4% NaOH and 6% urea (both w/r) are equimolar but not isotonic. Explain.

To understand why 4% NaOH and 6% urea are equimolar but not isotonic, we need to break down a few key concepts: molarity, isotonic solutions, and the properties of the solutes involved.

Equimolar Solutions

First, let's clarify what it means for two solutions to be equimolar. When we say that 4% NaOH and 6% urea are equimolar, we are indicating that they contain the same number of moles of solute per liter of solution. To determine this, we can use the formula for calculating molarity:

• Molarity (M) = moles of solute / liters of solution

For NaOH, a 4% solution means there are 4 grams of NaOH in 100 mL of solution. The molar mass of NaOH is approximately 40 g/mol. Therefore, the number of moles of NaOH in 4 grams is:

• 4 g / 40 g/mol = 0.1 moles

Since this is in 100 mL (or 0.1 L), the molarity of the NaOH solution is:

• 0.1 moles / 0.1 L = 1 M

Now, for urea, a 6% solution means there are 6 grams of urea in 100 mL of solution. The molar mass of urea (NH2CONH2) is about 60 g/mol. Thus, the number of moles of urea in 6 grams is:

• 6 g / 60 g/mol = 0.1 moles

Again, since this is in 100 mL (or 0.1 L), the molarity of the urea solution is:

• 0.1 moles / 0.1 L = 1 M

Both solutions are indeed equimolar at 1 M.

Understanding Isotonicity

Next, let’s delve into isotonic solutions. Two solutions are considered isotonic if they have the same osmotic pressure, which means they exert the same pressure on a semipermeable membrane. This is crucial in biological contexts, especially when dealing with cells, as isotonic solutions prevent net movement of water into or out of cells.

Osmotic pressure is influenced not just by the concentration of solute particles but also by the nature of those particles. NaOH is a strong electrolyte, meaning it dissociates completely in solution into sodium ions (Na⁺) and hydroxide ions (OH^-). Therefore, a 1 M NaOH solution actually contributes 2 osmoles of solute particles per liter (1 M Na⁺ + 1 M OH^-).

In contrast, urea is a non-electrolyte and does not dissociate in solution. Thus, a 1 M urea solution contributes only 1 osmole of solute particles per liter.

Comparing Osmotic Pressures

Now, let’s compare the osmotic pressures of the two solutions:

• NaOH: 1 M NaOH = 2 osmoles (due to dissociation)
• Urea: 1 M urea = 1 osmole (no dissociation)

Since the osmotic pressure of the NaOH solution is higher than that of the urea solution, they are not isotonic. This difference in osmotic pressure means that if these solutions were separated by a semipermeable membrane, water would move from the urea solution (lower osmotic pressure) to the NaOH solution (higher osmotic pressure) until equilibrium is reached.

Summary of Key Points

In summary, while 4% NaOH and 6% urea are equimolar, they are not isotonic due to the following reasons:

• NaOH dissociates into two ions, contributing more solute particles compared to urea.
• The difference in osmotic pressure leads to a net movement of water, indicating that the solutions are not isotonic.

Understanding these concepts is essential, especially in fields like biology and medicine, where osmotic balance is crucial for cell function and overall homeostasis.

To understand why 4% NaOH and 6% urea are equimolar but not isotonic, we need to break down a few concepts related to molarity, tonicity, and the properties of the solutes involved.

Equimolarity Explained

First, let's clarify what it means for two solutions to be equimolar. When we say that 4% NaOH and 6% urea are equimolar, we are referring to the number of moles of solute present in a given volume of solution. In this case, both solutions contain the same number of moles of solute per liter of solution.

• NaOH: Sodium hydroxide (NaOH) is a strong base that dissociates completely in water, producing Na⁺ and OH^- ions.
• Urea: Urea (NH₂

To determine the molarity of each solution, we can use the formula:

Molarity (M) = (mass of solute in grams) / (molar mass of solute in g/mol) / (volume of solution in liters)

Calculating Moles

For NaOH, the molar mass is approximately 40 g/mol. In a 4% solution, there are 4 grams of NaOH in 100 mL of solution, which translates to:

Moles of NaOH = 4 g / 40 g/mol = 0.1 moles

For urea, the molar mass is about 60 g/mol. In a 6% solution, there are 6 grams of urea in 100 mL of solution:

Moles of urea = 6 g / 60 g/mol = 0.1 moles

Since both solutions contain 0.1 moles of solute in 100 mL, they are indeed equimolar.

Understanding Tonicity

Tonicity, on the other hand, refers to the ability of a solution to affect the volume of cells through osmosis. A solution is isotonic to a cell if it has the same osmotic pressure as the cell's internal environment, meaning there is no net movement of water into or out of the cell.

Even though both solutions are equimolar, they differ significantly in their osmotic effects:

• NaOH: When NaOH dissolves, it dissociates into two ions (Na⁺ and OH^-), effectively doubling the number of particles in solution. This means that the osmotic pressure of the NaOH solution is higher than that of the urea solution.
• Urea: Urea does not dissociate in solution, so it remains as one particle per molecule. Therefore, the osmotic pressure of the urea solution is lower compared to the NaOH solution.

Comparing Osmotic Pressures

To summarize, while both solutions contain the same number of moles of solute, the presence of additional ions in the NaOH solution increases its osmotic pressure. This results in a situation where:

NaOH solution has a higher osmotic pressure than the urea solution.

As a result, 4% NaOH and 6% urea are equimolar but not isotonic, leading to different effects on cells when they are placed in these solutions. Cells in the NaOH solution may lose water and shrink, while those in the urea solution may not experience the same osmotic pressure, allowing for a more stable internal environment.