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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equivalence Point
In an acid-base titration, the equivalence point is the moment when the amount of acid equals the amount of base. It signifies that the acid and base have reacted completely. For a strong acid and strong base titration, this means all of the hydrogen ions ( H$^+$ ) have reacted with hydroxide ions ( OH$^-$ ), forming water (H$_2$O).
At the equivalence point, the solution is neutral if a strong acid and strong base are used, typically having a pH of 7. This is why indicators are essential to identify this point. They change color based on the pH of the solution, helping us recognize when the equivalence point has been reached.
The equivalence point is a critical part of titration because it informs us that titration has been successful and the substance being analyzed has reacted completely.
Strong Acid and Strong Base
When we talk about a strong acid and a strong base, we refer to substances that dissociate completely in water. A strong acid, like hydrochloric acid (HCl), yields a lot of H$^+$ ions, while a strong base, like sodium hydroxide (NaOH), provides OH$^-$ ions.
Their complete dissociation means that when they react with each other in a titration, they neutralize completely, producing water and usually a salt. For example, the reaction of HCl with NaOH gives water and sodium chloride (NaCl).
  • Complete dissociation makes calculations in titrations straightforward, as the initial concentration of each reactant equals the concentration of ions in solution.
  • This feature facilitates the identification of the equivalence point since the reactions proceed without leaving behind any of the original acid or base.
Acid-Base Indicators
Indicators are substances that undergo a noticeable color change when the pH of a solution changes. In titrations, they tell us when the equivalence point has been reached. Each indicator has its own unique pH range over which it changes color.
Choosing the right indicator is crucial, as the color change should occur near the solution's equivalence point. For a strong acid-strong base titration, phenolphthalein is a common indicator, turning from colorless to pink around pH 8-10.
  • Indicators work by changing color due to the concentration of hydrogen ions affecting the state of the indicator molecule.
  • The point at which the indicator changes color does not always coincide exactly with the equivalence point, but in strong acid-strong base titrations, they typically align closely.
Protonation and Deprotonation
Protonation and deprotonation describe the gain or loss of a proton (H\(^+\)) by a molecule.
Protonation involves a base gaining a proton, forming a conjugate acid, while deprotonation involves an acid losing a proton, forming a conjugate base. In a titration, this relates to the transformation of the indicator from one form to another, dictating its color.
For the indicator example, protonated form is HIn, which is in equilibrium with its deprotonated form In\(^-\) according to the reaction:\[ HIn \leftrightarrow H^{+} + In^{-} \]
At equilibrium, if the concentrations of HIn and In\(^-\) are equal, it indicates the equivalence point during a titration.
  • Protonation and deprotonation are crucial in understanding the color changes of indicators.
  • They highlight how an indicator responds to pH changes and provide insight into the underlying chemical mechanism guiding those changes.

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Most popular questions from this chapter

Use the following information to answer questions 14-16 The radius of atoms and ions is typically measured in Angstroms \((A),\) which is equivalent to \(1 * 10^{-10} \mathrm{m} .\) Below is a table of information for three different elements. TABLE NOT AVAILABLE The phosphorus ion is larger than a neutral phosphorus atom, yet a zinc ion is smaller than a neutral zinc atom. Which of the following statements best explains why? (A) The zinc atom has more protons than the phosphorus atom. (B) The phosphorus atom is paramagnetic, but the zinc atom is diamagnetic. (C) Phosphorus gains electrons when forming an ion, but zinc loses them. (D) The valence electrons in zinc are further from the nucleus than those in phosphorus.

After 44 minutes, a sample of \(_{19}^{44} \mathrm{K}\) is found to have decayed to 25 percent of the original amount present. What is the half-life of \(_{19}^{44} \mathrm{K}?\) (A) 11 minutes (B) 22 minutes (C) 44 minutes (D) 66 minutes

\(\begin{array}{ll}{\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{4}(g)} & {\Delta H^{\circ}=x} \\\ {\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=y} \\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=\mathrm{z}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (A) \(x+y+z\) (B) \(x+y-z\) (C) \(z+y-2 x\) (D) \(2 z+y-x\)

\(\mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \rightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)\) If the reaction above took place at standard temperature and pressure and 150 grams of \(\mathrm{CaCO}_{3}(\mathrm{s})\) were consumed, what was the volume of \(\mathrm{CO}_{2}(g)\) produced at STP? (A) 11 L (B) 22 L (C) 34 L (D) 45 L

1.50 g of \(\mathrm{NaNO}_{3}\) is dissolved into 25.0 \(\mathrm{mL}\) of water, causing the temperature to increase by \(2.2^{\circ} \mathrm{C}\) . The density of the final solution is found to be 1.02 \(\mathrm{g} / \mathrm{mL}\) . Which of the following expressions will correctly calculate the heat gained by the water as the NaNO, dissolves? Assume the volume of the solution remains unchanged. (A) \((25.0)(4.18)(2.2)\) (B) \(\frac{(26.5)(4.18)(2.2)}{1.02}\) (C) \(\frac{(1.02)(4.18)(2.2)}{1.50}\) (D) \((25.0)(1.02)(4.18)(2.2)\)

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