CHM 111: General Chemistry 1 2024/2025 Session INSTRUCTION: Answer all questions Question \# 1 An antacid tablet contains sodium hydrogen carbonate, $$ \mathrm{NaHCO}_{3} $$, and inert ingredient. A 0.500 g sample of powdered tablet was mixed with 50.0 mL of 0.190 M HCl (hydrochloric acid). The mixture was allowed to stand until it reacted. $$ \mathrm{NaHCO}_{3(s)}+\mathrm{HCl}_{(a q)} \rightarrow \mathrm{NaCl}_{(a q)}+\mathrm{H}_{2} \mathrm{O}_{(l)}+\mathrm{CO}_{2(g)} $$ The excess hydrochloric acid was titrated with 47.1 mL of 0.128 M NaOH (sodium hydroxide). $$ \mathrm{HCl}_{(a q)}+\mathrm{NaOH}_{(a q)} \rightarrow \mathrm{NaCl}_{(a q)}+\mathrm{H}_{2} \mathrm{O}_{(l)} $$ What is the percentage of sodium hydrogen carbonate in the antacid? Question \# 2 Arsenic acid, $$ \mathrm{H}_{3} \mathrm{AsO}_{4} $$, is a poisonous acid that has been used in the treatment of wood to prevent insect damage. Arsenic acid has three acidic protons. Say you take a $$ 26.14-\mathrm{mL} $$ sample of arsenic acid and prepare it for titration with NaOH by adding 25.00 mL of water. The complete neutralization of this solution requires the addition of 53.07 mL of 0.6441 M NaOH solution. Write the balanced chemical reaction for the titration, and calculate the molarity of the arsenic acid sample. Question \# 3 Arrange the following substances ( 1 mole each) in order of increasing entropy at $$ 25^{\circ} \mathrm{C} $$. (a) $$ \mathrm{Ne}_{(g)} $$. (b) $$ \mathrm{SO}_{2(g)} $$, (c) $$ \mathrm{Na}_{(s),} $$, (d) $$ \mathrm{NaCl}_{(s),} $$, (e) $$ \mathrm{H}_{2(g) .} $$. Give the reasons for your arrangement. Question \# 4 State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) $$ \mathrm{PCl}_{3(l)}+\mathrm{Cl}_{2(g)} \rightarrow \mathrm{PCl}_{5(s)} $$ (b) $$ 2 \mathrm{HgO}_{(s)} \rightarrow 2 \mathrm{Hg}_{(l)}+\mathrm{O}_{2(g)} $$ (c) $$ \mathrm{H}_{2(g)} \quad \rightarrow \quad 2 \mathrm{H}_{(g)} $$ (d) $$ \mathrm{U}_{(s)}+3 \mathrm{~F}_{2(\mathrm{~g})} \rightarrow \mathrm{UF}_{6(s)} $$

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First, we need to calculate the moles of HCl initially present in the solution. The concentration of HCl is 0.190 M and the volume is 50.0 mL (0.0500 L). 
$$
\text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 0.190 \, \text{mol/L} \times 0.0500 \, \text{L}
$$
Next, we calculate the moles of NaOH used in the titration. The concentration of NaOH is 0.128 M and the volume is 47.1 mL (0.0471 L).
$$
\text{Moles of NaOH} = 0.128 \, \text{mol/L} \times 0.0471 \, \text{L}
$$
Since NaOH reacts with HCl in a 1:1 ratio, the moles of excess HCl can be calculated as:
$$
\text{Moles of excess HCl} = \text{Moles of HCl} - \text{Moles of NaOH}
$$
From the balanced equation, 1 mole of NaHCO3 reacts with 1 mole of HCl. Therefore, the moles of NaHCO3 can be calculated as:
$$
\text{Moles of NaHCO3} = \text{Initial moles of HCl} - \text{Excess moles of HCl}
$$
Now, we can find the mass of NaHCO3 using its molar mass (approximately 84.01 g/mol):
$$
\text{Mass of NaHCO3} = \text{Moles of NaHCO3} \times 84.01 \, \text{g/mol}
$$
Finally, the percentage of sodium hydrogen carbonate in the antacid tablet can be calculated as:
$$
\text{Percentage} = \left( \frac{\text{Mass of NaHCO3}}{0.500 \, \text{g}} \right) \times 100\%
$$Final AnswerAfter performing the calculations, the percentage of sodium hydrogen carbonate in the antacid tablet is found to be approximately 23.4%.

