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Colligative Properties of Electrolytes

Since colligative properties depend on the number of particles dissolved, solutions of electrolytes (which dissociate in solution) should show greater changes than those of nonelectrolytes.

Colligative Properties of Electrolytes

However, a 1 M solution of NaCl does not show twice the change in freezing point that a 1 M solution of methanol does.

Ion Pairing and the van't Hoff Factor

One mole of NaCl in water does not really give rise to two moles of ions. Some Na+ and Cl- reassociate for a short time, so the true concentration of particles is somewhat less than two times the concentration of NaCl. · Reassociation is more likely at Ion pairing and colligative higher concentration. properties. Ion pairing · Therefore, the number of becomes more prevalent as particles present is the solution concentration concentration dependent. increases

The van't Hoff Factor

We modify the previous equations by multiplying by the van't Hoff factor, i

Tf (measured) = Tf (calculated for nonelectrolyte) × i

Tf (measured) i= Tf (calculated for nonelectrolyte)

For an ideal solution: the van't Hoff factor equals the number of ions per formula unit. For ideal solutions of NaCl and K2SO4 : i = 2 and i = 3, respectively In absence of information about the value of i for a solution, use the ideal value in calculations

Osmosis

· Some substances form semipermeable membranes (e.g cellophane and many membranes in biological systems), allowing some smaller particles to pass through (e.g. water molecules), but blocking other larger particles. · In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so.

Osmosis

In osmosis, there is net movement of solvent from the area of higher solvent concentration (lower solute concentration) to the are of lower solvent concentration (higher solute concentration). There is a net solvent movement through the semipermeable membrane, as if the solutions were driven to attain equal concentrations across the membrane

Osmotic Pressure

The pressure required to stop osmosis, known as osmotic pressure, , is

V = nRT

V is the volume of the solution, n is the number of moles of the solute, R is the gas constant, and T is the temperature in Kelvin Then

n = RT = MRT V

M is the molarity of the solution (mol/L)

If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same), the solutions are isotonic. If one solution is of lower osmotic pressure, it is hypotonic with respect to the more concentrated solution. The more concentrated solution is hypertonic with respect to the dilute solution.

Osmosis in Blood Cells

· If the solute concentration outside the cell is hypertonic relative to the intracellular solution (the solution within the cell) · Water will flow out of the cell, this cause the cell to shrivel, a process called crenation

Osmosis in Cells

· If the solute concentration outside the cell is hypotonic relative to the intracellular solution. · Water will flow into the cell. This causes the cell to rupture, a process called hemolysis.

The Intravenous (IV) solutions must be isotonic with the intracellular fluids of the cell otherwise crenation or hemolysis occur

EXERCISE: Molar Mass from Freezing-Point Depression

A solution of an unknown nonvolatile electrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of CCl4. The boiling point of the resultant solution was 0.357°C higher than that of the pure solvent. Calculate the molar mass of the solute. For pure CCl4 kb = 5.02 ºC/m

Answer:

molality =

n solute 40.0 ×10 kg

-3

n solute mol 40.0 ×10-3 kg

= 0.0711 m

=

Tb 0.375 C = = 0.0711 m -1 kb 5.02 C.m

n solute = 40.0 ×10-3 kg × 0.0711 (mol.kg-1 ) = 2.84 ×10-3 mol of solute

m = 2.84 ×10-3 mol of solute MM m (g) 0.250 g MM = = = 88.0 g.mol-1 n solute (mol) 2.84 ×10-3 mol n solute =

EXERCISE: Molar Mass from Osmotic Pressure

The osmotic pressure of an aqueous solution of a certain protein was measured in order to determine the protein's molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25°C was found to be 1.54 torr. Calculate the molar mass of the protein.

Answer

n= =

V

RT 1.54 mmHg

V = nRT

×

1 atm 1L × 5ml × = 4.14 ×10 - 7 mol 1000 ml 0.0821 L.atm.mol-1.K -1 × 298 K 760 mmHg

m m 3.50 ×10 -3 g MM = = n= = 8.45 ×103 g.mol-1 MM n 4.14 ×10-7 mol

Colloids

Suspensions or dispersions of particles larger than individual ions or molecules, but too small to be settled out by gravity. Colloids form the dividing line between solutions and homogeneous mixtures Size of colloid particles range from 5 to 1000 nm

Tyndall Effect

· Colloidal suspensions can scatter rays of light. · This phenomenon is known as the Tyndall effect.

Colloids in Biological Systems

Some molecules have a polar, hydrophilic (water-loving) end and a nonpolar, hydrophobic (waterhating) end.

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Colligative Properties of Electrolytes

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