Freezing Point Of Water Compared To A Salt Solution

When a solvent freezes, the particles of that solvent become more ordered. The intermolecular forces that act on these particles become more "permanent" since the particles are now closer together. For example, when water freezes into ice, the hydrogen bonds that give water many of its unique properties make a hexagonally shaped network of molecules inherent to the structure of ice.

So what happens when a solute is added to water or pure solvent? The addition of solute results in the ordering of the solvent molecules being disrupted. This means that more energy must be removed from the solution in order to freeze it.

For example, when salt is added to water, the resulting ions in the water disrupt the usual network of hydrogen bonds made upon freezing. As a result, the freezing point of the solution is lower than it is for the pure solvent. This is called freezing point depression.

The decrease in freezing point is directly proportional to the molality of the solute:

\(\Delta T_f=K_fmi\)

In this equation, K_f is the molal freezing point depression constant, and m is the molality of the solute. Remember, molality is the number of moles of solute per kg of solvent. The van't Hoff factor is i which relates to the number of ions in solution for each dissolved molecule of solute. For example, this would be 2 for NaCl.

Basically, this means that the more solute there is, the greater the decrease in freezing temperature.

Freezing point depression is defined by the freezing point of pure solvent minus the freezing point of the solution in question:

\(\Delta T_f=T^{\degree}_f-T_f\)

This allows you to find what the new freezing point is compared to the pure solvent.

The two most common real world applications of freezing point depression are antifreeze and salting roads in the winter.

Ethylene glycol is a compound often used in antifreeze because when it is added to water, the freezing point of water decreases. This can help ensure that water in the radiator of your car won't freeze.

When salt is added to the road in the winter, ice will melt at a lower temperature, thus making it safer because there won't be as much ice on the road.

Take a look at the following example, which shows how adding salt to water will result in a decrease in the freezing point of the solution.

What is the freezing temperature of a solution in which 100 grams of NaCl have been added to 1 kilogram of water? In other words, what is the freezing point of saltwater?

You can make use of the following equation:

\(\Delta T_f=K_fmi\)

The K_f for water is 1.86 °C/m. This number can be found in a table such as the one in the first reference. Since NaCl dissociates into two ions, the van't Hoff factor is 2. Finally, you need to calculate the molality of the solution.

To do this you will first need to convert the grams of NaCl to moles:

\(100g\enspace NaCl\Biggl(\dfrac{1mol\enspace NaCl}{58.44\enspace NaCl} \Biggr)=1.71mol\enspace NaCl\)

Now, you will need to divide the moles of NaCl by the mass of the solvent to find the molality:

\(\dfrac{1.71mol\enspace NaCl}{1kg} =1.71m\enspace NaCl\)

Next, you can plug this into the following equation:

\(\Delta T_f=K_fmi\)

So:

\(\Delta T_f=1.86 \dfrac{\degree C}{m} (1.71m)(2)\)

As such,

\(\Delta T_f=6.4 \degree C\)

Now, you can use the freezing point depression equation to find the new freezing point of the solution. (Remember, the freezing point of pure water is 0°C.)

\(\Delta T_f=T^{\degree}_f-T_f\)

So:

\(\Delta T_f=0\degree C-6.4 \degree C\)

Thus, adding 100 grams of salt to 1 kilogram of water would decrease the freezing point to -6.4°C.