Is the ionic strength of a solution simply the sum of the concentrations of the reactant and product ions?

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Is the ionic strength of a solution simply the sum of the concentrations of the reactant and product ions?

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Answer 1 · 2017-01-01T05:41:39

[New years answer! :D]

No, ionic strength is not necessarily just the sum of the concentrations of the ions that went into solution, and it is even less accurate if we ignore ion pairing. (Also, the reactants aren't ions; they're the original undissociated electrolyte.)

Ionic strength in terms of molarity is defined as follows:

$I_{c} = \frac{1}{2} \sum i z_{i}^{2} c_{i}$ $I_c = 1/2sum_i z_i^2 c_i$

where:

$z_{i}$ $z_i$ is the charge of the particular ion that came from the strong electrolyte (you can include the sign, but it doesn't matter because it's squared).
$c_{+} = ν_{+} c_{i}$ $c_(+) = nu_(+)c_i$ is the molarity of each cation.
$c_{-} = ν_{-} c_{i}$ $c_(-) = nu_(-)c_i$ is the molarity of each anion.
$c_{i}$ $c_i$ is the given molarity of the particular strong electrolyte $i$ $i$ in the solution.
$ν_{+}$ $nu_(+)$ is the stoichiometric coefficient of the cation.
$ν_{-}$ $nu_(-)$ is the stoichiometric coefficient of the anion.

This particular equation to calculate the ionic strength works best for strong electrolytes of low charge.

The molarity-scale ionic strength for a solution with only one strong electrolyte would be:

$I_{c} = \frac{1}{2} \sum i z_{i}^{2} c_{i}$ $color(blue)(I_c) = 1/2sum_i z_i^2 c_i$

$= \frac{1}{2} [z_{+}^{2} c_{+} + z_{-}^{2} c_{-}]$ $= 1/2[z_(+)^2c_(+) + z_(-)^2c_(-)]$

$= \frac{1}{2} [z_{+}^{2} ν_{+} c_{i} + z_{-}^{2} ν_{-} c_{i}]$ $= 1/2[z_(+)^2nu_(+)c_i + z_(-)^2nu_(-)c_i]$

$= \frac{1}{2} [| z_{+} z_{-} | ν_{+} c_{i} + | z_{+} z_{-} | ν_{-} c_{i}]$ $= 1/2[|z_(+)z_(-)|nu_(+)c_i + |z_(+)z_(-)|nu_(-)c_i]$

$= \frac{1}{2} | z_{+} z_{-} | (ν_{+} + ν_{-}) c_{i}$ $= color(blue)(1/2|z_(+)z_(-)|(nu_(+) + nu_(-))c_i)$

Here's how we might notate the general dissociation of a strong electrolyte:

$M_{ν_{+}} N_{ν_{-}} (a q) \to ν_{+} M_{+}^{z_{+}} (a q) + ν_{-} N_{-}^{z_{-}} (a q)$ $"M"_(nu_(+))"N"_(nu_(-))(aq) -> nu_(+)"M"^(z_(+))(aq) + nu_(-)"N"^(z_(-))(aq)$

Let's say we had a $0.050 M$ $"0.050 M"$ ${NaNO}_{3}$ $"NaNO"_3$ solution. The dissociation reaction is therefore:

${NaNO}_{3} (a q) \to {Na}^{+} (a q) + {NO}_{3}^{-} (a q)$ $"NaNO"_3(aq) -> "Na"^(+)(aq) + "NO"_3^(-)(aq)$

So, for this $0.050 M$ $"0.050 M"$ ${NaNO}_{3} (a q)$ $"NaNO"_3(aq)$ solution, the molarity-scale ionic strength would be:

$I_{c} = \frac{1}{2} ∣ ∣ z_{N a^{+}}^{+} z_{N O_{3}^{-}} ∣ ∣ (ν_{N a^{+}}^{+} + ν_{N O_{3}^{-}}) c_{N a N O_{3}}$ $color(blue)(I_c) = 1/2|z_(Na^(+))z_(NO_3^(-))|(nu_(Na^(+)) + nu_(NO_3^(-)))c_(NaNO_3)$

$= \frac{1}{2} | (1) (- 1) | (1 + 1) (0.050 M)$ $= 1/2|(1)(-1)|(1 + 1)("0.050 M")$

$=$ $=$ $0.050 M$ $color(blue)("0.050 M")$

Now, let's say we had a solution of $0.050 M$ $"0.050 M"$ ${Ca}_{3} {({PO}_{4})}_{2}$ $"Ca"_3("PO"_4)_2$ , which dissociates like this:

${Ca}_{3} {({PO}_{4})}_{2} (a q) \to 3 {Ca}^{2 +} (a q) + 2 {PO}_{4}^{3 -} (a q)$ $"Ca"_3("PO"_4)_2(aq) -> 3"Ca"^(2+)(aq) + 2"PO"_4^(3-)(aq)$

Then, when ignoring ion pairing, we would instead get:

$I_{c} = \frac{1}{2} ∣ ∣ z_{C a^{2 +}}^{2 +} z_{P O_{4}^{3 -}} ∣ ∣ (ν_{C a^{2 +}}^{2 +} + ν_{P O_{4}^{3 -}}) c_{C a_{3} {(P O_{4})}_{2}}$ $color(blue)(I_c) = 1/2|z_(Ca^(2+))z_(PO_4^(3-))|(nu_(Ca^(2+)) + nu_(PO_4^(3-)))c_(Ca_3(PO_4)_2)$

$= \frac{1}{2} | (2) (- 3) | (3 + 2) (0.050 M)$ $= 1/2|(2)(-3)|(3 + 2)("0.050 M")$

$=$ $=$ $0.75 M$ $color(blue)("0.75 M")$

In reality, there would be some ion pairing in this solution since the charges are larger than $1$ $1$ , so this ionic strength is not entirely accurate. The equations get very complicated, so I won't go into it...

But the point is, having more ion pairing decreases the ionic strength, because the ions associate a bit and form other charged species in solution.

So, this particular equation to calculate the ionic strength works best for strong electrolytes of low charge.

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Is the ionic strength of a solution simply the sum of the concentrations of the reactant and product ions?

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