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Chapter 11: Problem 115

Formic acid \(\left(\mathrm{HCO}_{2} \mathrm{H}\right)\) 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 approximately \(-0.194\) °C and the boiling point is approximately \(100.053\) °C.

Step by step solution

01

Calculate the Van't Hoff Factor

To calculate the Van't Hoff factor (i), we need to consider the percentage of ionization given. Since formic acid is a monoprotic acid, it can ionize to form one hydronium ion (H+) and one formate ion (HCO2-). Therefore, the Van't Hoff factor is calculated as follows: i = 1 + α Where α is the degree of ionization, which is given as 4.2%. We'll convert the percentage to a fractional value. α = 0.042 So, i = 1 + 0.042 = 1.042
02

Calculate the molality from molarity

Since we are given that the molarity and molality of the solution are equal, we can use the given molarity value as the molality: molality = 0.10 mol/kg
03

Calculate the freezing point depression

To calculate the freezing point depression, we will use the formula: ΔT_f = i * K_f * molality Where ΔT_f is the freezing point depression, K_f is the cryoscopic constant of water (1.86 °C kg/mol), and molality from step 2. ΔT_f = 1.042 * 1.860 * 0.10 ΔT_f = 0.19398 °C Now, we can calculate the freezing point of the formic acid solution by subtracting the freezing point depression from the freezing point of pure water (0 °C): Freezing point = 0 - 0.19398 = -0.19398 °C
04

Calculate the boiling point elevation

To calculate the boiling point elevation, we will use the formula: ΔT_b = i * K_b * molality Where ΔT_b is the boiling point elevation, K_b is the ebullioscopic constant of water (0.512 °C kg/mol), and molality from step 2. ΔT_b = 1.042 * 0.512 * 0.10 ΔT_b = 0.053344 °C Now, we can calculate the boiling point of the formic acid solution by adding the boiling point elevation to the boiling point of pure water (100 °C): Boiling point = 100 + 0.0533 = 100.0533 °C So, the freezing point of the 0.10 M formic acid solution is approximately -0.194 °C and the boiling point is approximately 100.053 °C.

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Most popular questions from this chapter

A solution is prepared by mixing \(0.0300 \mathrm{~mol} \mathrm{CH}_{2} \mathrm{Cl}_{2}\) and \(0.0500\) \(\mathrm{mol} \mathrm{CH}_{2} \mathrm{Br}_{2}\) at \(25^{\circ} \mathrm{C}\). Assuming the solution is ideal, calculate the composition of the vapor (in terms of mole fractions) 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.

From the following: pure water solution of \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(m=0.01)\) in water solution of \(\mathrm{NaCl}(m=0.01)\) in water solution of \(\mathrm{CaCl}_{2}(m=0.01)\) in water choose the one with the a. highest freezing point. d. lowest boiling point. b. lowest freezing point. e. highest osmotic pressure. c. highest boiling point.

Consider the following solutions: \(0.010 \mathrm{~m} \mathrm{Na}_{3} \mathrm{PO}_{4}\) in water \(0.020 \mathrm{~m} \mathrm{CaBr}_{2}\) in water \(0.020 \mathrm{~m} \mathrm{KCl}\) in water \(0.020 \mathrm{~m} \mathrm{HF}\) in water \((\mathrm{HF}\) is a weak acid. \()\) a. Assuming complete dissociation of the soluble salts, which solution(s) would have the same boiling point as \(0.040 \mathrm{~m}\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) in water? \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is a nonelectrolyte. b. Which solution would have the highest vapor pressure at \(28^{\circ} \mathrm{C}\) ? c. Which solution would have the largest freezing-point depression?

Patients undergoing an upper gastrointestinal tract laboratory test are typically given an X-ray contrast agent that aids with the radiologic imaging of the anatomy. One such contrast agent is sodium diatrizoate, a nonvolatile water-soluble compound. A \(0.378 \mathrm{~m}\) solution is prepared by dissolving \(38.4 \mathrm{~g}\) sodium diatrizoate (NaDTZ) in \(1.60 \times 10^{2} \mathrm{~mL}\) water at \(31.2^{\circ} \mathrm{C}\) (the density of water at \(31.2^{\circ} \mathrm{C}\) is \(\left.0.995 \mathrm{~g} / \mathrm{mL}\right)\). What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at \(31.2^{\circ} \mathrm{C}\) is \(34.1\) torr?

Using the following information, identify the strong electrolyte whose general formula is $$ \mathrm{M}_{x}(\mathrm{~A})_{y} \cdot z \mathrm{H}_{2} \mathrm{O} $$ Ignore the effect of interionic attractions in the solution. a. \(\mathrm{A}^{n-}\) is a common oxyanion. When \(30.0 \mathrm{mg}\) of the anhydrous sodium salt containing this oxyanion \(\left(\mathrm{Na}_{n} \mathrm{~A}\right.\), where \(n=1,2\), or 3 ) is reduced, \(15.26 \mathrm{~mL}\) of \(0.02313 M\) reducing agent is required to react completely with the \(\mathrm{Na}_{n}\) A present. Assume a \(1: 1\) mole ratio in the reaction. b. The cation is derived from a silvery white metal that is relatively expensive. The metal itself crystallizes in a body-centered cubic unit cell and has an atomic radius of \(198.4 \mathrm{pm}\). The solid, pure metal has a density of \(5.243 \mathrm{~g} / \mathrm{cm}^{3}\). The oxidation number of \(\mathrm{M}\) in the strong electrolyte in question is \(+3\). c. When \(33.45 \mathrm{mg}\) of the compound is present (dissolved) in \(10.0 \mathrm{~mL}\) of aqueous solution at \(25^{\circ} \mathrm{C}\), the solution has an osmotic pressure of 558 torr.
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