AN INTRODUCTION TO SATURATED VAPOUR PRESSURE

AN INTRODUCTION TO SATURATED VAPOUR PRESSURE This page looks at how the equilibrium between a liquid (or a solid) and its vapour leads to the idea of a saturated vapour pressure. It also looks at how saturated vapour pressure varies with temperature, and the relationship between saturated vapour pressure and boiling point. The origin of saturated vapour pressure The evaporation of a liquid The average energy of the particles in a liquid is governed by the temperature. The higher the temperature, the higher the average energy. But within that average, some particles have energies higher than the average, and others have energies lower than the average. Some of the more energetic particles on the surface of the liquid can be moving fast enough to escape from the attractive forces holding the liquid together. They evaporate. The diagram shows a small region of a liquid near its surface. Notice that evaporation only takes place on the surface of the liquid. That's quite different from boiling which happens when there is enough energy to disrupt the attractive forces throughout the liquid. That's why, if you look at boiling water, you see bubbles of gas being formed all the way through the liquid. If you look at water which is just evaporating in the sun, you don't see any bubbles. Water molecules are simply breaking away from the surface layer. Eventually, the water will all evaporate in this way. The energy which is lost as the particles evaporate is replaced from the surroundings. As the molecules in the water jostle with each other, new molecules will gain enough energy to escape from the surface. The evaporation of a liquid in a closed container Now imagine what happens if the liquid is in a closed container. Common sense tells you that water in a sealed bottle doesn't seem to evaporate - or at least, it doesn't disappear over time. But there is constant evaporation from the surface. Particles continue to break away from the surface of the liquid - but this time they are trapped in the space above the liquid. As the gaseous particles bounce around, some of them will hit the surface of the liquid again, and be trapped there. There will rapidly be an equilibrium set up in which the number of particles leaving the surface is exactly balanced by the number rejoining it. In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid. When these particles hit the walls of the container, they exert a pressure. This pressure is called the saturated vapour pressure (also known as saturation vapour pressure) of the liquid. Measuring the saturated vapour pressure It isn't difficult to show the existence of this saturated vapour pressure (and to measure it) using a simple piece of apparatus.
	Note:  This experiment is much easier to talk about than do, given the safety problems in handling mercury because of its poisonous vapour. This is particularly going to be a problem if you want to find the saturated vapour pressure of a liquid at a higher temperature. You would have to use a more complex bit of apparatus. That isn't a problem you need to worry about for UK A level purposes.
If you have a mercury barometer tube in a trough of mercury, at 1 atmosphere pressure the column will be 760 mm tall. 1 atmosphere is sometimes quoted as 760 mmHg ("millimetres of mercury"). If you squirt a few drops of liquid into the tube, it will rise to form a thin layer floating on top of the mercury. Some of the liquid will evaporate and you will get the equilibrium we've just been talking about - provided there is still some liquid on top of the mercury. It is only an equilibrium if both liquid and vapour are present. The saturated vapour pressure of the liquid will force the mercury level down a bit. You can measure the drop - and this gives a value for the saturated vapour pressure of the liquid at this temperature. In this case, the mercury has been forced down by a distance of 760 - 630 mm. The saturated vapour pressure of this liquid at the temperature of the experiment is 130 mmHg. You could convert this into proper SI units (pascals) if you wanted to. 760 mmHg is equivalent to 101325 Pa. A value of 130 mmHg is quite a high vapour pressure if we are talking about room temperature. Water's saturated vapour pressure is about 20 mmHg at this temperature. A high vapour pressure means that the liquid must be volatile - molecules escape from its surface relatively easily, and aren't very good at sticking back on again either. That will result in larger numbers of them in the gas state once equilibrium is reached. The liquid in the example must have significantly weaker intermolecular forces than water. The variation of saturated vapour pressure with temperature The effect of temperature on the equilibrium between liquid and vapour You can look at this in two ways. There is a common sense way. If you increase the temperature, you are increasing the average energy of the particles present. That means that more of them are likely to have enough energy to escape from the surface of the liquid. That will tend to increase the saturated vapour pressure. Or you can look at it in terms of Le Chatelier's Principle - which works just as well in this kind of physical situation as it does in the more familiar chemical examples.
	Note:  You could follow this link if you aren't sure about Le Chatelier's Principle. Use the BACK button on your browser to return to this page later.
When the space above the liquid is saturated with vapour particles, you have this equilibrium occurring on the surface of the liquid: The forward change (liquid to vapour) is endothermic. It needs heat to convert the liquid into the vapour. According to Le Chatelier, increasing the temperature of a system in a dynamic equilibrium favours the endothermic change. That means that increasing the temperature increases the amount of vapour present, and so increases the saturated vapour pressure. The effect of temperature on the saturated vapour pressure of water The graph shows how the saturated vapour pressure (svp) of water varies from 0°C to 100 °C. The pressure scale (the vertical one) is measured in kilopascals (kPa). 1 atmosphere pressure is 101.325 kPa. Saturated vapour pressure and boiling point A liquid boils when its saturated vapour pressure becomes equal to the external pressure on the liquid. When that happens, it enables bubbles of vapour to form throughout the liquid - those are the bubbles you see when a liquid boils. If the external pressure is higher than the saturated vapour pressure, these bubbles are prevented from forming, and you just get evaporation at the surface of the liquid. If the liquid is in an open container and exposed to normal atmospheric pressure, the liquid boils when its saturated vapour pressure becomes equal to 1 atmosphere (or 101325 Pa or 101.325 kPa or 760 mmHg). This happens with water when the temperature reaches 100°C. But at different pressures, water will boil at different temperatures. For example, at the top of Mount Everest the pressure is so low that water will boil at about 70°C. Depressions from the Atlantic can easily lower the atmospheric pressure in the UK enough so that water will boil at 99°C - even lower with very deep depressions. Whenever we just talk about "the boiling point" of a liquid, we always assume that it is being measured at exactly 1 atmosphere pressure. In practice, of course, that is rarely exactly true. Saturated vapour pressure and solids Sublimation Solids can also lose particles from their surface to form a vapour, except that in this case we call the effect sublimation rather than evaporation. Sublimation is the direct change from solid to vapour (or vice versa) without going through the liquid stage. In most cases, at ordinary temperatures, the saturated vapour pressures of solids range from low to very, very, very low. The forces of attraction in many solids are too high to allow much loss of particles from the surface. However, there are some which do easily form vapours. For example, naphthalene (used in old-fashioned "moth balls" to deter clothes moths) has quite a strong smell. Molecules must be breaking away from the surface as a vapour, because otherwise you wouldn't be able to smell it. Another fairly common example (discussed in detail on another page) is solid carbon dioxide - "dry ice". This never forms a liquid at atmospheric pressure and always converts directly from solid to vapour. That's why it is known as dry ice.

