Chapter 14: Problem 16
Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.
Chapter 14: Problem 16
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 best buffer has about equal quantities of weak acid and conjugate base present as well as having a large concentration of each species present. Explain.
Calculate the pH of each of the following solutions. a. \(0.100 M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) b. \(0.100 M\) sodium propanoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 M \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 M\) \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)
Consider a buffer solution where [weak acid] \(>\) [conjugate base]. How is the pH of the solution related to the \(\mathrm{p} K_{\mathrm{a}}\) value of the weak acid? If [conjugate base] \(>\) [weak acid], how is pH related to \(\mathrm{p} K_{\mathrm{a}} ?\)
Two drops of indicator HIn \(\left(K_{\mathrm{a}}=1.0 \times 10^{-9}\right),\) where HIn is yellow and \(\operatorname{In}^{-}\) is blue, are placed in \(100.0 \mathrm{mL}\) of \(0.10 \mathrm{M}\) HCl. 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?
Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. a. \(100.0 \mathrm{mL}\) of \(0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) titrated by 0.10 \(M \mathrm{NaOH}\) b. \(100.0 \mathrm{mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by 0.20 \(M \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{mL}\) of \(0.50 \mathrm{M}\) HCl titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)
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