Chapter 10 Source Answers and Examples
Chapter problems → · Original reading →
Reviewed solution to Problem 2
New reviewed response for this exact prompt.
Crushing increases exposed surface area; stirring transports dissolved species from the solid interface and brings unsaturated solvent toward it. These can speed approach to equilibrium without changing the equilibrium saturation amount. A changed solubility claim needs equilibrium data under specified conditions.
Reviewed solution to Problem 16
New reviewed response for this exact prompt.
Copper must be oxidized toCu⁺ orCu²⁺ by an identified oxidant; an electron-balanced reaction and actual products are needed. A mass increase does not uniquely determine a hydrated-salt composition or dissolved-ion inventory. A quantitative boiling model also needs solvent mass, dissolved species amounts, an appropriate dilute/real-solution model and its conditions. The original spontaneous hydrate and boiling inference are not justified.
Worked examples and reviewed revisions
Example Problem 2.1.1
Original example and steps
Example Problem 2.1.1
Is MgSO4 soluble or insoluble?
Okay, time to go through the rules until we find mention of the relevant ions.
Rule 1 discusses cations, but Magnesium is not mentioned. Next.
Rule 2 discusses Nitrates, not Sulfates. Next.
Rule 3 covers Silver, but we have Magnesium. Next.
Rule 4 refers to Acetic Acid, which is not present. Next.
Rule 5 covers a lot of anions, but no Sulfate. Next.
Rule 6 is irrelevant as we don’t have lead. Next.
Rule 7 is only useful if we had a halogen. Next.
Rule 8 finally mentions one of our ions. It tells us that Sulfate salts are going to be soluble, unless it has an anion that we don’t have. Therefore, we can safely assert that MgSO4 is soluble.
Example Problem 2.1.2
Worked example
Sodium carbonate has formula Na₂CO₃: two Na⁺ ions balance one CO₃²⁻ ion. Under the ordinary dilute aqueous solubility convention, it dissolves to supply those ions. NaCO₃ would not balance the charges of sodium(I) and carbonate.
This one’s easy. Right off the bat, rule 1 gives us our answer. As our cation is Sodium, we know that this salt will be soluble.
This one’s easy. Right off the bat, rule 1 gives us our answer. As our cation is Sodium, we know that this salt will be soluble.
Example Problem 2.1.3
Original example and steps
Example Problem 2.1.3
Is PbS soluble?
Time to go through our rules:
Rule 1 discusses cations, but Lead is not mentioned. Next.
Rule 2 discusses Nitrates, not Sulfur. Next.
Rule 3 covers Silver, but we have Lead. Next.
Rule 4 refers to Acetic Acid, which is not present. Next.
Rule 5 does indeed mention Sulfur as an anion. We note that this means our salt is insoluble.
Example Problem 2.2.1
Worked example
AP edition reviewed replacement.
Formal ionic-strength enrichment only: calculate I for final concentrations 1.00 M NaCl plus 0.750 M Mg(NO₃)₂, and for 2.00 M KOH. Assume complete dissociation. Does this alone rank copper salt solubility?
New independently computed worked answer.
For the mixture, I=½[1.00+1.00+4(0.750)+1.50]=3.25 M. For KOH, I=½[2.00+2.00]=2.00 M. The first has greater formal ionic strength. A copper solubility ranking does not follow: common ions, pH-dependent reactions, specific equilibria and activities need separate analysis.
Example Problem 2.2.2
Worked example
AP edition reviewed replacement.
Find the ideal molar and mass solubility of AgCl in 0.0100 M NaCl at a temperature where Ksp=1.80×10⁻¹⁰. Assume no other silver complexes, ideal concentrations and M(AgCl)=143.32 g mol⁻¹.
New independently computed worked answer.
AgCl(s)⇌Ag⁺+Cl⁻. Let s dissolve: [Ag⁺]=s and [Cl⁻]=0.0100+s. Thus s(0.0100+s)=1.80×10⁻¹⁰. Since s≪0.0100, s≈1.80×10⁻⁸ M; s/0.0100=1.80×10⁻⁶ confirms the approximation. Mass solubility is 2.58×10⁻⁶ g L⁻¹. The original concentrated MgCl₂ cubic and raster answers are superseded.
Example Problem 2.2.3
Worked example
AP edition reviewed replacement.
Dilute colligative enrichment model: dissolve 0.100 mol NaOH in 1.000 kg water. Use supplied ideal i=2, Kf=1.86 °C kg mol⁻¹ and Kb=0.512 °C kg mol⁻¹. Find the model freezing and boiling changes.
New independently computed worked answer.
Molality is 0.100 mol kg⁻¹. Solute mass is 4.00 g, solvent mass 1000 g and approximate solution mass 1004 g; they have distinct roles. ΔTf=2×1.86×0.100=0.372 °C; ΔTb=2×0.512×0.100=0.1024 °C. Relative to stated pure-water reference values 0 °C and 100 °C, the model gives −0.372 °C and 100.1024 °C. This dilute model supersedes the50wt%NaOH prediction and its old dependent raster formulas.
Example Problem 2.3.1
Original example and steps
Example Problem 2.3.1
What is the solubility of Oxygen from air to water at sea level, if the temperature is 15ºC?
Step 1: Identify the knowns and unknowns
We actually don’t know any of these factors. As we are solving for [O2], we can say that we need to figure out the partial pressure of O2 and the kH first.
Step 2: Determine the Partial Pressure of .
All we know for the partial pressure is that we are looking at air at sea level. We know that air has a pressure of 1atm at sea level, and you should know that air is 21% Oxygen. If you don’t remember that number, you should try to commit it to memory. Also know that Nitrogen composes 78% of air.
With the fraction of Oxygen in air and the total pressure, we can figure out the partial pressure of Oxygen:
Step 3: Identify the .
We know our gas is Oxygen and our temperature is 15ºC. Therefore we can use this equation:
We have the is 769 and is 1700, but we need to find out T in K, as we can’t use 15.
Now we can solve for kH:
Step 4: Solve for .
We now know both and . We can now plug these in to the Henry’s Law equation to solve for the solubility of O2.