Question

The majority of dissolved ions in seawater is NaCl, at roughly a 0.5 M NaCl concentration. The ionization constant for NaCl is 2. Calculate the solute potential for seawater if you know that the water is -5°Celsius.

a) -2.5 MPa
b) -5 MPa
c) -10 MPa
d) -20 MPa

Answer 1

The solute potential of seawater with 0.5 M NaCl concentration at -5°C is calculated using the formula ψs = -iCRT. With i = 2, R = 0.0831 liter·bar/mol·K, and T = 268.15 K, the solute potential is -2.23 MPa, rounded to -2.5 MPa, which is option (a).

Step-by-step explanation:

The question is asking to calculate the solute potential, or osmotic potential, of seawater at a temperature of -5°C, given the concentration of NaCl. The formula for the solute potential of a solution is ψs = -iCRT, where i is the ionization constant (the number of particles the compound dissociates into, which is 2 for NaCl), C is the molar concentration of the solute (0.5 M for NaCl), R is the pressure constant (0.0831 liter·bar/mol·K), and T is the temperature in Kelvin. To find T in Kelvin, add 273.15 to the Celsius temperature, which gives us 268.15 K for -5°C. Plugging in the known values, we get ψs = -(2)(0.5 M)(0.0831 liter·bar/mol·K)(268.15 K). To convert from bars to MPa, note that 1 bar = 0.1 MPa. Thus, the solute potential for seawater at -5°C is -2.23 MPa, which is closest to the provided answer (a) -2.5 MPa.