Chemistry - Ionic Equilibrium Question with Solution | TestHub

The lowering of vapor pressure ($\Delta P$) is 0.108 torr. The vapor pressure of pure solvent ($P^{\circ }$) is 300 torr.

Relative lowering of vapor pressure:

$$\frac{\Delta P}{P^{\circ }}=\frac{0.108}{300}=0.00036$$

For a dilute solution, the relative lowering of vapor pressure is approximately equal to the mole fraction of the solute.

For salt AB, which dissociates into A⁺ and B⁻, the van't Hoff factor $i=2$.

Let $s$ be the molar solubility of AB. The total moles of ions will be $2s$.

$$\frac{\Delta P}{P^{\circ }}=i\times molefractionofsolute=i\times \frac{n_{solute}}{n_{solvent}}$$

Assuming the solution is dilute, we can approximate the mole fraction of solute as $\frac{n_{solute}}{n_{solvent}}$.

Since the problem implies a dilute aqueous solution, we can assume the molarity is approximately equal to the molality, and the number of moles of solvent is large.

The mole fraction of solute is approximately $\frac{2s}{55.5}$ (where 55.5 is moles of water in 1 kg).

However, a more direct approach using Raoult's Law for relative lowering of vapor pressure for electrolytes is:

$$\frac{\Delta P}{P^{\circ }}=i\times X_{solute}$$

where $X_{solute}$ is the mole fraction of the solute particles.

For a dilute solution, $X_{solute}\approx \frac{n_{solute}}{n_{solvent}}$.

Given $\frac{\Delta P}{P^{\circ }}=0.00036$ and $i=2$.

So, $X_{solute}=\frac{0.00036}{2}=0.00018$.

This $X_{solute}$ represents the mole fraction of the dissolved salt AB.

For a dilute aqueous solution, the molality ($m$) is approximately equal to the molarity ($s$).

The mole fraction of solute can also be related to molality:

$$X_{solute}=\frac{m}{m+\frac{1000}{M_{solvent}}}$$

For water, $M_{solvent}=18g/mol$, so $\frac{1000}{18}\approx 55.5$.

$$X_{solute}\approx \frac{m}{55.5}$$

So, $m=X_{solute}\times 55.5=0.00018\times 55.5\approx 0.01mol/kg$.

Since the solution is dilute, we can assume molarity $s\approx 0.01M$.

For the sparingly soluble salt AB:

$$AB(s)\rightleftharpoons A^{+}(aq)+B^{-}(aq)$$

The solubility product $K_{sp}$ is given by:

$$K_{sp}=[A^{+}][B^{-}]$$

If $s$ is the molar solubility of AB, then $[A^{+}]=s$ and $[B^{-}]=s$.

$$K_{sp}=s\times s=s^2$$

Substituting $s=0.01$:

$$K_{sp}=(0.01{)}^2=(10^{-2}{)}^2=10^{-4}$$

The final answer is $\boxed{10^{-4}}$.