# Question d6d89

Sep 28, 2015

${m}_{\pm} = \text{0.211 molal}$
$I = 0.012$

#### Explanation:

So, lanthanum(III) bicarbonate, "La"("HCO"_3)_3, is an ionic compound that dissociates to form lanthanum(III) cations, ${\text{La}}^{3 +}$, and bicarbonate anions, ${\text{HCO}}_{3}^{-}$

${\text{La"("HCO"_3)_text(3(s]) -> "La"_text((aq])^(3+) + 3"HCO}}_{\textrm{3 \left(a q\right]}}^{-}$

Now, the mean molality of the solution can be thought of as the average molality of the two ions formed when the compound dissociates

${m}_{\pm} = {\left[{m}_{+}^{{\nu}_{+}} + {m}_{-}^{{\nu}_{-}}\right]}^{\frac{1}{\nu}} \text{ }$, where

${m}_{+}$ - the molality of the cations;
${m}_{-}$ - th molality of the anions;
${\nu}_{+}$ - the number of moles of cations produced in solution;
${\nu}_{-}$ - the number of moles of anions produced in solution;
$\nu$ - the total number of moles of ions.

Now, because no information was given about the mass of solvent, I will assume that the concentration you provided is actually 0.002 molal, not molar.

So, every formula unit of lanthanum(III) bicarbonate produces one lanthanum(III) cation and three bicarbonate anions, so you know that

${\nu}_{+} = 1 \text{ }$ $\text{ "nu_(-) = 3" }$ $\text{ } \nu = {v}_{+} + {\nu}_{-} = 4$

This means that the mean molality of the solution will be

m_(pm) = [(0.002)^1 + (3 xx 0.002)""^3]^(1/4)#

${m}_{\pm} = \left(0.002 + 0.006 \text{^3)""^(1/4) = color(green)("0.211 molal}\right)$

The ionic strenght of the solution is defined as

$I = \frac{1}{2} \cdot {\sum}_{i} \left({m}_{i} \cdot {z}_{i}^{2}\right) \text{ }$, where

${m}_{i}$ - the molality of the ion;
${z}_{i}$ - the charge of the ion;

In your case, the ionic strenght of the solution will be

$I = \frac{1}{2} \cdot \left[0.002 \cdot {\left(\text{3+")^(2) + 0.006 * ("1-}\right)}^{2}\right]$

$I = \frac{1}{2} \cdot \left(0.018 + 0.006\right) = \textcolor{g r e e n}{0.012}$