How do you calculate concentration of ions in a solution?

The of depends on the 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 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