂CO left. One mole OH- reacts with one mole of HCl; two moles of OH- reacts with one mole of (HO)₂CO. If all the HCl reacts, there will be 0.036-0.011 = 0.025 mole of OH- left. This is more than enough to react with 0.0026 mole of (HO)₂CO. So, at the end, both reactions 2.31 and 2.32 will have ‘reached completion’ and there will be CO₃^2-, and leftover OH^-, Na⁺, and Cl^-(from HCl).

(e) When OH^- is added, the ‘system’ responds in such a way as to it reacts with hydronium ion produced when (HO)₂CO(aq) reacts with water. The decrease in hydronium ion concentration means that the equilibrium system shown in 2.27 must shift right to produce more (HO)₂CO which, in turn, can react with water to replenish the lost hydronium.