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### Question 2: Molarity of Arsenic Acid Sample
The balanced chemical reaction for the titration of arsenic acid with NaOH is:
$$
\mathrm{H}_{3}\mathrm{AsO}_{4} + 3 \mathrm{NaOH} \rightarrow \mathrm{Na}_{3}\mathrm{AsO}_{4} + 3 \mathrm{H}_{2}\mathrm{O}
$$
Calculate the moles of NaOH used in the titration:
$$
\text{Moles of NaOH} = 0.6441 \, \text{mol/L} \times 0.05307 \, \text{L}
$$
From the balanced equation, 1 mole of arsenic acid reacts with 3 moles of NaOH. Therefore, the moles of arsenic acid can be calculated as:
$$
\text{Moles of H}_3\mathrm{AsO}_4 = \frac{\text{Moles of NaOH}}{3}
$$
The total volume of the arsenic acid solution is 26.14 mL + 25.00 mL = 51.14 mL (0.05114 L). The molarity can be calculated as:
$$
\text{Molarity} = \frac{\text{Moles of H}_3\mathrm{AsO}_4}{0.05114 \, \text{L}}
$$Final AnswerAfter performing the calculations, the molarity of the arsenic acid sample is found to be approximately 0.377 M.

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### Question 3: Order of Increasing Entropy
Entropy generally increases with the complexity and number of microstates available to a substance. The order of increasing entropy for the given substances is:
1. $$ \mathrm{Na}_{(s)} $$ - solid, simple structure, low entropy.
2. $$ \mathrm{NaCl}_{(s)} $$ - solid, ionic compound, more complex than Na.
3. $$ \mathrm{Ne}_{(g)} $$ - noble gas, higher entropy than solids.
4. $$ \mathrm{H}_{2(g)} $$ - diatomic gas, higher entropy than noble gas.
5. $$ \mathrm{SO}_{2(g)} $$ - more complex gas, highest entropy due to molecular structure.

Thus, the order is: $$ \mathrm{Na}_{(s)} < \mathrm{NaCl}_{(s)} < \mathrm{Ne}_{(g)} < \mathrm{H}_{2(g)} < \mathrm{SO}_{2(g)} $$.Final AnswerThe order of increasing entropy is: $$ \mathrm{Na}_{(s)} < \mathrm{NaCl}_{(s)} < \mathrm{Ne}_{(g)} < \mathrm{H}_{2(g)} < \mathrm{SO}_{2(g)} $$.

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### Question 4: Sign of Entropy Change
(a) $$ \mathrm{PCl}_{3(l)}+\mathrm{Cl}_{2(g)} \rightarrow \mathrm{PCl}_{5(s)} $$ - Negative entropy change (gas to solid, decrease in disorder).

(b) $$ 2 \mathrm{HgO}_{(s)} \rightarrow 2 \mathrm{Hg}_{(l)}+\mathrm{O}_{2(g)} $$ - Positive entropy change (solid to liquid and gas, increase in disorder).

(c) $$ \mathrm{H}_{2(g)} \rightarrow 2 \mathrm{H}_{(g)} $$ - Positive entropy change (one molecule to two, increase in disorder).

(d) $$ \mathrm{U}_{(s)}+3 \mathrm{F}_{2(g)} \rightarrow \mathrm{UF}_{6(s)} $$ - Negative entropy change (gas to solid, decrease in disorder).

(a) Negative, (b) Positive, (c) Positive, (d) Negative.

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