Question

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

The boiling point of a \(0.10 M\) NaHSO\(_4\) solution with a density of \(1.002 g/mL\) is approximately \(100.102 ^\circ C\).

Step-by-step solution

Determine the van't Hoff Factor (i) for NaHSO4

NaHSO4 dissociates into two ions in water as follows: NaHSO4 → Na+ + HSO4- Since there are two ions produced from each NaHSO4 formula unit, the van't Hoff factor (i) is 2.

Calculate the molality of the solution

We are given the molarity (M) of the solution, which is 0.10 M. To convert molarity to molality, we need to take into account the density of the solution (1.002 g/mL). First, let's find the mass of the solvent (water) in 1 L (1000 mL) solution: Mass = Density × Volume = 1.002 g/mL × 1000 mL = 1002 g Now, we can calculate the moles of the solute (NaHSO4), as we know the molarity: Moles of NaHSO4 = Molarity × Volume = 0.10 mol/L × 1 L = 0.10 mol Finally, let's calculate the molality of NaHSO4: Molality = moles of solute / mass of solvent (in kg) = 0.10 mol / (1002 g × (1 kg / 1000 g)) = 0.0998 mol/kg

Calculate the boiling point elevation (ΔT)

Now that we know the van't Hoff factor (i=2) and the molality of the NaHSO4 solution (0.0998 mol/kg), we can calculate the boiling point elevation, using the boiling point elevation constant for water (Kb = 0.512 °C kg/mol) and the formula: ΔT = Kb × molality × i = 0.512 °C kg/mol × 0.0998 mol/kg × 2 = 0.102 °C

Determine the boiling point of the NaHSO4 solution

Since the normal boiling point of the water is 100 °C, we can add the boiling point elevation (ΔT) to find the boiling point of the NaHSO4 solution: Boiling point = Normal boiling point + ΔT = 100 °C + 0.102 °C = 100.102 °C Thus, the boiling point of a 0.10 M NaHSO4 solution with a density of 1.002 g/mL is approximately 100.102 °C.

Key concepts

van't Hoff factor

In chemistry, the van’t Hoff factor, symbolized as \(i\), is a measure of the effect of solute particles on colligative properties. It counts how many individual particles (ions or molecules) are generated in a solution from a particular solute. For \(\text{NaHSO}_4\), which dissociates into \(\text{Na}^+\) and \(\text{HSO}_4^-\), there are two ions resulting from one unit of \(\text{NaHSO}_4\). Thus, the van’t Hoff factor for \(\text{NaHSO}_4\) is 2.Colligative properties, like boiling point elevation, depend not on the identity of the solute particles but only on the quantity of particles. Hence, knowing the van’t Hoff factor is crucial for predicting changes in properties when a solute is dissolved.

• For non-electrolytes, \(i = 1\) as they do not dissociate into ions.
• For strong electrolytes, \(i\) is equal to the total number of ions produced per formula unit dissolved.

molality

Molality, denoted as \(m\), is a measure of the concentration of a solute in a solution. It differs from molarity by using the mass of the solvent, rather than the total volume of the solution.The formula for molality is:\[m = \frac{\text{moles of solute}}{\text{kilograms of solvent}}\.\]Using the example of a \(0.10\, ext{M}\) \(\text{NaHSO}_4\) solution, to find molality:1. First, calculate the mass of the solvent (usually water) in grams per 1 L of solution.2. Use the given density (1.002 g/mL) to find the total mass. A 1 L solution would weigh 1002 g.3. Convert this mass to kilograms (1002 g × 1 kg/1000 g).4. Calculate moles of solute, given as 0.10 moles of \(\text{NaHSO}_4\).5. Finally, compute molality using the equation above, which yields \(0.0998\, ext{mol/kg}\).Note that molality is not affected by temperature changes, unlike molarity.

molarity

Molarity is a widely-used unit of concentration, denoted as \(\text{M}\), which measures the number of moles of solute per liter of solution.The formula for calculating molarity is:\[\text{Molarity} (M) = \frac{\text{moles of solute}}{\text{liters of solution}}.\]This unit of concentration allows for easy calculation and is utilized mainly in quantitative chemical experiments. For instance, in the given problem:

• A solution with a molarity of 0.10 M contains 0.10 moles of \(\text{NaHSO}_4\) in 1 L of solution.
• It assumes that this concentration applies uniformly across the entire solution volume.

However, it is sensitive to temperature changes since liquid volumes expand or contract with temperature shifts.

density of solution

Density plays a key role in the calculation of molality and other solution properties. It is defined as the mass per unit volume and is often expressed in grams per milliliter (g/mL). When the density of a solution is known, it helps convert between volume and mass, which is vital when transitioning from molarity (volume based) to molality (mass-based). For example, with a solution density of 1.002 g/mL:

• 1 L of the solution would weigh 1002 g.
• This weight includes both the mass of the solute and the solvent together.
• Subtracting the solute mass, you can determine the solvent mass needed for the molality calculations.

Thus, density assists in accurately determining the solvent's mass, which is crucial in solution chemistry when different concentration units must be interconverted.