Problem 126 - General Chemistry: Principles and Modern Applications

Question

NaCl(aq) isotonic with blood is \(0.92 \%\) NaCl (mass/volume). For this solution, what is (a) \(\left[\mathrm{Na}^{+}\right]\) (b) the total molarity of ions; (c) the osmotic pressure at \(37^{\circ} \mathrm{C} ;\) (d) the approximate freezing point? (Assume that the solution has a density of 1.005 g/mL.)

Step-by-Step Solution

Answer

The molarity of Na+ is 0.157 M, the total molarity of ions in the solution is 0.314 M, the osmotic pressure at 37 °C is 19.1 atm, and the approximated freezing point will be lower than 0°C and can be calculated based on the formula for freezing point depression and the given parameters in the problem statement.

Step 1: Calculate Molarity of Na+

Molarity is calculated as moles of solute per liter of solution. First calculate the moles of NaCl. The solution is 0.92% NaCl by mass/volume. So, 100 mL of this solution contains 0.92g of NaCl. Moles = Mass/Molar mass: Moles of NaCl = 0.92 g / 58.44 g/mol = 0.0157 mol. Then divide this by the volume of the solution in liters: Molarity of NaCl = 0.0157 mol / 0.1 L = 0.157 M.

Step 2: Calculate Total Molarity of Ions

Each NaCl unit dissociates into one Na+ ion and one Cl- ion in solution. Since the molarity of NaCl is 0.157 M, the total molarity of ions will be 2 * 0.157 M = 0.314 M.

Step 3: Calculate Osmotic Pressure

Since we know the total molarity of ions and if we take R = 0.0821 L.atm/(K.mol), the osmotic pressure can be calculated using the formula: 'osmotic pressure = i * n * R * T'. Here i = 2 (ionisation factor for NaCl), n is the molarity of ions we calculated in the previous step, and T is the absolute temperature in Kelvin: Osmotic Pressure = 2 * 0.314 mol/L * 0.0821 L.atm/K.mol * (37 °C + 273 K) = 19.1 atm.

Step 4: Calculate Freezing Point

NaCl is a strong electrolyte and will fully dissociate into its ions in a solution. In this case, the van’t Hoff factor (i) is 2. Using the equation for freezing point depression, which is ΔT = i * Kf * m, where ΔT is the depression in freezing point, Kf is the cryoscopic constant (which is 1.86 °C/kg for water), and m is the molality of the solution. We'll use the calculated moles of NaCl and the mass of the solution (weight of 100mL water is approx 100g, or 0.1kg) to find molality, then substitute all values into the equation to calculate ΔT. It's worth noting that the freezing point of a solution will always be lower than that of the pure solvent.

Key Concepts

Molarity

Molarity is a term used to describe the concentration of a solute in a solution. It is expressed as the number of moles of solute present in one liter of solution. When you deal with solutions like NaCl in water, molarity becomes an essential factor to measure. For instance, in a solution that is 0.92% NaCl by mass, you would calculate how many moles of NaCl are present in 100 ml of solution. Given the molar mass of NaCl, which is 58.44 g/mol, you can find that there are 0.0157 moles of NaCl in this amount of solution. To calculate molarity, you divide the moles by the volume of the solution in liters, giving the molarity of NaCl at 0.157 M.

• Molarity tells you how much solute is in a specific volume of the solution.
• It can help determine other properties like boiling and freezing points.

Osmotic Pressure

Osmotic pressure is an important concept when studying properties of solutions. It is the pressure required to stop water from flowing across a semipermeable membrane due to osmosis. For example, in a solution containing NaCl, it splits into two ions: Na⁺ and Cl⁻, adding to the total molarity of ions. This is crucial in calculating osmotic pressure. You use the formula \[\pi = i \times n \times R \times T\]where \(\pi\) is the osmotic pressure, \(i\) is the ionization factor (which is 2 for NaCl), \(n\) is the molarity of ions, \(R\) is the gas constant (0.0821 L.atm/mol.K), and \(T\) is the temperature in Kelvin. At body temperature, this means an osmotic pressure of 19.1 atm.

• Osmotic pressure helps in understanding the movement of water in biological systems.
• It is affected by temperature and solute concentration.

Freezing Point Depression

Freezing point depression is a phenomenon where the addition of a solute to a solvent results in lowering the freezing point of the solvent. This is particularly relevant when dealing with electrolyte solutions like NaCl in water. The freezing point depression can be calculated using the equation \[\Delta T = i \times K_f \times m\]where \(\Delta T\) is the change in freezing point, \(i\) is the van’t Hoff factor (2 for NaCl as it dissociates into two ions), \(K_f\) is the cryoscopic constant (1.86 °C/kg for water), and \(m\) is the molality of the solution. This results in a lower freezing point, ensuring the solution remains liquid at temperatures where pure water would freeze.

• Freezing point depression is used in de-icing roads with salt.
• It explains why seawater remains liquid at temperatures below 0 °C.