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Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.

Short Answer

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The "magic" behind acid-base indicators is due to the equilibrium between their protonated (HIn) and deprotonated (In-) forms, each displaying a distinct color. As the pH of the solution changes during titration, the ratio of HIn to In- also changes, causing a visible color change when their concentrations are equal. The exact pH at which this occurs depends on the indicator's pKa value, following the Henderson-Hasselbalch equation. Factors affecting the color change include the specific indicator chosen, its concentration, and the pH range of the titration.

Step by step solution

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1. Acid-base indicators and their role in titration

Acid-base indicators are weak acids or bases that display different colors in their ionized and unionized forms. In titrations, they are used to determine the end point – the point at which the reaction between the analyte and the titrant is complete. At the end point, the indicator changes its color, signaling the completion of the reaction.
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2. How color change in indicators occur

Acid-base indicators are composed of molecules that exist in equilibrium between their protonated (HIn) and deprotonated (In-) forms. Each form has a distinct color. When the pH of the solution changes during titration, the ratio of HIn to In- also changes, which eventually causes a visible color change. The equation representing this equilibrium is: \[HIn(aq) \rightleftharpoons H^+(aq) + In^-(aq)\] The color change occurs when the concentrations of the protonated HIn and deprotonated In- forms are equal. The exact pH at which this occurs depends on the indicator's pKa value, as given by the Henderson-Hasselbalch equation: \[\text{pH} = \text{p}K_a + \log_{10}\Big(\frac{[\text{In}^-]}{[\text{HIn}]}\Big)\]
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3. Factors affecting the color change

The color change of an acid-base indicator depends on: 1. The specific indicator chosen: Different indicators have different pKa values and distinctive color changes that occur at different pH levels. 2. The concentration of the indicator: Higher concentrations can make the color change more intense and easier to see. 3. The pH range of the titration: An appropriate indicator should be chosen based on the pH range of the titration. For instance, if the titration is expected to have a pH range of 4-6, an indicator with a pKa value close to 5 should be chosen. In summary, the "magic" behind acid-base indicators lies in the equilibrium between their protonated and deprotonated forms along with their distinctive colors in different forms and pH-dependent color changes, which allow them to signal the end point of a titration.

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Most popular questions from this chapter

One method for determining the purity of aspirin \(\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\right)\) is to hydrolyze it with NaOH solution and then to titrate the remaining \(\mathrm{NaOH}\). The reaction of aspirin with \(\mathrm{NaOH}\) is as follows: \(\mathrm{C}_{?} \mathrm{H}_{3} \mathrm{O}_{4}(s)+2 \mathrm{OH}^{-}(a q)\) Aspirin $$ \begin{array}{c} \text { Bail } \mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{3}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \\ \text { Salicylate ion Acetate ion } \end{array} $$ A sample of aspirin with a mass of \(1.427 \mathrm{~g}\) was boiled in \(50.00 \mathrm{~mL}\) of \(0.500 M \mathrm{NaOH}\). After the solution was cooled, it took \(31.92 \mathrm{~mL}\) of \(0.289 \mathrm{M} \mathrm{HCl}\) to titrate the excess \(\mathrm{NaOH}\). Calculate the purity of the aspirin. What indicator should be used for this titration? Why?

Sketch two pH curves, one for the titration of a weak acid with a strong base and one for a strong acid with a strong base. How are they similar? How are they different? Account for the similarities and the differences.

What volumes of \(0.50 \mathrm{M} \mathrm{HNO}_{2}\) and \(0.50 \mathrm{M} \mathrm{NaNO}_{2}\) must be mixed to prepare \(1.00 \mathrm{~L}\) of a solution buffered at \(\mathrm{pH}=3.55\) ?

Consider a solution containing \(0.10 M\) ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\right.\) ), \(0.20 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{3}^{+}\), and \(0.20 \mathrm{M} \mathrm{Cl}^{-}\) a. Calculate the \(\mathrm{pH}\) of this solution. b. Calculate the \(\mathrm{pH}\) after \(0.050 \mathrm{~mol} \mathrm{KOH}(s)\) is added to \(1.00 \mathrm{~L}\) of this solution. (Ignore any volume changes.)

A buffer is made using \(45.0 \mathrm{~mL}\) of \(0.750 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{a}=1.3 \times\right.\) \(10^{-5}\) ) and \(55.0 \mathrm{~mL}\) of \(0.700 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} .\) What volume of \(0.10 M\) NaOH must be added to change the \(\mathrm{pH}\) of the original buffer solution by \(2.5 \% ?\)

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