Question

What's the solubility (in grams per liter) of $La{{F}_{3}}$ in pure water?

Answer 1

We know that to solve this question we have to first write the dissociation of salt into its two individual ions. Then find the concentration of the ions in terms of molar solubility. Calculate the solubility product using the concentration of ions.

Complete answer:
In order to solve this problem, you would need the value of the solubility product constant, Ksp, for lanthanum trifluoride, $La{{F}_{3}}$which is usually given to you with the problem. In this case, We'll pick
$Ksp=2.0\times {{10}^{-19}}.$Since you're dealing with an insoluble ionic compound, equilibrium will be established between the undissolved solid and the dissolved ions.

| $La{{F}_{3\left( s \right)}}$| $La_{\left( aq \right)}^{3+}$ $+$| $3F_{\left( aq \right)}^{-}$
---|---|---|---
I| \\_| $0$| $0$
C| \\_| $(+s)$| $(+3s)$
E| \\_| $s$| $3s$

Initially, the concentrations of the $L{{a}^{3+~}}$ and ${{F}^{-~}}$ ions are equal to zero; the solid was not yet placed in water. Keep in mind that the solid's concentration is presumed to be either unknown or constant, which is why it's not relevant here. By definition, the solubility product constant for this equilibrium will be $Ksp=\left[ L{{a}^{3+}} \right]\cdot {{\left[ {{F}^{-}} \right]}^{3}}$ This will be equivalent to $Ksp=s\cdot {{\left( 3s \right)}^{3}}$
Substituting the values in equation; $\Rightarrow 2.0\cdot {{10}^{-19}}=27{{s}^{4}}$ , You will thus have
$s=\sqrt{\dfrac{2.0\cdot {{10}^{-19}}}{27}}=9.3\cdot {{10}^{-6}}$Since s represents the molar solubility of the salt, i.e. how many moles of lanthanum trifluoride can be dissolved in a liter of water, you will have $s=9.3\cdot {{10}^{-6}}mol\text{ }{{L}^{-}}$
In order to express the solubility in grams per liter, $g\text{ }{{L}^{-1}}$, use lanthanum trifluoride molar mass
$\left( 9.3\cdot {{10}^{-6}}\dfrac{mol}{L} \right)\cdot \left( \dfrac{195.9g}{1mol} \right)=1.8\cdot {{10}^{-3}}g\text{ }{{L}^{-1}}$

Note:
Remember that the solubility product is calculated using the concentration of the ions in which the salt has dissociated. Solubility factor depends on various factors such as temperature, pressure and nature of the electrolyte. The concentration of ions is affected by these factors and thus, the solubility product gets affected.