Osmotic Pressure - Chemistry LibreTexts

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1. Introduction
   1. Exercise \(\PageIndex{1}\)
   2. Exercise \(\PageIndex{2}\)
   3. Exercise \(\PageIndex{3}\)

Introduction

Semipermiable membranes do not let the solute pass through (Think of the sugar example). A solvent will move to the side that is more concentrated to try to make each side more similar! Since there is a flow of solvents, the height of each side changes, which is osmotic pressure. When we work with aqueous solutions, we use mm of H2O to describe the difference.

Osmosis is the diffusion of a fluid through a semipermeable membrane. When a semipermeable membrane (animal bladders, skins of fruits and vegetables) separates a solution from a solvent, then only solvent molecules are able to pass through the membrane. The osmotic pressure of a solution is the pressure difference needed to stop the flow of solvent across a semipermeable membrane. The osmotic pressure of a solution is proportional to the molar concentration of the solute particles in solution.

\[\Pi = i \dfrac{n}{V}RT = i M RT \label{eq1}\]

where

• \(\Pi\) is the osmotic pressure,
• \(R\) is the ideal gas constant (0.0821 L atm / mol K),
• \(T\) is the temperature in Kelvin,
• \(i\) is the van 't Hoff factor
• \(n\) is the number of moles of solute present,
• \(V\) is the volume of the solution, and
• \(M\) is the molar concentration of added solute (the \(i\) factor accounts for how many species in solution are generated)

Exercise \(\PageIndex{1}\)

Calculate molarity of a sugar solution in water (300 K) has osmotic pressure of 3.00 atm.

Answer

Since it is sugar, we know it doesn't dissociate in water, so \(i\) is 1. Then we use Equation \ref{eq1} directly

\[M = \dfrac{\Pi}{RT} = \dfrac{3.00\, atm}{(0.0821\, atm.L/mol.K)(300\,K)} = 0.122\,M \nonumber\]

Exercise \(\PageIndex{2}\)

Calculate osmotic pressure for 0.10 M \(\ce{Na3PO4}\) aqueous solution at 20°C.

Answer

Since \(\ce{Na3PO4}\) ionizes into four particles (3 Na+1 + \(PO_4^{-3}\)), then \(i = 4\). We can then calculate the osmotic pressure via Equation \ref{eq1}

\[\Pi = iMRT = (0.40)(0.0821)(293) = 9.6\, atm \nonumber\]

Exercise \(\PageIndex{3}\)

Hemoglobin is a large molecule that carries oxygen in human blood. A water solution that contains 0.263 g of hemoglobin (Hb) in 10.0 mL of solution has an osmotic pressure of 7.51 torr at \(25 ^oC\). What is the molar mass of the hemoglobin?

Answer

\(6.51 \times 10^4 \; g/mol\)