2.5 The Concentration of Ions in Solutions

Last updated
Save as PDF

Page ID: 32265

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

The solutions we work with in chemistry often involve ionic compounds and acids. Both of these produce electrolytic solutions, meaning they conduct an electrical current due to the production of ions in solution.

It will often be important for us to be able to determine the concentration of these ions, not just the overall concentration of the solution.

To begin, it will be important for you to be able to write balanced reactions that show how these substances break down into ions. If you are given the the formula of the compound (either ionic or acid) you will need to be able to determine:

which ions are produced and
in what mole ratio.

For example, sodium carbonate, Na₂CO₃ dissociates into ions as:

Na₂CO_{3 (s)} → 2 Na⁺_(aq)+ CO₃^2-_(aq)

Notice that two mole of Na⁺ ions are produced for every one mole of Na₂CO₃. The total volume of the solution, however, remains unchanged.

If we have a 0.20 M solution of Na₂CO₃, what will be the concentration of our two ions, Na⁺ and CO₃^2-? This can easily be determined from the coefficients in our balanced equation:

[Na⁺] = 2 × [Na₂CO₃] = 2 × 0.20 M = 0.40 M

[CO₃^2-] = 1 × [Na₂CO₃] = 1 × 0.20 M = 0.20 M