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The freezing point of \(0.10 \mathrm{M} \mathrm{KHSO}_{3}\) is \(-0.38^{\circ} \mathrm{C}\). Which of the following equations best represents what happens when \(\mathrm{KHSO}_{3}\) dissolves in water? (a) \(\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{KHSO}_{3}(a q)\) (b) \(\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{HSO}^{3-}(a q)\) (c) \(\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{SO}_{3}{ }^{2-}(a q)+\mathrm{H}^{+}(a q)\)

Short Answer

Expert verified
Answer: The correct equation representing the dissolution of KHSO₃ in water is KHSO₃(s) → K⁺(aq) + HSO₃⁻(aq).

Step by step solution

01

Calculate the van 't Hoff factor

We'll start by finding the expected van 't Hoff factor for the given freezing point depression. The formula for freezing point depression is: \(\Delta T = iK_{f} m\) Where \(\Delta T\) is the change in freezing point, \(i\) is the van 't Hoff factor, \(K_{f}\) is the freezing point depression constant of water (\(1.86\, ^\circ\mathrm{C \cdot kg/mol}\)), and \(m\) is molality. We're given the freezing point depression (\(\Delta T = 0.38\, ^\circ \mathrm{C}\)) and the molality (\(m = 0.10 \, \mathrm{mol/kg}\)). Thus, we have: \(0.38 = i(1.86)(0.10)\) Now we'll solve for the van 't Hoff factor (i): \(i = \frac{0.38}{(1.86)(0.10)} = 2.04 \approx 2\)
02

Analyze each given equation

Now that we have the expected van 't Hoff factor, we will analyze each equation given and determine the number of ions produced in each equation to find the one that matches the calculated van 't Hoff factor of 2. (a) \(\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{KHSO}_{3}(a q)\) This equation produces only one ion: \(\mathrm{KHSO}_{3}\). Van 't Hoff factor = 1 (doesn't match) (b) \(\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{HSO}^{3-}(a q)\) This equation produces two ions: \(\mathrm{K}^{+}\) and \(\mathrm{HSO}^{3-}\). Van 't Hoff factor = 2 (matches) (c) \(\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{SO}_{3}{ }^{2-}(a q)+\mathrm{H}^{+}(a q)\) This equation produces three ions: \(\mathrm{K}^{+}\), \(\mathrm{SO}_{3}{ }^{2-}\), and \(\mathrm{H}^{+}\). Van 't Hoff factor = 3 (doesn't match)
03

Determine the correct equation

Comparing the van 't Hoff factors of each given equation, we find that only equation (b) has a matching van 't Hoff factor of 2: \(\mathrm{KHSO}_{3}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{HSO}^{3-}(a q)\) Thus, the correct equation representing the dissolution of \(\mathrm{KHSO}_{3}\) in water is given by option (b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

van 't Hoff factor
Understanding the van 't Hoff factor is crucial when delving into the properties of solutions, particularly with freezing point depression. This factor, represented by the symbol i, tells us how the presence of solute particles affects the colligative properties of a solution. In simple terms, it reflects the number of particles a compound dissociates into when it dissolves. For example, if a substance breaks into three ions upon dissolution, the van 't Hoff factor would be 3.

The impact of this factor on the freezing point can be significant - the more particles that are released into the solution, the more it will interfere with the formation of ice crystals, leading to a lower freezing point. This effect is captured by the formula for freezing point depression:
\[\Delta T = iK_{f} m\]
Here, \(\Delta T\) represents the amount by which the freezing point is lowered, K_{f} is a constant specific to the solvent (for water, it's 1.86 oC⋅kg/mol), and m is molality. By rearranging this formula and using the known values of \(\Delta T\) and m, we can solve for the van 't Hoff factor, helping us understand the behavior of our solute in solution.
solubility of ionic compounds
The solubility of ionic compounds plays a pivotal part in determining the behavior of salts and similar substances in water. Solubility refers to the maximum amount of solute that can dissolve in a solvent at a given temperature and pressure. Ionic compounds dissociate into ions when they dissolve, which is a process governed by factors such as the lattice energy of the solid and the hydration energy of the ions.

