Question

A 0.1000 m aqueous solution of a weak acid, HA, is 1.5% ionized. At what temperature does it freeze? Kf for water = 1.86°C/m.

a. -0.0764°C
b. -0.189°C
c. -0.372°C
d. -0.564°C
e. -0.721°C

Answer 1

To determine the freezing temperature of the 0.1000 m aqueous solution of the weak acid, HA, we need to calculate the freezing point depression. The freezing point depression can be calculated using the equation ΔT = Kf * molality. By substituting the given values, we find that the solution freezes at -0.189 °C. The closest option is -0.189 °C.

Step-by-step explanation:

To determine the temperature at which the 0.1000 m aqueous solution of the weak acid, HA, freezes, we need to calculate the freezing point depression.

The freezing point depression is given by the equation:

ΔT = Kf * molality

First, we need to calculate the molality of the solution.

Since the solution is 1.5% ionized, the concentration of the undissociated acid is (100% - 1.5%) = 98.5% of the total concentration:

molality = (0.985 * 0.1000 m) / (1 kg solvent)

ΔT = (1.86 °C/m) * (0.0985 m)

ΔT = -0.18381 °C

The temperature at which the solution freezes is the freezing point of the pure solvent minus the freezing point depression:

Freezing point = 0 °C - (-0.18381 °C) = 0.1838 °C.

Therefore, the solution freezes at -0.1838 °C.

The closest option is -0.189 °C.

Answer 2

To determine the freezing point of the 0.1000 m aqueous solution of weak acid HA, calculate the molality and the freezing point depression. The correct answer is -0.372°C.

Step-by-step explanation:

To determine the temperature at which the 0.1000 m aqueous solution of the weak acid HA freezes, we need to calculate the freezing point depression using the equation:

AT = -Kf × m

Where AT is the change in temperature, Kf is the freezing point depression constant for water (1.86°C/m), and m is the molality of the solution. Molality (m) is calculated as:

m = moles of solute / mass of solvent (in kg)

In this case, the solution is 1.5% ionized, which means that 1.5% of the acid HA has dissociated into ions. So the concentration of the ionized form (A-) can be calculated as 0.015 × 0.1000 m. The concentration of the undissociated form (HA) is 0.1000 m - (0.015 × 0.1000 m).

Using these concentrations, we can calculate the moles of solute. The mass of the solvent will be approximately equal to its volume (assuming a density of 1 g/mL).

Finally, the molality and freezing point depression can be calculated, giving us the change in temperature.

Calculating the freezing point depression, we get:

AT = -1.86°C/m × (moles of solute / mass of solvent)

Based on the given options, the correct answer is c. -0.372°C.