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Chapter 4: Problem 86

How does an acid-base indicator work?

Short Answer

Expert verified
Acid-base indicators work by exhibiting a color change in response to the concentration of hydrogen (H+) or hydroxide (OH-) ions in an aqueous solution. The color change occurs because the molecular structure of the indicator changes when it donates or accepts a proton, resulting in a change in color observed.

Step by step solution

01

- Define an Acid-Base Indicator

An acid-base indicator is a weak acid or weak base that exhibits a color change as the concentration of hydrogen (H+) or hydroxide (OH-) ions changes in an aqueous solution.
02

- Explain the Functioning of an Indicator

Acid-base indicators work by donating or accepting protons. The specific color observed in a solution of an indicator depends on the molecular structure of the indicator, which changes if it donates or accepts a proton.
03

- Show How Color is Impacted

Since the color observed depends on the molecular structure, an indicator may appear one color (color A) in its original form and another color (color B) in its protonated or deprotonated form. Therefore, the color of an acid-base indicator will change depending on whether it is in an acidic or basic solution.
04

- Detail the Changes

In an acidic solution, an indicator donates protons and tends to be in its original (usually color A) form, while in a basic solution, the indicator tends to accept protons and is in its deprotonated (usually color B) form. The observed color is either color A, color B or a combination of both if the solution is neutral.

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Most popular questions from this chapter

Sodium reacts with water to yield hydrogen gas. Why is this reaction not used in the laboratory preparation of hydrogen?

Describe in each case how you would separate the cations or anions in an aqueous solution of: (a) \(\mathrm{NaNO}_{3}\) and \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2},\) (b) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{KNO}_{3}\), (c) \(\mathrm{KBr}\) and \(\mathrm{KNO}_{3}\), (d) \(\mathrm{K}_{3} \mathrm{PO}_{4}\) and \(\mathrm{KNO}_{3}\), (e) \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) and \(\mathrm{NaNO}_{3}\).

Describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.707 M \mathrm{NaNO}_{3}\) solution.

Classify the following reactions according to the types discussed in the chapter. (a) \(\mathrm{Cl}_{2}+2 \mathrm{OH}^{-} \longrightarrow \mathrm{Cl}^{-}+\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{Ca}^{2+}+\mathrm{CO}_{3}^{2-} \longrightarrow \mathrm{CaCO}_{3}\) (c) \(\mathrm{NH}_{3}+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_{4}^{+}\) (d) \(2 \mathrm{CCl}_{4}+\mathrm{CrO}_{4}^{2-} \longrightarrow\) \(2 \mathrm{COCl}_{2}+\mathrm{CrO}_{2} \mathrm{Cl}_{2}+2 \mathrm{Cl}^{-}\) (e) \(\mathrm{Ca}+\mathrm{F}_{2} \longrightarrow \mathrm{CaF}_{2}\) (f) \(2 \mathrm{Li}+\mathrm{H}_{2} \longrightarrow 2 \mathrm{LiH}\) (g) \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4} \longrightarrow 2 \mathrm{NaNO}_{3}+\mathrm{BaSO}_{4}\) (h) \(\mathrm{CuO}+\mathrm{H}_{2} \longrightarrow \mathrm{Cu}+\mathrm{H}_{2} \mathrm{O}\) (i) \(\mathrm{Zn}+2 \mathrm{HCl} \longrightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\) (j) \(2 \mathrm{FeCl}_{2}+\mathrm{Cl}_{2} \longrightarrow 2 \mathrm{FeCl}_{3}\) (k) \(\mathrm{LiOH}+\mathrm{HNO}_{3} \longrightarrow \mathrm{LiNO}_{3}+\mathrm{H}_{2} \mathrm{O}\)

A quantity of \(25.0 \mathrm{~mL}\) of a solution containing both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions is titrated with \(23.0 \mathrm{~mL}\) of \(0.0200 M \mathrm{KMnO}_{4}\) (in dilute sulfuric acid). As a result, all of the \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\) ions. Next, the solution is treated with Zn metal to convert all of the \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ions. Finally, the solution containing only the \(\mathrm{Fe}^{2+}\) ions requires \(40.0 \mathrm{~mL}\) of the same \(\mathrm{KMnO}_{4}\) solution for oxidation to \(\mathrm{Fe}^{3+}\). Calculate the molar concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the original solution. The net ionic equation is \(\mathrm{MnO}_{4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} \longrightarrow\) \(\mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H}_{2} \mathrm{O}\)
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