Question

A $0.100-\mathrm{g}$ sample of the weak acid $\mathrm{HA}$ (molar mass $=$ 100.0 $\mathrm{g}/\mathrm{mol})$ is dissolved in 500.0 $\mathrm{g}$ water. The freezing point of the resulting solution is $-{0.0056}^{\circ }\mathrm{C}$ . Calculate the value of $K_{\mathrm{a}}$ for this acid. Assume molality equals molarity in this solution.

Short answer

The value of $K_a$ for the weak acid is approximately 1.1 × 10^-4.

Step-by-step solution

Calculate molality of the solution

First, we need to determine the number of moles of the weak acid (HA) in the solution by using the given mass and molar mass: n(HA) = mass of HA / molar mass of HA = $0.100\mathrm{g}/100.0\mathrm{g}·{\mathrm{mol}}^{-1}$ = $0.00100\mathrm{mol}$ Now we will determine the molality of the solution, using the mass of water, which assuming molality equals molarity: molality(HA) = moles of HA / mass of water (in kg) = $0.00100\mathrm{mol}/0.500\mathrm{kg}$ = $0.00200\mathrm{mol}·{\mathrm{kg}}^{-1}$

Find the molality of the solute

Next, we will use the freezing point depression formula to find the molality of the solute: ΔT = Kf × m × i where ΔT is the freezing point depression, Kf is the cryoscopic constant for water (1.86 °C·kg/mol), m is molality, and i is the van't Hoff factor. In this problem, i = 2, because the weak acid dissociates into two ions: H+ and A-. Rearrange the formula to find the molality of the solute: m = ΔT / (Kf × i) = $-0.0056°\mathrm{C}/(1.86°\mathrm{C}·\mathrm{kg}·{\mathrm{mol}}^{-1}\times 2)$ = $0.001505\mathrm{mol}·{\mathrm{kg}}^{-1}$

Set up the ion concentration ratios

Now that we have the molality of the solute, we can set up equilibrium expressions for the dissociation of the weak acid into ions: $HA\rightleftharpoons H^{+}+A^{-}$ The equilibrium concentrations for the ions should be the same, assuming complete dissociation of the weak acid: $$H^{+}=A^{-}=x$$ While the concentration of the weak acid is: $$HA=0.00200-x$$

Calculate Ka using equilibrium concentrations

Finally, we will set up an equilibrium expression for the acid dissociation reaction and solve for Ka: $$K_a=\frac{[H^{+}][A^{-}]}{[HA]}$$ Substitute the equilibrium concentrations into the equation: $$K_a=\frac{x^2}{0.00200-x}$$ Since the value of x (concentration of the ions) is relatively small and can be neglected when compared to 0.00200, we can simplify the equation: $$K_a\approx \frac{x^2}{0.00200}$$ Now, substitute the molality of the solute (0.001505 mol·kg^-1) for x in the equation: $$K_a\approx \frac{(0.001505{)}^2}{0.00200}=1.1\times {10}^{-4}$$ Thus, the value of Ka for this weak acid is approximately 1.1 × 10^-4.

Key concepts

Freezing Point Depression

Freezing point depression is a colligative property of solutions, which means it depends on the number of solute particles in a solvent, not their identity. When a solute is dissolved in a solvent, the freezing point of the solution becomes lower than that of the pure solvent. This phenomenon occurs because the presence of solute particles disrupts the formation of the solid phase, requiring a lower temperature to freeze.

The formula for freezing point depression is $$\mathrm{\Delta }{T}_f=K_f\cdot m\cdot i$$ where:

• $\mathrm{\Delta }{T}_f$ is the change in freezing point,
• $K_f$ is the cryoscopic constant specific to each solvent, and for water, it is 1.86 °C·kg/mol,
• $m$ is the molality of the solution,
• $i$ is the van’t Hoff factor, which accounts for the number of particles the solute dissociates into.

In the case of weak acids like HA, they partially dissociate in solution, so the calculation of $i$ depends on the extent of dissociation. It's crucial to understand this concept when solving problems related to freezing point depression.

Molality and Molarity

Molality and molarity are important concentration units used in chemistry. Understanding the difference between these terms is essential, especially when dealing with colligative properties like freezing point depression.

Molality $(m)$ is defined as the moles of solute per kilogram of solvent. Its formula is:$$m=\frac{\text{moles of solute}}{\text{mass of solvent in kg}}$$Unlike molarity, molality does not change with temperature since it only depends on mass. It’s especially useful in freezing or boiling point calculations.

Molarity $(M)$ is defined as the moles of solute per liter of solution:$$M=\frac{\text{moles of solute}}{\text{volume of solution in L}}$$Since molarity depends on volume, it can change with temperature variations as the solution expands or contracts.

• Molality is more commonly used in scenarios where temperature change is involved, such as in freezing point depression.
• In dilute solutions, particularly aqueous ones, molality and molarity can be assumed to be approximately the same.

Weak Acid Solutions

Weak acids are only partially ionized in water, which makes their behavior in solutions more complex than strong acids. When a weak acid like $\text{HA}$ dissociates, it forms ${\text{H}}^{+}$ and ${\text{A}}^{-}$ ions with $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$. This incomplete dissociation creates a dynamic equilibrium between the molecules and ions.

The acid dissociation constant $K_a$ is used to quantify the extent of dissociation. It is defined as:$$K_a=\frac{[{\text{H}}^{+}][{\text{A}}^{-}]}{[\text{HA}]}$$where square brackets denote concentration.

• A larger $K_a$ value indicates a stronger acid with more dissociation.
• A smaller $K_a$ value signifies a weaker acid with less dissociation.

Calculating $K_a$ from experimental data involves measuring changes due to partial dissociation, such as effects on freezing points. This requires setting up equilibrium expressions and assuming negligible ion concentrations for simplification when necessary.

For instance, in the solution given, by calculating the decrease in freezing point and using the initial molality, you can estimate the $K_a$ of the weak acid.