How do you calculate concentration of ions in a solution?

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How do you calculate concentration of ions in a solution?

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Answer 1 · 2015-01-25T12:19:28

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:

$N a C l_{(a q)} \to N a_{(a q)}^{+} + C l_{(a q)}^{-}$ $NaCl_((aq)) -> Na_((aq))^(+) + Cl_((aq))^(-)$

Sodium chloride dissociates into $N a^{+}$ $Na^(+)$ cations and $C l^{-}$ $Cl^(-)$ anions when dissolved in water. Notice that 1 mole of $N a C l$ $NaCl$ will produce 1 mole of $N a^{+}$ $Na^(+)$ and 1 mole of $C l^{-}$ $Cl^(-)$ .

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

Let's take another example. Assume you have a $1.0 M$ $"1.0 M"$ $N a_{2} S O_{4}$ $Na_2SO_4$ solution

$N a_{2} S O_{4 (a q)} \to 2 N a_{(a q)}^{+} + S O_{4 (a q)}^{2 -}$ $Na_2SO_(4(aq)) -> 2Na_((aq))^(+) + SO_(4(aq))^(2-)$

Notice that the mole ratio between $N a_{2} S O_{4}$ $Na_2SO_4$ and $N a^{+}$ $Na^(+)$ is $1 : 2$ $1:2$ , which means that 1 mole of the former will produce 2 moles of the latter in solution.

This means that the concentration of the $N a^{+}$ $Na^(+)$ ions will be

$1.0 M \cdot \frac{{2 moles Na}^{+}}{{1 mole Na}_{2} {SO}_{4}} = 2.0 M$ $"1.0 M" * ("2 moles Na"^(+))/("1 mole Na"_2"SO"_4) = "2.0 M"$

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:

$C_{compound} = \frac{n_{Compound}}{V} \Rightarrow V = \frac{n_{compound}}{C_{compound}}$ $C_("compound") = n_("Compound")/V => V = n_("compound")/C_("compound")$

$C_{ion} = \frac{n_{ion}}{V} = n_{ion} \cdot \frac{1}{V} = n_{ion} \cdot \frac{C_{compound}}{n_{compound}}$ $C_("ion") = n_("ion")/V = n_("ion") * 1/V = n_("ion") * C_("compound")/n_("compound")$

$C_{ion} = C_{compound} \cdot \frac{n_{ion}}{n_{compound}}$ $C_("ion") = C_("compound") * n_("ion")/n_("compound")$

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

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How do you calculate concentration of ions in a solution?

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