 Hi students, welcome to Year 12 Chemistry and Module 5 Equilibrium and Acid Reactions. This is video number 16 and we're going to now look at acid and base solutions and their relationship to the equilibrium constant. And yes, you found it, the actual acid reaction section of the equilibrium and acid reactions topic. It's interesting because we're not really even looking at any reactions of acids except for the fact that they do some interesting things when we put them in water. We will be looking at acids and bases in a lot more detail in the next module, but for now we can look at the fact that when acids are dissolved or at least added to water they can ionize and interact with water molecules in order to change the nature of the solution. And just like ions in solution we can work out what the equilibrium expression is for these processes. So here I have an example of a weak acid and a weak acid is an acid which does not fully ionize as opposed to a strong acid which does. So I've used an example like hydrochloric acid because it's not going to set up an equilibrium. That acid completely ionizes in water and so therefore would be a single directional arrow rather than the two headed arrow that we use for equilibrium expressions. And for weak acids you find that in the solution exist both the molecular form of the acid which is the H2CO3 molecule as well as the ionic forms after it has been ionized in water. What you can actually see is we are transferring the hydrogen ion from the acid into the water molecule. So we produce these molecules called hydronium ions, sometimes also written as H+, just be aware of that and from that we do calculate things like the pH and also the anion which is what is left over from the acid once the hydrogen ions have been removed. So in this case that would be the carbonate ion. The expression for the equilibrium constant as always is products, concentration of products over concentration of reactants and therefore in this case we would have our H3O+, raised to the power of 2 multiplied by our concentration of CO32-, and that's our products and our reactants are concentration of H2CO3 multiplied by water which because there's a 2 in front we would raise to the power of 2. Of course from previous examples we know that because water is existing in its liquid form this part is a constant and so therefore even if we have changed the volume of water we also changed the number of moles of water and so therefore its overall concentration doesn't change and when we multiply a constant by a constant we just get another constant. So the expression for this particular acid becomes the equilibrium is equal to the concentration of H3O+, and as we'll look at later on this can actually be termed from the pH of the solution and then we look at H2CO3 down the bottom here. This type of equilibrium that involves acids is often given a specific name called the acid ionization constant and it also often has a subscript of Ka indicating that this is equilibrium specifically related to acid solutions. If you assume that all acids are going to ionize in water in the same way and therefore the water will not become part of this particular reaction then we can write an expression that is a general expression that is the concentration of hydrogen ions and I'm just reducing the hydrogen ion concentration down multiplied by the concentration of anions assuming a one to one ratio for the concentration divided by the concentration of HA where HA is the actual acid itself. This is a general form of the acid ionization constant but it works for weak acids.