Ion concentrations in solution

Themes > Science > Chemistry > General Chemistry > Solution and Solubility > Solubility of Solutes and Aqueous Solutions > Solution Index > Ion concentrations in solution

The molarity of a solute is the number of moles of that solute divided by the volume of the solvent in liters. The molarity of the ions of that solute in the solution may be different from the molarity of the solute due to the stoichiometry of the reaction between the solute and the solvent.

For example, consider what happens when MgCl₂ dissolves in water

MgCl₂(s) -> Mg⁺²(aq) + 2Cl^-(aq)

If we have 2 moles of MgCl₂, we will form 2 moles of Mg⁺² ion but 4 moles of Cl^- ion. If the volume of the solution is 1.00 L, then the solution is

2 moles MgCl₂/1.00 L = 2 M MgCl₂

2 moles Mg⁺²/1.00 L = 2 M Mg^Cl₂+2

4 moles Cl^-/1.00 L = 4 M Cl^-

It is important to keep track of the amount of the various ions, since if you work solution stoichiometry problems you must start with the correct number of moles.

Since the volume in the above problem is the same before and after the reaction, we can work directly with molarity if we so choose:

2 M MgCl₂ * 2 Cl^-/1 MgCl₂ = 4 M Cl^-

Example: What is the concentration of sodium and sulfide ions in a 0.450 M solution of Na₂S?

Solution: The reaction when sodium sulfide is added to water is

Na₂S(s) -> 2Na⁺(aq) + S^-2(aq)

Thus, if the concentration of Na₂S is 0.450 M, we have

0.450 M Na₂S * 2 Na⁺/1 Na₂S = 0.900 M Na⁺

0.450 M Na₂S * 1 S^-2/1 Na₂S = 0.450 M S^-

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