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Chapter 10: Problem 127

Formic acid (HCO_A) is a monoprotic acid that ionizes only partially in aqueous solutions. A 0.10-M formic acid solution is \(4.2 \%\) ionized. Assuming that the molarity and molality of the solution are the same, calculate the freezing point and the boiling point of 0.10 \(M\) formic acid.

Short Answer

Expert verified
The freezing point of the 0.10 M formic acid solution is -0.372 °C, and the boiling point is 100.1024 °C.

Step by step solution

01

Calculate the molality of the solution

We are given that the molarity and molality are the same for this solution, so the molality is also 0.10 mol/kg.
02

Determine the molality of the ionized formic acid

Since the solution is 4.2% ionized, we can calculate the molality of the ionized formic acid with the following formula: Molality of ionized formic acid = (molality of formic acid) * (% ionization) = 0.10 mol/kg * 0.042 = 0.0042 mol/kg
03

Calculate the total molality of the solution

Since the solution is partially ionized, both the unionized formic acid and its ions contribute to the changes in freezing and boiling points. We can find the total molality by adding the molality of ionized formic acid to the molality of unionized formic acid: Molality of unionized formic acid = initial molality - molality of ionized formic acid = 0.10 mol/kg - 0.0042 mol/kg = 0.0958 mol/kg Total molality = molality of unionized formic acid + molality of ionized formic acid = 0.0958 mol/kg + 0.0042 mol/kg = 0.1 mol/kg
04

Calculate the freezing point depression and boiling point elevation

We will use the following formulas for calculating freezing point depression and boiling point elevation: ΔTf = Kf * m * i ΔTb = Kb * m * i Where ΔTf is the freezing point depression, ΔTb is the boiling point elevation, Kf is the freezing point depression constant, Kb is the boiling point elevation constant, m is the molality, and i is the van't Hoff factor (which is 2 for formic acid, as it forms 1 H+ ion and 1 HCO2- ion upon ionization) For water, Kf = 1.86 °C kg/mol and Kb = 0.512 °C kg/mol. We have the total molality and van't Hoff factor, so we can plug these values into the formulas: ΔTf = 1.86 °C kg/mol * 0.1 mol/kg * 2 ΔTf = 0.372 °C ΔTb = 0.512 °C kg/mol * 0.1 mol/kg * 2 ΔTb = 0.1024 °C
05

Calculate the freezing and boiling points of the solution

Now that we have the freezing point depression and boiling point elevation, we can find the freezing point and boiling point of the formic acid solution by subtracting ΔTf and adding ΔTb to the normal freezing and boiling points of water: Freezing point = normal freezing point of water - ΔTf = 0 °C - 0.372 °C = -0.372 °C Boiling point = normal boiling point of water + ΔTb = 100 °C + 0.1024 °C = 100.1024 °C Therefore, the freezing point of the 0.10 M formic acid solution is -0.372 °C, and the boiling point is 100.1024 °C.

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Most popular questions from this chapter

a. Calculate the freezing-point depression and osmotic pressure at \(25^{\circ} \mathrm{C}\) of an aqueous solution containing \(1.0 \mathrm{g} / \mathrm{L}\) of a protein (molar mass \(=9.0 \times 10^{4} \mathrm{g} / \mathrm{mol}\) ) if the density of the solution is \(1.0 \mathrm{g} / \mathrm{cm}^{3} .\) b. Considering your answer to part a, which colligative property, freezing- point depression or osmotic pressure, would be better used to determine the molar masses of large molecules? Explain.

Which of the following will have the lowest total vapor pressure at \(25^{\circ} \mathrm{C} ?\) a. pure water (vapor pressure \(=23.8\) torr at \(25^{\circ} \mathrm{C}\) ) b. a solution of glucose in water with \(\chi_{\mathrm{C}_{\mathrm{s}} \mathrm{H}_{\mathrm{l} 2} \mathrm{O}_{\mathrm{s}}}=0.01\) c. a solution of sodium chloride in water with \(\chi_{\mathrm{NaCl}}=0.01\) d. a solution of methanol in water with \(\chi_{\mathrm{CH_{3}}, \mathrm{OH}}=0.2\) (Consider the vapor pressure of both methanol \([143\) torr at \(\left.25^{\circ} \mathrm{C}\right]\) and water.

A solution is prepared by mixing 0.0300 mole of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and 0.0500 mole of \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) at \(25^{\circ} \mathrm{C}\). Assuming the solution is ideal, calculate the composition of the vapor (in terms of molefractions) at \(25^{\circ} \mathrm{C} .\) At \(25^{\circ} \mathrm{C},\) the vapor pressures of pure \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) and pure \(\mathrm{CH}_{2} \mathrm{Br}_{2}\) are 133 and 11.4 torr, respectively.

What mass of sodium oxalate \(\left(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is needed to prepare \(0.250 \mathrm{L}\) of a \(0.100-M\) solution?

What volume of \(0.25 \mathrm{M}\) HCl solution must be diluted to prepare 1.00 L of \(0.040 M\) HCl?
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