Consider two compartments separated by a semipermeable membrane. The membrane allows solvent molecules to pass through but prevents solute molecules from crossing.
Suppose the left side contains pure solvent and the right side contains a solution. Which direction will the solvent move?
The answer can be understood by examining the chemical potential of the solvent. For a pure solvent,
\[ \mu_A=\mu_A^* \]while for a solvent in an ideal solution,
\[ \mu_A=\mu_A^*+RT\ln x_A \]Because the mole fraction of the solvent is less than one,
\[ x_A<1 \]and therefore
\[ \ln x_A<0 \]meaning that the chemical potential of the solvent in the solution is lower than that of the pure solvent.
A spontaneous process tends to move material from regions of higher chemical potential to regions of lower chemical potential. Consequently, solvent molecules move through the membrane from the pure solvent side into the solution.
This spontaneous flow of solvent through a semipermeable membrane is called osmosis.
As solvent enters the solution, the liquid level on the solution side rises. If pressure is applied to the solution, the chemical potential of the solvent in the solution increases.
Eventually, enough pressure can be applied that the chemical potential of the solvent in the solution becomes equal to that of the pure solvent:
\[ \mu_A(\text{solution}) = \mu_A(\text{pure solvent}) \]At this point, there is no longer a thermodynamic driving force for net solvent flow.
The pressure required to stop osmosis is called the osmotic pressure, denoted by \(\pi\).
For dilute solutions, the osmotic pressure is given by
\[ \pi = iMRT \]where:
This equation closely resembles the ideal gas law. In fact, osmotic pressure is often described as the solution analogue of gas pressure because both arise from the tendency of particles to spread throughout the available volume.
If a pressure larger than \(\pi\) is applied, solvent can be forced to flow out of the solution. This process is known as reverse osmosis and is widely used for water purification and desalination.
Big picture: Osmosis occurs because dissolved solutes lower the chemical potential of the solvent. The osmotic pressure is the pressure required to counteract this reduction and restore equilibrium. Like freezing-point depression and boiling-point elevation, osmotic pressure is a colligative property that depends primarily on the number of dissolved particles.
A physiological saline solution contains 0.9% (m/v) NaCl. This means that the solution contains 0.9 g of NaCl per 100 mL of solution. Calculate the osmotic pressure of the solution at 37 °C. Assume complete dissociation of NaCl.
The osmotic pressure of a dilute solution is given by
\[ \pi=iMRT \]where \(i\) is the van't Hoff factor, \(M\) is the molar concentration, \(R\) is the gas constant, and \(T\) is the absolute temperature.
First determine the molar concentration of NaCl.
A 0.9% (m/v) solution contains
\[ 0.9\ \text{g NaCl} \]per
\[ 100\ \text{mL solution} \]which corresponds to
\[ 9.0\ \text{g NaCl} \]per liter of solution.
The molar mass of NaCl is
\[ 58.44\ \text{g mol}^{-1} \]so the molarity is
\[ M = \frac{9.0\ \text{g L}^{-1}} {58.44\ \text{g mol}^{-1}} = 0.154\ \text{mol L}^{-1} \]Assuming complete dissociation,
\[ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \]and therefore
\[ i=2 \]The temperature is
\[ T = 37+273.15 = 310.15\ \text{K} \]Substituting into the osmotic-pressure equation:
\[ \pi = (2)(0.154\ \text{mol L}^{-1}) (0.08206\ \text{L atm mol}^{-1}\text{K}^{-1}) (310.15\ \text{K}) \] \[ \pi = 7.83\ \text{atm} \]Therefore,
\[ \boxed{\pi=7.83\ \text{atm}} \]Physical interpretation: The osmotic pressure of physiological saline is surprisingly large—nearly eight times atmospheric pressure. This illustrates how strongly dissolved particles influence the chemical potential of water. The concentration of saline used in hospitals is chosen because its osmotic pressure is close to that of blood plasma, minimizing net water movement into or out of cells.
| Practice Problem |
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| Hard water contains dissolved metal ions, especially calcium and magnesium ions. Estimate the osmotic pressure of hard water at 25 °C, assuming that the hardness comes entirely from dissolved \( \text{CaCl}_2 \). The calcium ion concentration is \(30\ \text{ppm Ca}^{2+}\), which may be interpreted as \(30\ \text{mg Ca}^{2+}\) per liter of solution. |
Show AnswerBegin by converting the calcium ion concentration into molarity. \[ 30\ \text{ppm Ca}^{2+} = 30\ \text{mg Ca}^{2+}\ \text{L}^{-1} = 0.030\ \text{g Ca}^{2+}\ \text{L}^{-1} \] \[ [\text{Ca}^{2+}] = \frac{0.030\ \text{g L}^{-1}} {40.08\ \text{g mol}^{-1}} = 7.49\times10^{-4}\ \text{mol L}^{-1} \]If the calcium comes from dissolved \( \text{CaCl}_2 \), then \[ \text{CaCl}_2(aq) \rightarrow \text{Ca}^{2+}(aq) + 2\text{Cl}^-(aq) \]For every mole of \( \text{Ca}^{2+} \), the solution also contains two moles of \( \text{Cl}^- \). Therefore, the total particle concentration is \[ C_{\text{particles}} = 3(7.49\times10^{-4}\ \text{mol L}^{-1}) = 2.25\times10^{-3}\ \text{mol L}^{-1} \]The osmotic pressure is \[ \pi = C_{\text{particles}}RT \]At 25 °C, \[ T=298.15\ \text{K} \]so \[ \pi = (2.25\times10^{-3}\ \text{mol L}^{-1}) (0.08206\ \text{L atm mol}^{-1}\text{K}^{-1}) (298.15\ \text{K}) \] \[ \pi = 5.50\times10^{-2}\ \text{atm} \]Therefore, \[ \boxed{\pi=5.50\times10^{-2}\ \text{atm}} \]This is about \[ 42\ \text{Torr} \]Physical interpretation: The osmotic pressure is much smaller than that of physiological saline because the dissolved ion concentration in hard water is much lower. Even so, the value is not zero: dissolved mineral ions still lower the chemical potential of water and produce a measurable osmotic pressure. |
Big picture: Osmosis provides one of the clearest demonstrations of chemical potential as a driving force for spontaneous change. The osmotic pressure measures how strongly dissolved particles lower the chemical potential of a solvent and serves as another example of how colligative properties arise from the number of particles present in solution.