Question

A Minnesota gardener notes that the plants immediately bordering a walkway are stunted compared with those farther away. Suspecting that the soil near the walkway may be contaminated from salt added to the walkway in winter, the gardener tests the soil. The composition of the soil near the walkway is identical to that farther away except that it contains an additional 50 mM NaCl. Assuming that the NaCl is completely ionized, calculate how much it will lower the solute potential of the soil at 20°C using the solute potential equation:

Ψ_S = -iCRT

where i is the ionization constant (2 for NaCl), C is the molar concentration (in mol/L), R is the pressure constant [R = 0.00831 (L · MPa)/(mol · K)], and T is the temperature in Kelvin (273 + °C). How would this change in the solute potential of the soil affect the water potential of the soil? In what way would the change in the water potential of the soil affect the movement of water in or out of the roots?

Short answer

The solute potential of the soil would be -0.243 MPa.

The change in solute potential would lower the water potential of the soil.

Lower soil water potential would result in water movement out of the roots, causing water loss from the root system.

Step-by-step solution

Water potential and solute potential

The soil's water potential is the potential energy per unit mass of soil water with reference to pure water. The solute potential is osmotic potential or the pressure required to prevent the inward flow of water through a semi-permeable membrane. When solute concentration increases, it leads to a decrease in solute potential.

The solute potential of the soil

To calculate the solute potential for NaCl, we know,

${\Psi }_S=-iCRT$, where ${\Psi }_S$is the solute potential, i is the ionization constant, C is the molar concentration, R is the pressure constant, and T is the temperature in Kelvin.

Given:

The ionization constant for NaCl is, $i=2$

The molar concentration of NaCl is:

$$\begin{gathered} C=50mM \\ =0.05mol/L \end{gathered}$$

Pressure constant is$R=0.00831LMPa/molK$

Temperature:

$T=20°C=(273+20°C)K=293K$

Therefore, the solute potential of the soil is:

${\Psi }_S=-iCRT=-2\times (0.05mol/L)\times (0.00831LMPa/molK)\times \left(293K\right)=-0.243MPa$

The solute potential of soil at $20°C$or$293K$is calculated as -0.243 MPa.

Effect on water potential of the soil

The solute potential of the soil is lower close to the walkway. The expression$\Psi ={\Psi }_S+{\Psi }_P$ gives water potential. In this expression, is the solute potential of the soil. Therefore, water potential is directly dependent on solute potential.

As the solute potential of the soil is lower, the water potential of the soil would be lower too.

Water movement

Water moves from a region of higher water potential to a region with lower water potential. The decrease in the water potential of the soil would result in the movement of water or water loss from the roots.