Skip to main content
Chemistry LibreTexts

1.12: Practice Problems for Bronsted-Lowry Acid-Base Chemistry Answers

  • Page ID
    218646
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    1.

    2.

    3. 

    4.

    5. The numbers represent approximate pKa values for the substances acting as acids

    6. A

    7. Methanesulfonic acid is the stronger acid. The lower the pKa, the stronger the acid. A lower pKa is associated with a larger Ka which signifies greater dissociation. The large relative difference in acidity in this case can be most easily seen by gauging the relative basicities of the conjugate bases. The weaker the base, the stronger the corresponding conjugate acid. Methanesulfonate, CH3SO3-, is considerably stabilized by resonance delocalization which is not found in ethoxide, CH3CH2O-. This effect greatly reduces the basicity of methanesulfonate relative to ethoxide. Draw the Lewis formula for methanesulfonate and the resonance forms for practice.

    8. Trifluoromethanesulfonic acid is a stronger acid. Compare the strengths of the conjugate bases and remember that the weaker the base, the stronger the conjugate acid. Both bases are stabilized by resonance, but in the case of the trifluoro derivative, the presence of the highly electronegative fluorine atoms serves to delocalize the negative charge to an even greater extent by the inductive effect of fluorine. This additional delocalization makes trifluoromethanesulfonate a weaker base.

    9. CH3COO- < CH3O- < NH2-

    first factor to consider is the nature of the atom which bears the negative charge. The more electronegative the atom that bears the negative charge, the more stable the anion. Stable anions are less reactive and are hence weaker bases. Since O is more electronegative than N, the NH2- is the strongest base in the set. In the remaining two species, the negative charge is on the O, but in the case of CH3COO-, the negative charge is also delocalized by resonance.

    10. D

    11. See class notes, or p. 417 of the Wade textbook, for all the resonance structures of the phenoxide ion shown below.

    12. CH4 < NH3 < CH3OH < HCl (periodic trend)

    13. H2O < NH3 < CH3O- < H2N- Negatively charged ions are stronger bases than neutral counterparts. A table of pKa values will further aid in deciding the final order of basicity.

    14. B

    15. B

    16. 


    This page titled 1.12: Practice Problems for Bronsted-Lowry Acid-Base Chemistry Answers is shared under a not declared license and was authored, remixed, and/or curated by Sergio Cortes.

    • Was this article helpful?