Problem 64 - General Chemistry: Principles and Modern Applications

Question

The molecular mass of hemoglobin is \(6.86 \times 10^{4} \mathrm{u}\) What mass of hemoglobin must be present per \(100.0 \mathrm{mL}\) of a solution to exert an osmotic pressure of \(7.25 \mathrm{mmHg}\) at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Approximately 2.67 grams of hemoglobin must be present in 100.0 mL of the solution to exert the given osmotic pressure at 25 degrees Celsius.

Step 1: Understand the Relation Between Osmotic Pressure and Concentration

According to the ideal gas law and the concept of osmotic pressure, the osmotic pressure \( P \) of a solution is given by \( P = C R T \), where \( c \) is the molar concentration of the solute, \( R \) is the ideal gas constant, and \( T \) is the absolute temperature. Here, we need to find \( c \), knowing that \( P = 7.25 \, mmHg \), \( R = 0.0821 \, L.atm/K.mol \) (in appropriate units), and \( T = 25^{\circ}C = 298 \, K \) (converting from Celsius to K).

Step 2: Convert pressure to appropriate units

Before we directly put the values into our equation, it is important that the units match. Therefore, convert pressure from mmHg to atm. Knowing that 1 atmosphere is equivalent to 760 mmHg, the pressure in atmospheres is \( P = 7.25 \, mmHg \times \frac{1 \, atm}{760 \, mmHg} = 0.00954 \, atm \).

Step 3: Calculate the concentration

Now, rearrange the ideal gas law to solve for the molar concentration \( c \). After rearranging, we get \( c = \frac{P}{RT} \). Substituting our known values, we find \( c = \frac{0.00954 \, atm}{0.0821 \, L.atm/K.mol \times 298 \, K} = 0.00039 \, moles/L \). This is the molar concentration of hemoglobin that we need.

Step 4: Calculate the mass of Hemoglobin

We know that hemoglobin has a molar mass of \(6.86 \times 10^{4} \, u\) or \(6.86 \times 10^{4} \, g/mol\). To find out the mass of hemoglobin that must be present to produce this concentration in 100.0 mL of solution, we need to multiply the molar concentration by the volume (in liters) and then by the molar mass. Doing so gives us \(0.00039 \, moles/L \times 0.1 \, L \times 6.86 \times 10^{4} \, g/mol = 2.67 \, g\). So, we need about 2.67 grams of hemoglobin.

Key Concepts

Hemoglobin

Hemoglobin is a vital protein found in red blood cells. It plays a crucial role in transporting oxygen from the lungs to the rest of the body and returning carbon dioxide back to the lungs for exhalation. This complex protein consists of four subunits, each with an iron-containing heme group that binds oxygen.
Hemoglobin’s efficiency in oxygen transport is due to its ability to change shape as it binds or releases oxygen. This characteristic makes it highly effective under various physiological conditions.
The molecular mass of hemoglobin is significant when calculating how much would be needed to exert a certain osmotic pressure. By understanding its mass and behavior in solutions, scientists and medical professionals can deduce concentrations essential for various physiological processes.
The molecular weight of hemoglobin can also give insights into how it behaves in physiological settings, like blood viscosity and pressure.

Molecular Mass

Molecular mass is the sum of the masses of all the atoms within a molecule. It's a critical property that allows chemists to understand how different molecules interact under various conditions.

• For hemoglobin, its molecular mass is given as \(6.86 \times 10^{4} \mathrm{u}\) or equivalently, \(6.86 \times 10^{4} \mathrm{g/mol}\).
• Molecular mass helps determine how much of a substance is needed to reach a certain concentration in a solution.

Knowing the molecular mass is essential for performing conversions in chemical calculations. For example, we need to know how many grams of hemoglobin are required for solutions exerting specific osmotic pressures. Multiply the molar concentration by the volume and the molecular mass to find this mass.
This understanding is essential for tasks like drug formulation, where precise concentrations can significantly impact efficacy and safety.

Ideal Gas Law

Although often associated with gases, the ideal gas law can also apply to solutions. It’s given by the equation: \( PV = nRT \), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
In the context of osmotic pressure, the formula adapts slightly to \( P = C R T \), where \(C\) refers to the molar concentration of the solute.
This equation becomes valuable when determining factors such as how much solute is necessary to achieve a particular osmotic pressure in a solution.

• The Ideal Gas Law aids in these calculations by providing a relationship between pressure, volume, and temperature.
• This understanding helps in interpreting physiological phenomena like blood plasma interactions and cellular osmosis.

By rearranging the equation, you can find the concentration if given the osmotic pressure, allowing for vital applications in various scientific and medical fields.