Question

Vapour pressure of a saturated solution of a sparingly soluble salt,${A_2}{B_2}$ is $31.8mm\;of\;Hg$ at$40^\circ C$. If vapour pressure of pure water is $31.9mm\;of\;Hg$ at $40^\circ C$. Calculate the ${K_{sp}}$ of ${A_2}{B_2}$ at $40^\circ C$.

Answer 1

Hint: We know, Raoult’s law:
At the same temperature, the vapor pressure of a solvent is equal to the vapor pressure of the pure solvent.
Mathematically expressed as,
${P_{solvent}} = {X_{solvent}}P_{solvent}^0$
Where, x is the mole fraction.
The vapor pressure of the solvent and the solution is $P\& {P^ \circ }$ respectively.

Complete step by step answer:
Now, we see a solubility product.
The equilibrium constant of a solid substance dissolved in aqueous solution is termed as the solubility product ${K_{sp}}$.
The mathematical expression is,
${K_{sp}} = {\left[ C \right]^c}{\left[ D \right]^d}$
To calculate the solubility product the concentration of each ion in molarity is needed.
Let’s consider the dissociation reaction,
${{\text{A}}_{\text{2}}}{{\text{B}}_{\text{2}}}\xrightarrow{{}}{\text{2}}{{\text{A}}^{\text{ + }}}{\text{ + 2}}{{\text{B}}^{\text{ - }}}$
${K_{sp}} = {\left[ {2s} \right]^2}{\left[ {2s} \right]^2}$
${K_{sp}} = 16{s^4}$……..(1)
We have to calculate the number of moles in one liter of water.
The mole can be given by,
${\text{Mole = }}\dfrac{{{\text{weight}}}}{{{\text{Molecular}}\,{\text{weight}}}}$
We know, the molecular weight of water is $18\,g$
${\text{Mole = }}\dfrac{{{\text{1000}}}}{{{\text{18}}}}{\text{ = 55}}{\text{.5}}\,{\text{moles}}$
We can derive a new equation by combining Raoult’s law and Dalton’s law. The equation is,
$X = \dfrac{{{P^ \circ } - P}}{P}$
Where, x is the mole fraction.
Given,
The vapor pressure of the solvent is $31.8mmHg$ .
The vapor pressure of the solution is $31.9mmHg$ .
We use this equation when we have an ideal solution with pure substance with different vapour pressure.
Now, use the modified Raoult’s law,
$\dfrac{{{{\text{P}}^{\text{o}}}{\text{ - P}}}}{{\text{P}}}{\text{ = }}\dfrac{{\text{n}}}{{\text{N}}}$
Now we can substituting the known given values we get,
$\dfrac{{{\text{31}}{\text{.9 - 31}}{\text{.8}}}}{{31.8}}{\text{ = }}\dfrac{{4s}}{{55.55}}$
$s = 0.043moles/liter$
Substitute the value of s in ${K_{sp}}$ formula of equation (1) we get,
$ \Rightarrow {K_{sp}} = 16 \times {\left( {0.043} \right)^4}$
On simplification we get,
${K_{sp}} = 5.47 \times {10^ - }^5$
$\therefore $ The solubility of the substance is $5.47 \times {10^{ - 5}}$ .

Note: We must know that the Raoult’s law is not valid for all types of solution it is only applicable for an ideal gas. Therefore, Mole fraction only gives the number of molecules of a component in the mixture divided by total number of moles in the mixture.
\[{\text{Mole fraction(x)}} = \dfrac{{{\text{Moles of solute}}}}{{{\text{Moles of solute}} + {\text{Moles of solvent}}}}\]