In the previous post, we talked about molarity which is the ratio of the moles of solute over the volume of the solution.
For example, if we dissolve 43.6 g K2SO4 in 1 L of water, the molarity of the salt is:
\[{\rm{n}}\;{\rm{(}}{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{)}}\;{\rm{ = }}\;{\rm{43}}{\rm{.6}}\;{\rm{g}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{174}}{\rm{.3}}\;{\rm{g}}}}{\rm{ = }}\;{\rm{0}}{\rm{.250}}\;{\rm{mol}}\]
\[{\rm{M}}\,{\rm{(}}{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{) = }}\;\frac{{\rm{n}}}{{\rm{V}}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.250}}\;{\rm{mol}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.250}}\,M\]
However, we know that when a strong electrolyte is dissolved in water, it dissociates into ions and the salt does not exist in the solution in its molecular form.
The concentration of the ions can be calculated from the concentration of the salt. For this, we need to identify how many of each ion appears in one molecule of the salt. So, for K2SO4, there are two K+ and one SO42- ion. You can also see this by writing the dissociation equation:
K2SO4(aq) → 2K+(aq) + SO42-(aq)
Therefore, 0.250 M K2SO4 will produce two times more K+ ions and the same concentration of SO42- ions. The total concentration of all the ions will be 0.500 M K+ and 0.250 M SO42- = 0.750 M.
The concentration of an ion or the total ionic concentration can also be done using molar conversions.
For the concentration of K+ ions, we will have:
\[{\rm{M}}\;{\rm{(}}{{\rm{K}}^{\rm{ + }}}{\rm{)}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.250}}\;\cancel{{{\rm{mol}}\;{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ \times }}\,\frac{{{\rm{2}}\;{\rm{mol}}\;{{\rm{K}}^{\rm{ + }}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}\, = \;0.500\;M\]
For the total concentration of all ions:
\[{\rm{M}}\;{\rm{(Ions)}}\;{\rm{ = }}\;\frac{{{\rm{0}}{\rm{.250}}\;\cancel{{{\rm{mol}}\;{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}{{{\rm{1}}\;{\rm{L}}}}\;{\rm{ \times }}\,\frac{{{\rm{3}}\;{\rm{mol}}\;{\rm{ions}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\rm{K}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}}}\, = \;0.750\;M\]
Check Also
- Solutions
- Strong and Weak Electrolytes
- Dissociation of Ionic Compounds
- Molecular, Ionic, and Net Ionic Equations
- Molarity
- Dilution
- Precipitation Reactions
- Definitions of Acids and Bases
- Acid-Base Reactions
- Stoichiometry of Reactions in Aqueous Solutions
- Acid-Base Titrations
- Oxidation State
- Oxidation-Reduction (Redox) Reactions
- Reactions in Aqueous Solutions Practice Problems
Practice
What is the total concentration of ions in a solution prepared by dissolving 60.08 g of AlCl3 in enough water to make 600. mL of solution?
How many potassium ions are present in 350. mL of 0.150 M K3PO4 solution? Consider K3PO4 as a strong electrolyte and ignore the reaction of phosphate ion with water. The molar mass of potassium phosphate is 212.3 g/mol.
Calculate the concentration of each ion and their total concentration in a solution prepared by dissolving 0.2500 mol of Ca(NO3)2 in 150.0 mL of water.