SOLUBILITY PRODUCT and THE COMMON ION EFFECT

SOLUBILITY PRODUCT and THE COMMON ION EFFECT This page looks at the common ion effect related to solubility products, including a simple calculation. You need to know about solubility products and calculations involving them before you read this page. What is the common ion effect? I need to look again at a simple solubility product calculation, before we go on to the common ion effect. The solubility of lead(II) chloride in water Lead(II) chloride is sparingly soluble in water, and this equilibrium is set up between the solid and its ions in solution: If you just shook up some solid lead(II) chloride with water, then the solution would obviously contain twice as many chloride ions as lead(II) ions. The expression for the solubility product and its value are given by:
	Note:  I have no confidence in the accuracy of this value. There are serious discrepancies between the values from different sources. But that makes no difference to the discussion.
For comparison purposes later, I need to work out the lead(II) ion concentration in this saturated solution. If the concentration of dissolved lead(II) chloride is s mol dm-3, then: [Pb2+] = s mol dm-3 [Cl- ] = 2s mol dm-3 Put these values into the solubility product expression, and do the sum. So the concentration of lead(II) ions in the solution is 1.62 x 10-2 mol dm-3 (or 0.0162 mol dm-3 if you prefer). What happens if you add some sodium chloride to this saturated solution? Now we are ready to think about the common ion effect. Sodium chloride shares an ion with lead(II) chloride. The chloride ion is common to both of them. This is the origin of the term "common ion effect". Look at the original equilibrium expression again: What would happen to that equilibrium if you added extra chloride ions? According to Le Chatelier, the position of equilibrium would shift in order to counter what you have just done. In this case, it would tend to remove the chloride ions by making extra solid lead(II) chloride.
	Note:  Actually, of course, the concentration of lead(II) ions in the solution is so small to start with, that only a tiny proportion of the extra chloride ions can be converted into solid lead(II) chloride.
The lead(II) chloride will become even less soluble - and, of course, the concentration of lead(II) ions in the solution will decrease. Something similar happens whenever you have a sparingly soluble substance. It will be less soluble in a solution which contains any ion which it has in common. This is the common ion effect. A simple calculation to show this Suppose you tried to dissolve some lead(II) chloride in some 0.100 mol dm-3 sodium chloride solution instead of in water. What would the concentration of the lead(II) ions be this time? As before, let's call the concentration of the lead(II) ions s. [Pb2+] = s mol dm-3 Now the sum gets different. This time the concentration of the chloride ions is governed by the concentration of the sodium chloride solution. The number of ions coming from the lead(II) chloride is going to be tiny compared with the 0.100 mol dm-3 coming from the sodium chloride solution. In calculations like this, you can always assume that the concentration of the common ion is entirely due to the other solution. This makes the maths a lot easier. In fact if you don't make this assumption, the maths of this can become impossible to do at this level. So we assume: [Cl- ] = 0.100 mol dm-3 The rest of the sum looks like this: Finally, compare that value with the simple saturated solution we started with: Original solution: [Pb2+] = 0.0162 mol dm-3 Solution in 0.100 mol dm-3 NaCl solution: [Pb2+] = 0.0017 mol dm-3 The concentration of the lead(II) ions has fallen by a factor of about 10. If you tried the same sum with more concentrated solutions of sodium chloride, the solubility would fall still further. Try it yourself with chloride ion concentrations of 0.5 and 1.0 mol dm-3.

