Skip to main content
Chemistry LibreTexts

Solution to Exercise 7-2-6

  • Page ID
    43264
    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    We use the Henderson-Hasselbalch equation and let the base to acid ratio be X.

    For pH = 2.8:

    2.8 = 4.8 + log X

    X = 0.01 to 1

    pH 3.8, the ratio is 0.10 to 1

    pH 4.8, the ratio is 1.0 to 1

    pH 5.8, the ratio is 10 to 1

    pH 6.8, the ratio is 100 to 1


    Solution to Exercise 7-2-6 is shared under a not declared license and was authored, remixed, and/or curated by .


    Solution to Exercise 7-2-6 is shared under a not declared license and was authored, remixed, and/or curated by Tim Soderberg.

    • Was this article helpful?