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What is the boiling point of a \(0.10 \mathrm{M}\) solution of \(\mathrm{NaHSO}_{4}\) if the solution has a density of \(1.002 \mathrm{~g} / \mathrm{mL}\) ?

Short Answer

Expert verified
The boiling point of the 0.10 M NaHSO4 solution is approximately 100.1 掳C.

Step by step solution

01

Calculate the molality of the solution

We are given the molarity of NaHSO4 solution, which is 0.10 M. To convert it to molality, we need the mass of the solvent. We can calculate it using the density formula and considering a 1-liter solution. Since the density of the solution is given as 1.002 g/mL or 1002 g/L, we can find the mass of the solute and solvent as follows: Mass of 1 liter solution = 1002 g Since Molarity = moles of solute / volume of solution in liters, 0.10 M NaHSO4 solution contains 0.10 moles of NaHSO4 in 1 liter of solution. Mass of solute (NaHSO4) = moles * molar mass = 0.10 moles * 120 g/mol (rounded) Mass of solute (NaHSO4) = 12 g Now we can calculate the mass of the solvent (water): Mass of solvent (water) = Mass of solution - Mass of solute Mass of solvent (water) = 1002 g - 12 g = 990 g Molality = moles of solute / mass of solvent in kg Molality = 0.10 moles / (990 g * 1 kg/1000 g) = 0.101 mol/kg
02

Calculate the molal boiling point elevation constant (Kb) for water

The molal boiling point elevation constant (Kb) for water is generally given and has a value of 0.512 掳C/mol/kg. We can use this value in the boiling point elevation formula.
03

Apply the boiling point elevation formula

The boiling point elevation formula is given by: 螖Tb = Kb 脳 molality 脳 i where 螖Tb = boiling point elevation Kb = molal boiling point elevation constant of solvent molality = molality of the solution i = Van't Hoff factor (number of particles produced by solute in solution) We know that NaHSO4 will dissociate into Na+ and HSO4- ions. So, the Van't Hoff factor (i) will be 2. Now we can calculate the boiling point elevation: 螖Tb = 0.512 掳C/mol/kg 脳 0.101 mol/kg 脳 2 = 0.1035 掳C
04

Calculate the boiling point of the solution

Since we have calculated the boiling point elevation, we can now find the boiling point of the NaHSO4 solution. The normal boiling point of pure water is 100 掳C. Adding the boiling point elevation, we get the boiling point of the solution: Boiling point of solution = normal boiling point of water + 螖Tb Boiling point of solution = 100 掳C + 0.1035 掳C 鈮 100.1 掳C Therefore, the boiling point of the 0.10 M NaHSO4 solution is approximately 100.1 掳C.

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Key Concepts

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

Molality
Molality is a measure of the concentration of a solute in a solution. It's expressed as moles of solute per kilogram of solvent. This is different from molarity, which is moles of solute per liter of solution. To understand molality, it's important to recognize why we use it, especially in boiling point elevation problems. It's useful because it depends only on the amount of solute and solvent, not the temperature or pressure, providing a very stable measure of concentration.
To calculate molality, you need:
  • Moles of solute: This is given by the amount of solute, often found using its chemical formula and molar mass.
  • Mass of solvent in kilograms: This might require converting from grams, making sure to subtract any solute mass from the total solution mass.
For example, if you dissolve NaHSO鈧 in water, you find the number of moles of NaHSO鈧 and the mass of water in kilograms, then divide the two to find molality. In the provided exercise, the molality of NaHSO鈧 was found to be 0.101 mol/kg, showing how these calculations come about in real-world problems.
Van't Hoff Factor
The Van't Hoff factor, commonly designated as "i", indicates the number of particles a solute generates when it dissolves. It's crucial in colligative properties, such as boiling point elevation and freezing point depression, because these depend on the number of particles in a solution, not the type of particles.
For electrolytes like NaHSO鈧, which dissociate in solution, the Van't Hoff factor helps us understand how many ions are produced. NaHSO鈧 dissociates into two ions: Na鈦 and HSO鈧勨伝. Thus, the Van't Hoff factor for NaHSO鈧 is 2.
The Van't Hoff factor plays a significant role in the formula for boiling point elevation:
  • 螖T鈧 = Kb 脳 molality 脳 i
Here, "i" directly affects the boiling point elevation. The more particles produced, the higher the boiling point elevation. Understanding this concept is essential for applying colligative properties in solutions where electrolyte dissociation occurs.
Molarity
Molarity is another way to express the concentration of a solution, often denoted as M. It is the number of moles of solute per liter of solution. In contrast to molality, molarity depends on the entire volume of the solution, which can vary with temperature and pressure.
Calculating molarity involves knowing:
  • The amount of solute in moles: Usually derived from the solute's weight and molar mass.
  • The volume of the solution in liters: This includes both solute and solvent.
For the NaHSO鈧 solution in the example, the initial concentration was given as 0.10 M. Molarity was important here as a starting point before converting to molality for the boiling point elevation calculation. While molarity is easy to calculate and widely used, it can be less precise than molality in certain scientific calculations, especially where temperature changes are involved.

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Most popular questions from this chapter

Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid, or a species with negligible acidity: \((\mathbf{a}) \mathrm{F}^{-}(\mathbf{b}) \mathrm{Br}^{-}(\mathbf{c}) \mathrm{HS}^{-}(\mathbf{d}) \mathrm{ClO}_{4}^{-}(\mathbf{e}) \mathrm{HCOO}^{-}\)

Predict the stronger acid in each pair: (a) \(\mathrm{HCl}\) or HF; (b) \(\mathrm{H}_{3} \mathrm{PO}_{4}\) or \(\mathrm{H}_{3} \mathrm{AsO}_{4} ;\) (c) \(\mathrm{HBrO}_{3}\) or \(\mathrm{HBrO}_{2}\) (d) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) or \(\mathrm{HC}_{2} \mathrm{O}_{4} \overline{;} ;(\mathbf{e})\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) or phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right) .\)

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}\right),\) a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) What is the \(\mathrm{pH}\) of a solution obtained by dissolving one regular aspirin tablet, containing \(100 \mathrm{mg}\) of acetylsalicylic acid, in \(200 \mathrm{~mL}\) of water?

(a) Given that \(K_{a}\) for cyanic acid is \(3.5 \times 10^{-4}\) and that for hydrofluoric acid is \(6.8 \times 10^{-4},\) which is the stronger acid? (b) Which is the stronger base, the cyanate ion or the fluoride ion? (c) Calculate \(K_{b}\) values for \(\mathrm{NCO}^{-}\) and \(\mathrm{F}^{-}\).

A \(0.100 \mathrm{M}\) solution of cyanic acid (HCNO) is \(5.9 \%\) ionized. Using this information, calculate [CNO^ \(\left.^{-}\right],\left[\mathrm{H}^{+}\right],[\mathrm{HCNO}],\) and \(K_{a}\) for cyanic acid.

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