SOLUBILITY PRODUCT CALCULATIONS

SOLUBILITY PRODUCT CALCULATIONS This page is a brief introduction to solubility product calculations. These are covered in more detail in my chemistry calculations book. Calculating solubility products from solubilities I am going to assume that you are given the solubility of an ionic compound in mol dm-3. If it was in g dm-3, or any other concentration units, you would first have to convert it into mol dm-3. Example 1 The solubility of barium sulphate at 298 K is 1.05 x 10-5 mol dm-3. Calculate the solubility product. The equilibrium is: Notice that each mole of barium sulphate dissolves to give 1 mole of barium ions and 1 mole of sulphate ions in solution. That means that: [Ba2+] = 1.05 x 10-5 mol dm-3 [SO42-] = 1.05 x 10-5 mol dm-3 All you need to do now is to put these values into the solubility product expression, and do the simple sum. Don't forget to work the units out.
	Important:  Get your calculator and work this out! Students frequently mis-enter numbers like 1.05 x 10-5. If you try this sum, and get a different answer, then you are probably misusing the EXP button. To enter this number, you would enter 1.05, press the EXP button, and then enter -5 (probably by entering 5 and then pressing the +/- button). People often try to enter x 10 in the middle of this process as well. The EXP button includes this.
Example 2 These calculations are very simple if you have a compound in which the numbers of positive and negative ions are 1 : 1. This next example shows you how to cope if the ratio is different. The solubility of magnesium hydroxide at 298 K is 1.71 x 10-4 mol dm-3. Calculate the solubility product. The equilibrium is: For every mole of magnesium hydroxide that dissolves, you will get one mole of magnesium ions, but twice that number of hydroxide ions. So the concentration of the dissolved magnesium ions is the same as the dissolved magnesium hydroxide: [Mg2+] = 1.71 x 10-4 mol dm-3 The concentration of dissolved hydroxide ions is twice that: [OH-] = 2 x 1.71 x 10-4 = 3.42 x 10-4 mol dm-3 Now put these numbers into the solubility product expression and do the sum. Calculating solubilities from solubility products Reversing the sums we have been doing isn't difficult as long as you know how to start. We will take the magnesium hydroxide example as above, but this time start from the solubility product and work back to the solubility. If the solubility product of magnesium hydroxide is 2.00 x 10-11 mol3 dm-9 at 298 K, calculate its solubility in mol dm-3 at that temperature. The trick this time is to give the unknown solubility a symbol like x or s. I'm going to choose s, because an x looks too much like a multiplication sign. If the concentration of dissolved magnesium hydroxide is s mol dm-3, then: [Mg2+] = s mol dm-3 [OH-] = 2s mol dm-3 Put these values into the solubility product expression, and do the sum.