When an ionic compound dissolves in water, we expect it to separate into its constituent ions. Taking the example in our exercise, potassium hydrogen sulfate (KHSO3) dissolves into potassium ions (K+) and hydrogen sulfate ions (HSO3). This process of dissolution is crucial when calculating the van 't Hoff factor as it ultimately influences the extent of freezing point depression. The greater the solubility, the more ions available to disrupt the freezing process, leading to a larger depression of the freezing point.
colligative properties
Last but not least, let's explore the concept of colligative properties. These properties depend solely on the ratio of solute particles to solvent molecules in a solution, rather than the identity of the solute. They include phenomena like vapor pressure lowering, boiling point elevation, osmotic pressure, and, pertinent to our exercise, freezing point depression.

Freezing point depression is particularly interesting because it tells us that the more particles we dissolve in a liquid (like water), the lower the temperature has to be for the liquid to freeze. This is why we use salt to melt ice on roads in the winter: the salt dissolves, increasing the number of particles in the water and thus lowering its freezing point. It's important to understand that freezing point depression is a colligative property because it exemplifies how properties of solutions can vary based on the quantity, not the type, of solute particles present. In a way, colligative properties are like democracy in solutions: every particle gets an equal vote in defining the properties, no matter what element or compound it comes from.

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Most popular questions from this chapter

Acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}\), is the main ingredient of nail polish remover. A solution is made up by adding \(35.0 \mathrm{~mL}\) of acetone \((d=0.790 \mathrm{~g} / \mathrm{mL})\) to \(50.0 \mathrm{~mL}\) of ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}(d=0.789 \mathrm{~g} / \mathrm{mL})\). Assuming volumes are additive, calculate (a) the mass percent of acetone in the solution. (b) the volume percent of ethyl alcohol in the solution. (c) the mole fraction of acetone in the solution.

When water is added to a mixture of aluminum metal and sodium hydroxide, hydrogen gas is produced. This is the reaction used in commercial drain cleaners: $$ 2 \mathrm{Al}(s)+6 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{Al}(\mathrm{OH})^{4-}(a q)+3 \mathrm{H}_{2}(g) $$ A sufficient amount of water is added to \(49.92 \mathrm{~g}\) of \(\mathrm{NaOH}\) to make \(0.600 \mathrm{~L}\) of solution; \(41.28 \mathrm{~g}\) of \(\mathrm{Al}\) is added to this solution and hydrogen gas is formed. (a) Calculate the molarity of the initial \(\mathrm{NaOH}\) solution. (b) How many moles of hydrogen were formed? (c) The hydrogen was collected over water at \(25^{\circ} \mathrm{C}\) and \(758.6 \mathrm{~mm} \mathrm{Hg}\). The vapor pressure of water at this temperature is \(23.8 \mathrm{~mm} \mathrm{Hg}\). What volume of hydrogen was generated?

A certain gaseous solute dissolves in water, evolving \(12.0 \mathrm{~kJ}\) of heat. Its solubility at \(25^{\circ} \mathrm{C}\) and \(4.00\) atm is \(0.0200 M .\) Would you expect the solubility to be greater or less than \(0.0200 M\) at (a) \(5^{\circ} \mathrm{C}\) and 6 atm? (b) \(50^{\circ} \mathrm{C}\) and 2 atm? (c) \(20^{\circ} \mathrm{C}\) and 4 atm? (d) \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm} ?\)

The freezing point of \(0.20 \mathrm{~m}\) HF is \(-0.38^{\circ} \mathrm{C}\). Is HF primarily nonionized in this solution (HF molecules), or is it dissociated to \(\mathrm{H}^{+}\) and \(\mathrm{F}^{-}\) ions?

What is the osmotic pressure of a \(0.250 \mathrm{M}\) solution of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) at \(25^{\circ} \mathrm{C}\) ? (Assume complete dissociation.)

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