Question

How do you calculate concentration of ions in a solution?

Answer 1

The concentration of ions in solution depends on the mole ratio between the dissolved substance and the cations and anions it forms in solution.

So, if you have a compound that dissociates into cations and anions, the minimum concentration of each of those two products will be equal to the concentration of the original compound. Here's how that works:

`NaCl_((aq)) -> Na_((aq))^(+) + Cl_((aq))^(-)`

Sodium chloride dissociates into `Na^(+)` cations and `Cl^(-)` anions when dissolved in water. Notice that 1 mole of `NaCl` will produce 1 mole of `Na^(+)` and 1 mole of `Cl^(-)`.

This means that if you have a `NaCl` solution with a concentration of `"1.0 M"`, the concentration of the `Na^(+)` ion will be `"1.0 M"` and the concentration of the `Cl^(-)` ion will be `"1.0 M"` as well.

Let's take another example. Assume you have a `"1.0 M"` `Na_2SO_4` solution

`Na_2SO_(4(aq)) -> 2Na_((aq))^(+) + SO_(4(aq))^(2-)`

Notice that the mole ratio between `Na_2SO_4` and `Na^(+)` is `1:2`, which means that 1 mole of the former will produce 2 moles of the latter in solution.

This means that the concentration of the `Na^(+)` ions will be

`"1.0 M" * ("2 moles Na"^(+))/("1 mole Na"_2"SO"_4) = "2.0 M"`

Think of it like this: the volume of the solution remains constant, but the number of moles doubles; automatically, this implies that the concentration will be two times bigger for that respective ion.

Here's how that would look mathematically:

`C_("compound") = n_("Compound")/V => V = n_("compound")/C_("compound")`

`C_("ion") = n_("ion")/V = n_("ion") * 1/V = n_("ion") * C_("compound")/n_("compound")`

`C_("ion") = C_("compound") * n_("ion")/n_("compound")`

As you can see, the mole ratio between the original coumpound and an ion it forms will determine the concetration of the respective ion in solution.

Here's a link to another answer on this topic:

http://socratic.org/questions/how-do-you-calculate-the-number-of-ions-in-a-solution?source=search