AN INTRODUCTION TO SOLUBILITY PRODUCTS

AN INTRODUCTION TO SOLUBILITY PRODUCTS This page looks at how solubility products are defined, together with their units. It also explores the relationship between the solubility product of an ionic compound and its solubility. What are solubility products, Ksp? Solubility products are equilibrium constants Barium sulphate is almost insoluble in water. It isn't totally insoluble - very, very small amounts do dissolve. That's true of any so-called "insoluble" ionic compound. if you shook some solid barium sulphate with water, a tiny proportion of the barium ions and sulphate ions would break away from the surface of the solid and go into solution. Over time, some of these will return from solution to stick onto the solid again. You get an equilibrium set up when the rate at which some ions are breaking away is exactly matched by the rate at which others are returning. The position of this equilibrium lies very far to the left. The great majority of the barium sulphate is present as solid. In fact, if you shook solid barium sulphate with water you wouldn't be aware just by looking at it that any had dissolved at all. But it is an equilibrium, and so you can write an equilibrium constant for it which will be constant at a given temperature - like all equilibrium constants. The equilibrium constant is called the solubility product, and is given the symbol Ksp. To avoid confusing clutter, solubility product expressions are often written without the state symbols. Even if you don't write them, you must be aware that the symbols for the ions that you write are for those in solution in water. Why doesn't the solid barium sulphate appear in the equilibrium expression? For many simple equilibria, the equilibrium constant expression has terms for the right-hand side of the equation divided by terms for the left-hand side. But in this case, there is no term for the concentration of the solid barium sulphate. Why not? This is a heterogeneous equilibrium - one which contains substances in more than one state. In a heterogeneous equilibrium, concentration terms for solids are left out of the expression.
	Note:  The simplest explanation for this is that the concentration of a solid can be thought of as a constant. Rather than have an expression with two constants in it (the equilibrium constant and the concentration of the solid), the constants are merged to give a single value - the solubility product.
Solubility products for more complicated solids Here is the corresponding equilibrium for calcium phosphate, Ca3(PO4)2: And this is the solubility product expression: Just as with any other equilibrium constant, you raise the concentrations to the power of the number in front of them in the equilibrium equation. There's nothing new here. Solubility products only apply to sparingly soluble ionic compounds You can't use solubility products for normally soluble compounds like sodium chloride, for example. Interactions between the ions in the solution interfere with the simple equilibrium we are talking about. The units for solubility products The units for solubility products differ depending on the solubility product expression, and you need to be able to work them out each time. Working out the units in the barium sulphate case Here is the solubility product expression for barium sulphate again: Each concentration has the unit mol dm-3. So the units for the solubility product in this case will be: (mol dm-3) x (mol dm-3) = mol2 dm-6 Working out the units in the calcium phosphate case Here is the solubility product expression for calcium phosphate again: The units this time will be: (mol dm-3)3 x (mol dm-3)2 = (mol dm-3)5 = mol5 dm-15 If you are asked to calculate a solubility product in an exam, there will almost certainly be a mark for the correct units. It isn't very hard - just take care! Solubility products apply only to saturated solutions Let's look again at the barium sulphate case. Here is the equilibrium expression again: . . . and here is the solubility product expression: Ksp for barium sulphate at 298 K is 1.1 x 10-10 mol2 dm-6. In order for this equilibrium constant (the solubility product) to apply, you have to have solid barium sulphate present in a saturated solution of barium sulphate. That's what the equilibrium equation is telling you. If you have barium ions and sulphate ions in solution in the presence of some solid barium sulphate at 298 K, and multiply the concentrations of the ions together, your answer will be 1.1 x 10-10 mol2 dm-6. What if you mixed incredibly dilute solutions containing barium ions and sulphate ions so that the product of the ionic concentrations was less than the solubility product? All this means is that you haven't got an equilibrium. The reason for that is that there won't be any solid present. If you lower the concentrations of the ions enough, you won't get a precipitate - just a very, very dilute solution of barium sulphate. So it is possible to get an answer less than the solubility product when you multiply the ionic concentrations together if the solution isn't saturated. Can you get an answer greater than the solubility product if you multiply the ionic concentrations together (allowing for any powers in the solubility product expression, of course)? No! The solubility product is a value which you get when the solution is saturated. If there is any solid present, you can't dissolve any more solid than there is in a saturated solution.
	Note:  In the absence of any solid, a few substances produce unstable supersaturated solutions. As soon as you add any solid, or perhaps just scratch the glass to give a rough bit that crystals can form on, all the excess solid precipitates out to leave a normal saturated solution.
If you mix together two solutions containing barium ions and sulphate ions and the product of the concentrations would exceed the solubility product, you get a precipitate formed. Enough solid is produced to reduce the concentrations of the barium and sulphate ions down to a value which the solubility product allows.