Skip to main content
Chemistry LibreTexts

Solution to Exercise 7-2-1

  • Page ID
    43257
    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Remember that the stronger conjugate acid has the weaker conjugate base. From the table of approximate pKa values given in this section, CH3SH (pKa ~ 10) is a stronger acid than CH3OH (pKa ~ 15) , so CH3O- is the stronger base.

    Following the same reasoning and using the given approximate pKa values of the conjugate acids, ammonia (NH3) is a stronger base than acetate (CH2COO-), and hydroxide (OH-) is a stronger base than acetate.


    Solution to Exercise 7-2-1 is shared under a not declared license and was authored, remixed, and/or curated by .


    Solution to Exercise 7-2-1 is shared under a not declared license and was authored, remixed, and/or curated by Tim Soderberg.

    • Was this article helpful?