What is the water potential of a .20 M solution of sucrose in an open beaker at 23 degrees Celsius?
The correct answer and explanation is:
Sure! Let’s calculate the water potential of a 0.20 M sucrose solution at 23°C.
Step 1: Understanding Water Potential (Ψ)
Water potential (Ψ) is a measure of the potential energy of water in a system compared to pure water, and it determines the direction of water movement. It is expressed in units of pressure (usually megapascals, MPa).
The formula to calculate the water potential of a solution is: Ψ=Ψs+Ψp+Ψg+Ψm\Psi = \Psi_s + \Psi_p + \Psi_g + \Psi_m
where:
- Ψs\Psi_s = solute potential (osmotic potential)
- Ψp\Psi_p = pressure potential
- Ψg\Psi_g = gravitational potential
- Ψm\Psi_m = matric potential
For an open beaker exposed to the atmosphere:
- Pressure potential (Ψp\Psi_p) is 0 (atmospheric pressure balanced)
- Gravitational potential (Ψg\Psi_g) is 0 (assuming negligible height difference)
- Matric potential (Ψm\Psi_m) is 0 (no soil or matrix effect)
Thus, Ψ≈Ψs\Psi \approx \Psi_s
Step 2: Calculate Solute Potential (Ψs\Psi_s)
Solute potential is calculated by the formula derived from the van’t Hoff equation: Ψs=−CRT\Psi_s = -CRT
where:
- CC = molar concentration of the solute (mol/L)
- RR = universal gas constant = 0.00831 L·MPa/(mol·K)
- TT = absolute temperature in Kelvin (K)
Step 3: Convert Temperature to Kelvin
T=23∘C+273=296KT = 23^\circ C + 273 = 296 K
Step 4: Plug values into the equation
Given:
- C=0.20C = 0.20 M
- R=0.00831R = 0.00831 L·MPa/(mol·K)
- T=296T = 296 K
Calculate: Ψs=−(0.20)(0.00831)(296)=−0.4919 MPa\Psi_s = -(0.20)(0.00831)(296) = -0.4919 \text{ MPa}
Final Answer:
Ψ=−0.49 MPa\boxed{ \Psi = -0.49 \text{ MPa} }
Explanation (300 words):
Water potential is an important concept in plant physiology, environmental science, and chemistry. It represents the free energy of water per unit volume relative to pure water under reference conditions, influencing how water moves across membranes or within ecosystems.
In a solution like 0.20 M sucrose, dissolved sugar molecules reduce the water potential compared to pure water because they bind water molecules and reduce their free energy. Water naturally moves from areas of higher (less negative) to lower (more negative) water potential.
The main component affecting water potential in an open solution is the solute potential (Ψs\Psi_s), which is always negative. This negativity reflects that solutes reduce water’s free energy.
The solute potential is calculated using the van’t Hoff equation, which relates the osmotic pressure exerted by solutes to concentration, temperature, and gas constant. Here, sucrose is a non-ionizing solute (does not dissociate), so its molarity directly corresponds to the concentration term.
At 23°C (296 K), the negative solute potential for 0.20 M sucrose is about -0.49 MPa, indicating that water in this solution has 0.49 MPa less free energy than pure water.
In practice, this means if a plant cell is placed in this sucrose solution, water will tend to move out of the cell (where water potential is higher, closer to zero) into the sucrose solution (lower water potential), potentially causing the cell to lose water and shrink.
Thus, calculating water potential helps understand osmosis, water transport, and cellular water balance in biological systems.