Chapter 1

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. At the equivalence point, is the solution acidic, basic, or neutral? Why? (A) Acidic; the strong acid dissociates more than the weak base (B) Basic; the only ion present at equilibrium is the conjugate base (C) Basic; the higher concentration of the base is the determining factor (D) Neutral; equal moles of both acid and base are present

Directions: Questions 1-3 are long free-response questions that require about 23 minutes each to answer and are worth 10 points each. Write your response in the space provided following each question. Examples and equations may be included in your responses where appropriate. For calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Pay attention to significant figures. The unbalanced reaction between potassium permanganate and acidified iron (II) sulfate is a redox reaction that proceeds as follows: $$\mathrm{H}^{+}(a q)+\mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-(a q)} \rightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$$ (a) Provide the equations for both half-reactions that occur below: (i) Oxidation half-reaction (ii) Reduction half-reaction (b) What is the balanced net ionic equation? A solution of 0.150 M potassium permanganate is placed in a buret before being titrated into a flask containing 50.00 mL of iron (II) sulfate solution of unknown concentration. The following data describes the colors of the various ions in solution: $$\begin{array}{|c|c|}\hline \text { Ion } & {\text { Color in solution }} \\\ \hline \mathrm{H^{+ }} & {\text { Colorless }} \\ \hline \mathrm{Fe}^{2+} & {\text { Pale Green }} \\ \hline \mathrm{MnO}_{4^{-}} & {\text {Dark Purple }} \\ \hline \mathrm{Mn}^{2+} & {\text { Colorless }} \\ \hline \mathrm{Fe}^{3+} & {\text { Yellow }} \\ \hline \mathrm{K}^{+} & {\text {Colorless }} \\ \hline \mathrm{SO}_{4}^{2-} & {\text { Colorless }} \\\ \hline\end{array}$$ (c) Describe the color of the solution in the flask at the following points: (i) Before titration begins (ii) During titration prior to the endpoint (iii) At the endpoint of the titration (d) (i) If 15.55 mL of permanganate are added to reach the endpoint, what is the initial concentration of the iron (II) sulfate? (ii) The actual concentration of the \(\mathrm{FeSO}_{4}\), is 0.250 \(M\) . Calculate the percent error. (e) Could the following errors have led to the experimental result deviating in the direction that it did? You must justify your answers quantitatively. (i) 55.0 \(\mathrm{mL}\) of \(\mathrm{FeSO}_{4}\) was added to the flask prior to titration instead of 50.0 mL . (ii) The concentration of the potassium permanganate was actually 0.160 \(M\) instead of 0.150 \(M\) .

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) If reaction 2 were repeated at a higher temperature, how would the reaction's value for \(\Delta G\) be affected? (A) It would become more negative because entropy is a driving force behind this reaction. (B) It would become more positive because the reactant molecules would collide more often. (C) It would become more negative because the gases will be at a higher (D) It will stay the same; temperature does not affect the value for \(\Delta G\) .

When some LiCl is dissolved in water, the temperature of the water increases. This means that: (A) the strength of the intermolecular forces between the water molecules is stronger than the bond energy within the LiCl lattice (B) the attraction of the lithium ions to the negative dipoles of the water molecules is weaker than the attraction of the chloride ions to the positive dipoles of the water molecules (C) breaking the bonds between the lithium and chloride ions is an exothermic process (D) the strength of the ion-dipole attractions that are formed exceeds the lattice energy in LiCl

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. If the formic acid were replaced with a strong acid such as HCl at the same concentration (2.0 M), how would that change the volume needed to reach the equivalence point? (A) The change would reduce the amount as the acid now fully dissociates. (B) The change would reduce the amount because the base will be more strongly attracted to the acid. (C) The change would increase the amount because the reaction will now go to completion instead of equilibrium. (D) Changing the strength of the acid will not change the volume needed to reach equivalence.

Why can a molecule with the structure of NBr_ not exist? (A) Nitrogen only has two energy levels and is thus unable to expand its octet. (B) Bromine is much larger than nitrogen and cannot be a terminal atom in this molecule. (C) It is impossible to complete the octets for all six atoms using only valence electrons. (D) Nitrogen does not have a low enough electronegativity to be the central atom of this molecule.

A multi-step reaction takes place with the following elementary steps: $\begin{array}{ll}{\text { Step I. }} & {A+B=C} \\ {\text { Step II. }} & {C+A \rightarrow D} \\ {\text { Step III. }} & {C+D \rightarrow B+E}\end{array}$ If step II is the slow step for the reaction, what is the overall rate law? (A) Rate = k[A]2[B] (B) Rate = k[A][C] (C) Rate = k[A][B] (D) Rate = k[A]/[D]

An electron from which peak would have the greatest velocity after ejection? (A) The peak at 104 \(\mathrm{MJ} / \mathrm{mol}\) (B) The peak at 6.84 \(\mathrm{MJ} / \mathrm{mol}\) (C) The peak at 4.98 \(\mathrm{MJ} / \mathrm{mol}\) (D) The peak at 1.76 \(\mathrm{MJ} / \mathrm{mol}\)

If equimolar solutions of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NaCl}\) are mixed, which ion will not be present in significant amounts in the resulting solution after equilibrium is established? (A) \(\mathrm{Pb}^{2+}\) (B) \(\mathrm{NO}_{3}^{-}\) (C) \(\mathrm{Na}^{+}\) (D) \(\mathrm{Cl}^{-}\)

A multi-step reaction takes place with the following elementary steps: $\begin{array}{ll}{\text { Step I. }} & {A+B=C} \\ {\text { Step II. }} & {C+A \rightarrow D} \\ {\text { Step III. }} & {C+D \rightarrow B+E}\end{array}$ Why would increasing the temperature make the reaction rate go up? (A) It is an endothermic reaction that needs an outside energy source to function. (B) The various molecules in the reactions will move faster and collide more often. (C) The overall activation energy of the reaction will be lowered. (D) A higher fraction of molecules will have the same activation energy.