Come si calcola la concentrazione di ioni in una soluzione?

The concentrazione of ions in solution dipende dal rapporto molare 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^(+)# cationi e #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^(+)# lo sarà #"1.0 M"# and the concentration of the #Cl^(-)# lo sarà #"1.0 M"# come pure.

Let's take another example. Assume you have a #"1.0 M"# #Na_2SO_4# soluzione

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

Notare che la Talpa ratio between #Na_2SO_4# e #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.

Ecco un link ad un'altra risposta su questo argomento:

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