Chapter 15: Problem 18
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 18
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 freeA \(10.00-\mathrm{g}\) sample of the ionic compound \(\mathrm{NaA},\) where \(\mathrm{A}^{-}\) is the anion of a weak acid, was dissolved in enough water to make \(100.0 \mathrm{~mL}\) of solution and was then titrated with \(0.100 \mathrm{M}\) HCl. After \(500.0 \mathrm{~mL}\) HCl was added, the \(\mathrm{pH}\) was 5.00 . The experimenter found that \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) was required to reach the stoichiometric point of the titration. a. What is the molar mass of NaA? b. Calculate the \(\mathrm{pH}\) of the solution at the stoichiometric point of the titration.
Two drops of indicator HIn \(\left(K_{2}=1.0 \times 10^{-9}\right),\) where HIn is yellow and In - is blue, are placed in 100.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{MCl}\) . a. What color is the solution initially? b. The solution is titrated with 0.10\(M\) NaOH. At what pH will the color change (yellow to greenish yellow) occur? c. What color will the solution be after 200.0 \(\mathrm{mL}\) NaOH has been added?
In the titration of 50.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) methylamine, CH_{3} \mathrm { NH } _ { 2 } \(\left(K_{\mathrm{b}}=4.4 \times 10^{-4}\right),\) with 0.50 \(\mathrm{M} \mathrm{HCl}\) , calculate the pH under the following conditions. a. after 50.0 \(\mathrm{mL}\) of 0.50\(M \mathrm{HCl}\) has been added b. at the stoichiometric point
Calculate the pH of a solution prepared by mixing \(250 . \mathrm{mL}\) of 0.174 \(\mathrm{m}\) aqueous \(\mathrm{HF}\) (density \(=1.10 \mathrm{g} / \mathrm{mL} )\) with 38.7 \(\mathrm{g}\) of an aqueous solution that is 1.50\(\% \mathrm{NaOH}\) by mass (density \(=\) 1.02 \(\mathrm{g} / \mathrm{mL} ) .\left(K_{\mathrm{a}} \text { for } \mathrm{HF}=7.2 \times 10^{-4} .\right)\)
A certain acetic acid solution has \(\mathrm{pH}=2.68\) . Calculate the volume of 0.0975 \(\mathrm{M} \mathrm{KOH}\) required to reach the equivalence point in the titration of 25.0 \(\mathrm{mL}\) of the acetic acid solution.
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