Extra Credit 10

Q15.1.X

Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Table E3 for K_sp values.)

1. KClO₄: [K⁺] = 0.01 M, \(\ce{[ClO4- ]}\) = 0.01 M
2. K₂PtCl₆: [K⁺] = 0.01 M, \(\ce{[PtCl6^2- ]}\) = 0.01 M
3. PbI₂: [Pb²⁺] = 0.003 M, [I^–] = 1.3 × 10^–3 M
4. Ag₂S: [Ag⁺] = 1 × 10^–10 M, [S^2–] = 1 × 10^–13 M

Solutions:

a.

\[K_{sp}=1.05*10^{-2}=[K^{+}][ClO_{4}^{-}]\]

We also know that

\[\frac{KClO_{4}\hspace{5px}moles\hspace{5px}dissolved}{1\hspace{5px}L}=[K^{+}]=[ClO_{4}^{-}]\]

\[\sqrt{K_{sp}}=[K^{+}]_{max}=\frac{KClO_{4}\hspace{5px}maximum\hspace{5px}moles\hspace{5px}dissolved}{1\hspace{5px}L}\approx0.10\]

\[[K^{+}]<[K^{+}]_{max}\]

KClO₄ will not precipitate because the maximum number of moles has not dissolved into the solution

For b-d, we can repeat this process:

b.

Q16.3.6

From the following information, determine \(ΔS^\circ_{298}\) for the following:

1. \(\ce{N}(g)+\ce{O}(g)⟶\ce{NO}(g) \hspace{20px} ΔS^\circ_{298}=\,?\)
2. \(\ce{N2}(g)+\ce{O2}(g)⟶\ce{2NO}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{24.8\: J/K}\)
3. \(\ce{N2}(g)⟶\ce{2N}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{115.0\: J/K}\)
4. \(\ce{O2}(g)⟶\ce{2O}(g) \hspace{20px} ΔS^\circ_{298}=\mathrm{117.0\: J/K}\)

Solution:

Because ΔS is a state variable, we can flip equations c and d

\[\ce{2N}(g)⟶\ce{N2}(g) \hspace{20px} ΔS^\circ_{1}=\mathrm{-115.0\: J/K}\]

\[2O(g)⟶O_{2}(g) \hspace{20px} ΔS^\circ_{2}=\mathrm{-117.0\: J/K}\]

We can then sum the 3 equations as well as the ΔS because it is a state variable

\[2N(g) + 2O(g) + N_{2}(g) + O_{2}(g)⟶N_{2}(g) + O_{2}(g) + 2NO(g) \hspace{20px} ΔS^\circ_{t}=\SigmaΔS^\circ_{n}=24.8+(-117)+(-115)=-207.2 \frac{J}{K}\]

We can then eliminate like terms

\[2N(g) + 2O(g) + \cancel{N_{2}(g)} + \cancel{O_{2}(g)}⟶\cancel{N_{2}(g)} + \cancel{O_{2}(g)} + 2NO(g) \hspace{20px} ΔS^\circ_{t}=-207.2 \frac{J}{K}\]

\[2N(g) + 2O(g)⟶2NO(g) \hspace{20px} ΔS^\circ_{t}=-207.2 \frac{J}{K} \]

\[\frac{1}{2}*2N(g) + \frac{1}{2}*2O(g)⟶\frac{1}{2}*2NO(g) \hspace{20px} ΔS^\circ_{t}=\frac{1}{2}*(-207.2)=-103.6 \frac{J}{K}\]

So the ΔS of a.) is

\[\ce{N}(g)+\ce{O}(g)⟶\ce{NO}(g) \hspace{20px} ΔS^\circ_{298}=-103.6 \frac{J}{K}\]

Q5.4.11

How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed?

\[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l)\hspace{20px}ΔH^\circ_{298}=\mathrm{−58\:kJ}\]

If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation?

Solution:

To find moles of both substances, we must multiply their molarities by their volumes

\[0.250 M * \frac{100 mL}{1000\frac{mL}{L}} = 0.0250\hspace{5px}mol\hspace{5px} HCl\]

\[0.150 M * \frac{200 mL}{1000\frac{mL}{L}} = 0.0300\hspace{5px}mol\hspace{5px} NaOH\]

We then must assume that the reaction goes to completion and that there is a limiting reactant HCl so

\[\Delta H^\circ_{298} = q = 0.0250*-58\hspace{5px}kJ*1000\frac{J}{kJ}=-1450\hspace{5px}J\]

We then assume the heat is fully transferred and not lost to the solvent

\[-q_{rxn}=q_{cal}=1450\hspace{5px}J\]

So we can then use the equation

\[q = mC_{p}\Delta T\]

To find mass we can use the density and volume

\[(100\hspace{5px}mL*1\hspace{5px}\frac{g}{mL})+(200\hspace{5px}mL*1\hspace{5px}\frac{g}{mL}) = 300\hspace{5px}g\]

\[1450\hspace{5px}J = (300\hspace{5px}g)(4.19\hspace{5px}\frac{J}{gC^\circ})\Delta T\]

\[\Delta T = \frac{1450\hspace{5px}J}{(300\hspace{5px}g)(4.19\hspace{5px}\frac{J}{gC^\circ})}\approx 1.1535\hspace{5px}C^\circ\]

So the temperature of the solution will increase by 1.1535 degrees

Q10.3.15

The enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why.

Solution:

Because enthalpy of fusion only slightly increases the separation between particles, there is still a strong interaction between them. This small separation, moving from solid to liquid, requires very little enthalpy because few interactions are actually being broken, and breaking an interaction always requires an input of energy. However, when we move from liquid to gas, we are essentially breaking all intermolecular interactions between particles in the liquid state. This requires a tremendously large input of energy to overcome, essentially, every intermolecular interaction in a liquid.

Q14.2.6

What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?

Solution:

Because HCl ionizes completely

\[[HCl]_{initial} = [H_{3}O^{+}]_{final}\]

Also, because most of the protons will come from HCl and not water, we can neglect water or mathematically

\[\sqrt{K_{a}*[HCl]}>>\sqrt{Kw}=[H_{3}O^{+}]_{water}\]

So we can assume

\[[H_{3}O^{+}] \approx 2.0 M\]

We also know that, under dilute conditions

\[pH\approx-log[H_{3}O^{+}]=-log[2]=-0.3010\]

To find pOH we will then use the relation

\[pH + pOH = pK_{w} = 14\]

\[pOH = 14 - pH = 14 - (-0.3010) = 14.3010\]