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Chapter 1: Problem 27

A strong acid/strong base titration is completed using an indicator which changes color at the exact equivalence point of the titration. The protonated form of the indicator is HIn, and the deprotonated form is In. At the equivalence point of the reaction: (A) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]\) (B) \([\mathrm{HIn}]=1 /\left[\mathrm{In}^{-}\right]\) (C) \([\mathrm{HIn}]=2\left[\mathrm{In}^{-}\right]\) (D) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]^{2}\)

Short Answer

Expert verified
The correct answer is option (A) \([HIn]=[In^-]\), because at the equivalence point the concentration of the protonated form of the indicator is equal to the concentration of the deprotonated form.

Step by step solution

01

Analyze the reaction

In a strong acid/strong base titration, the reaction between the acid and the base goes to completion. The indicator changes color at the equivalence point where the moles of acid equal the moles of base. The equilibrium of the color change reaction is: \[HIn \leftrightarrow H^{+} + In^{-}\]
02

Understand the equivalency condition

At the equivalence point, all the acid and base have reacted. The indicator can exist in two forms, the protonated form HIn and the deprotonated form In-. The color change occurs when an equal amounts of acid and base has been added, which means that the concentration of HIn equals the concentration of In-, that is, \([HIn]=[In^-]\).
03

Match with the given options

Looking at the given options, option (A) suggests that \([HIn]=[In^-]\), which is in line with the analysis stated above. None of the other options reflect this equilibrium condition at the equivalence point.

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Most popular questions from this chapter

Use the following information to answer questions 12-15. When heated in a closed container in the presence of a catalyst, potassium chlorate decomposes into potassium chloride and oxygen gas via the following reaction: \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) If 12.25 g of potassium chlorate decomposes, how many grams of oxygen gas will be generated? (A) 1.60 g (B) 3.20 g (C) 4.80 g (D) 18.37 g

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following occurs at the cathode? (A) \(\mathrm{Cr}^{3+}\) is reduced to \(\mathrm{Cr}(\mathrm{s})\) (B) \(\mathrm{Pb}^{2+}\) is reduced to \(\mathrm{Pb}(\mathrm{s})\) (C) \(\mathrm{Cr}(s)\) is oxidized to \(\mathrm{Cr}^{3+}\) (D) \(\quad \mathrm{Pb}(s)\) is oxidized to \(\mathrm{Pb}^{2+}\)

\(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu} \quad E^{\circ}=+0.3 \mathrm{V}\) \(\mathrm{Fe}^{2+}+2 e^{-} \rightarrow \mathrm{Fe} \quad E^{\circ}=-0.4 \mathrm{V}\) Based on the reduction potentials given above, what is the reaction potential for the following reaction? \(\mathrm{Fe}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Fe}+\mathrm{Cu}^{2+}\) (A) \(-0.7 \mathrm{V}\) (B) \(-0.1 \mathrm{V}\) (C) \(+0.1 \mathrm{V}\) (D) \(+0.7 \mathrm{V}\)

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ What is the required voltage to make this cell function? (A) 0.34 V (B) 0.42 V (C) 0.76 V (D) 1.10 V

Which expression below should be used to calculate the mass of copper that can be plated out of a 1.0 \(\mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) , solution using a current of 0.75 A for 5.0 minutes? (A) \(\frac{(5.0)(60)(0.75)(63.55)}{(96500)(2)}\) (B) \(\frac{(5.0)(60)(63.55)(2)}{(0.75)(96500)}\) (C) \(\frac{(5.0)(60)(96500)(0.75)}{(63.55)(2)}\) (D) \(\frac{(5.0)(60)(96500)(63.55)}{(0.75)(2)}\)
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