Questions
\(C_{9}H_{8}O_{4}(aq)+NaOH(aq) \rightleftharpoons C_9H_7O_4^{-}(aq)+Na^{+}(aq)+H_{2}O(l)\)
The reaction represented above occurs when\( 2.00 × 10^{–4}\) mol of pure acetylsalicylic acid, \(C_9H_8O_4\) , is completely dissolved in 15.00 mL of water in a flask and titrated to the equivalence point with 0.100 M NaOH(aq). Which of the following statements about the titration is true at the equivalence point?
▶️Answer/Explanation
Ans: C
At the equivalence point of a titration, the moles of acid and base are stoichiometrically equivalent. The reaction provided indicates a 1:1 mole ratio between \(C_9H_80_4\) and \(C_9H_70_4^-\).
Given that 2.00 x \(10^{-4}\) mol of pure acetylsalicylic acid, \(C_9H_80_4\), is titrated with NaOH, which has a molarity of 0.100 M, we can determine the volume of NaOH required to reach the equivalence
point using the equation \(n_{acid} = n_{base}\).
\(n_{acid} = 2.00 x 10^{-4}\) mol (since it’s given)
\(n_{base}=(0.100\,\text{mol/L}) \times (V_{NaOH} \, \text{L})\)
Equating the two:
\(2.00x 10^{-4}\)mol=\(0.100\,\text{mol/L} \times V_{NaOH}\)
\(V_{NaOH}=\frac{2.00 x 10^{-4}\,\text{mol}}{0.100 |,\text{mol/L}}\)
\(V_{NaOH} = 0.002 |, text{L} = 2.00 |, \text{mL}\)
So, at the equivalence point, 2.00 mL of \(0.100 \, \text{M}\) NaOH has been added to the flask.
Now, let’s consider the stoichiometry of the reaction. At the equivalence point, all of the acetylsalicylic acid, \(C_9H_80_4\), has reacted to form the \(C_9H_70_4^-\) ion. Since the reaction proceeds in a 1:1
mole ratio, the concentration of \(C_9H_80_4\) is equal to the concentration of \(C_9H_70_4^-\) at the equivalence point.
Therefore, the correct answer is:
(C)\(\left[C_9H_80_4\right]\) is equal to\(\left[C_9H_70_4^-\right]\).
Question
A 20. mL sample of 0.50 M \(HC_{2}H_{3}O_{2}\)(aq) is titrated with 0.50 M NaOH(aq). Which of the following best represents the species that react and the species produced in the reaction?
(A) \(H^{+}(aq) + OH^{−}(aq) → H_{2}O(l)\)
(B) \(H^{+}(aq) + C_{2}H_{3}O_{2}^{−}(aq) + Na^{+}(aq) + OH^{−}(aq)\) →\( H_{2}O(l) + NaC_{2}H_{3}O_{2}\)(aq)
(C) \(HC_{2}H_{3}O_{2}(aq)\) + OH^{−}(aq) → \(C_{2}H_{3}O_{2}^{−}(aq) + H_{2}O(l)\)
(D) \(HC_{2}H_{3}O_{2}(aq) + NaOH(aq) \)→ \(H_{2}O(l) + Na+(aq) + C_{2}H_{3}O_{2}^{−}\)(aq)
▶️Answer/Explanation
Ans:C
In this titration, acetic acid (\(HC_{2}H_{3}O_{2}\)) reacts with sodium hydroxide (\(NaOH\)). The reaction can be represented as follows:
\[HC_{2}H_{3}O_{2}(aq) + OH^{-}(aq) \rightarrow C_{2}H_{3}O_{2}^{-}(aq) + H_{2}O(l) \]
This reaction involves the transfer of a hydrogen ion (\(H^+\)) from the acetic acid to the hydroxide ion (\(OH^-\)), forming water (\(H_2O\)) and the acetate ion (\(C_2H_3O_2^-\)).
So, the best representation of the species that react and the species produced in the reaction is option:
(C) \(HC_{2}H_{3}O_{2}(aq)+ OH^{-}(aq)\) → \(C_{2}H_{3}O_{2}^{-}(aq) + H_{2}O(l)\)
Questions
A particle view of a sample of \(H_{2}O_{2}(aq)\) is shown above. The \(H_{2}O_{2}(aq)\) is titrated with \(KMnO_{4}(aq)\), as represented by the equation below.
\(2MnO_{4}^{-}(aq)+5H_{2}O_{2}(aq)+6H^{+}(aq)\rightarrow 2Mn^{2+}(aq)+5O_{2(g)}+8H_{2}O(l)\)
Which of the following particle views best represents the mixture when the titration is halfway to the equivalence point? \(H_{2}O\) molecules and H+ ions are not shown.)
▶️Answer/Explanation
Ans: C
Image (C) represents the mixture at the halfway point of the titration between \( \text{H}_2\text{O}_2(\text{aq}) \) and \( \text{KMnO}_4(\text{aq}) \) accurately.
Explanation:
\( \text{Mn}^{2+} \) ions, indicating the reduction of some \( \text{MnO}_4^- \) by \( \text{H}_2\text{O}_2 \).
Remaining \( \text{H}_2\text{O}_2 \) molecules, suggesting that the reaction is partially complete.
Absence of \( \text{MnO}_4^- \) ions or \( \text{O}_2 \) gas molecules.
At the halfway point, half of the initial \( \text{H}_2\text{O}_2 \) would have reacted with half of the initial \( \text{KMnO}_4 \), resulting in the presence of \( \text{Mn}^{2+} \) ions. The presence of remaining \( \text{H}_2\text{O}_2 \) molecules and the absence of \( \text{MnO}_4^- \) ions or \( \text{O}_2 \) gas molecules align with this halfway state.
On the other hand, images (A), (B), and (D) do not accurately represent the halfway point because they depict the presence of either unreacted \( \text{KMnO}_4^- \) ions or \( \text{O}_2 \) gas molecules, which would indicate that the reaction has progressed beyond the halfway mark.
Question
The pH versus volume data for the titration of 0.10M \(HNO_2(aq)\) with 0.10M \(KOH(aq)\) is plotted on the graph above. Based on the data, which of the following species is present in the greatest concentration after 6.0mL of \(KOH(aq)\) has been added to the solution of \(HNO_2(aq)\) ?
A \(H^+(aq)\)
B \(HNO_2(aq)\)
C \(NO_2^−(aq)\)
D \(OH^−(aq)\)
▶️Answer/Explanation
Ans:C
At the half-equivalence point in a titration of a weak acid with a strong base, \([HA]=[A^−]\). Based on the data, the half-equivalence point is reached after 5mL of \(KOH(aq)\) has been added. After this point, \([NO_2^−]>[HNO_2]\). Because \(HNO_2\) is a weak acid and the majority of the acid has reacted at this point, \([H^+]\) is very low. Before the equivalence point, \(OH^− \) added to the solution is consumed by the reaction with \(HNO_2(aq)\). Therefore, the species in the greatest concentration after 6.0mL of \(KOH(aq)\) has been added to the solution of \(HNO_2(aq)\) is \(NO_2^−(aq)\).