Chapter 15: Problem 14
Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.
Chapter 15: Problem 14
Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.
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Get started for freeOne 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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