Question

The ${{K}_{sp}}\text{ (2}{{\text{5}}^{\circ }}\text{C)}$ of sparingly soluble salt $X{{Y}_{2}}$ is $3.56\text{ x 1}{{\text{0}}^{-5}}\text{ mo}{{\text{l}}^{3}}{{L}^{-3}}$ and at ${{30}^{\circ }}C$, the vapor pressure of its saturated solution in water is 31.78 mm Hg.
$X{{Y}_{2}}+Aq.\rightleftharpoons {{X}^{2+}}(aq)+2{{Y}^{-}}(aq)$ (100% ionization)
The enthalpy change of the reaction if the vapor pressure of ${{H}_{2}}O$ at ${{30}^{\circ }}C$ is 31.82 mm in KJ (divide by 25 and write an answer to the nearest integer) is __________

Answer 1

We can use many formula like $\dfrac{{{p}^{0}}-{{p}_{s}}}{{{p}_{s}}}=i\dfrac{{{n}_{2}}}{{{n}_{1}}}$ where ${{p}^{0}}$ is the vapor pressure of water, ${{p}_{s}}$ is the vapor pressure of solution, i is ionization constant, and ${{n}_{1}}\text{ and }{{n}_{2}}$ are the number of moles, $\log \dfrac{{{K}_{sp1}}}{{{K}_{sp2}}}=\dfrac{\Delta H}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]$, where ${{K}_{sp}}$ are the solubility products, ${{T}_{2}}\text{ and }{{T}_{1}}$ are the temperatures and $\Delta H$ is the enthalpy change.

Complete step by step solution: We are given the value of solubility product at ${{25}^{\circ }}C$ is $3.56\text{ x 1}{{\text{0}}^{-5}}\text{ mo}{{\text{l}}^{3}}{{L}^{-3}}$.
The value of vapor pressure of water is 31.78 mm Hg that is ${{p}^{0}}$ and the value of vapor pressure of the solution is 31.82 mm Hg that is ${{p}_{s}}$.
The given reaction is:
$X{{Y}_{2}}(s)\rightleftharpoons {{X}^{2+}}+2{{Y}^{-}}$
Before equilibrium, the concentration of $X{{Y}_{2}}$ will be a and the concentrations of the products will be 0. After equilibrium, the concentration of ${{X}^{2+}}$ will be s and the concentration is $2{{Y}^{-}}$ will be 2s. Therefore, total ions forms are 3 and the value of i will be three. We know that:
$\dfrac{{{p}^{0}}-{{p}_{s}}}{{{p}_{s}}}=i\dfrac{{{n}_{2}}}{{{n}_{1}}}$
where ${{p}^{0}}$ is the vapor pressure of water, ${{p}_{s}}$ is the vapor pressure of solution, i is ionization constant, and ${{n}_{1}}\text{ and }{{n}_{2}}$ are the number of moles of solute and solution.
$\dfrac{{{n}_{2}}}{{{n}_{1}}}={{x}_{2}}$
We can write:
$\dfrac{{{p}^{0}}-{{p}_{s}}}{{{p}_{s}}}=i\text{ }{{x}_{2}}$
Putting the values, we get:
$\dfrac{31.82-31.78}{31.78}=3\text{ x }{{x}_{2}}$
${{x}_{2}}=0.004$
Therefore, $\dfrac{{{n}_{2}}}{{{n}_{1}}}=0.004$, in which ${{n}_{1}}$is the number of moles of water and it is equal to 55.55. The value of ${{n}_{2}}$ will be:
$\dfrac{{{n}_{2}}}{55.55}=0.004$
${{n}_{2}}=0.0233$
This is the value of moles of solute and it is the solubility. So, we can write s = 0.0233 mol /L.
The ${{K}_{sp}}$ at ${{30}^{{}^\circ }}C$ will be = $s\text{ x }{{s}^{2}}$
${{K}_{sp}}$ at ${{30}^{{}^\circ }}C$ = $0.0233\text{ x (0}\text{.0233}{{\text{)}}^{2}}$
${{K}_{sp}}$ at ${{30}^{{}^\circ }}C$ = $5.05\text{ x 1}{{\text{0}}^{-5}}\text{ (mol / L}{{\text{)}}^{3}}$
To calculate the change in enthalpy, we use the formula:
$\log \dfrac{{{K}_{sp1}}}{{{K}_{sp2}}}=\dfrac{\Delta H}{2.303R}\left[ \dfrac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]$
Where ${{K}_{sp}}$ are the solubility products, ${{T}_{2}}\text{ and }{{T}_{1}}$ are the temperatures and $\Delta H$ is the enthalpy change.
Now, putting the values, we get:
$\log \dfrac{5.05\text{ x 1}{{\text{0}}^{-5}}}{3.56\text{ x 1}{{\text{0}}^{-5}}}=\dfrac{\Delta H}{2.303\text{ x 8}\text{.314 }}\left[ \dfrac{5}{303\text{ x 298}} \right]$
$\Delta H=52.5\text{ KJ /mol}$
Now divide this value by 25 and the nearest integer value will be 2.

Note: The value of the number of moles of water is taken as 55.55 because the volume of water is 1000 ml and 18 is the molecular mass. The value of temperature must be taken always in Kelvin, not in Celsius.