Equilibrium Systems - Concept

Hi guys, so we're back talking about conjugate acid base equilibrium systems so before we talk about acid base specifically, we should discuss a little more about equilibrium with which you should already be familiar but a few pointers here so that we can get through the acid base application. So remember that k sub c is the equilibrium expression, c meaning that it's based on concentrations as opposed to partial pressures so k sub c the equilibrium expression is equal to the concentration of products over the concentration of reactants for the most basic reaction where we just have reactants in equilibrium with products. We're calling it the double headed arrow means the rate of the forward reaction is equal to the rate of the backward reaction so when we have an equilibrium expression much much greater than 1 that means that the equilibrium lies to the right and out products are favored and simple Math will tell us that right, if our numerator has a greater concentration than our denominator then our kc should be greater than 1. On the other side of that is if our equilibrium expression is much much less than 1 then our equilibrium lies to the left and then our reactants are favored right, if our denominator is greater than our numerator then our value should be less than 1 so let's make it a little more specific now for acid base.

So remember that acid base involves protons transfer. Okay so here then the proton transfer reactions can be thought of being governed by the ability of two bases to abstract a proton. Remember an acid is a proton donor and a base is a proton acceptor so here from our generic example we have H sub x sorry Hx denoting our acid and here water acting as a base, our double headed arrow where black indicates the forward direction and the blue arrow indicates the backwards direction and we have our hydronium ion H3O+ as our conjungate acid and x- as our conjugate base. Okay so in this case because water is a stronger base then x- which here is just some generic acid the equilibrium will lie to the right, right because the tendency here if this is the strong base is for water to extract this proton from Hx and to make the hydronium ion an x- okay so if in fact x- happens to be the stronger base if it's stronger than water then that means that the tendency will be for the back reaction to be favored so that x- would abstract the proton from the hydronium ion to make water an Hx and that would mean that the equilibrium will lie to the left.

So in every acid base reaction, the position of the equilibrium favors the transfer of the proton to the stronger base as I just indicated in the last example so then the equilibrium position favors the reaction of the stronger acid and the stronger base to form the weaker acid base pair.

So now it's time for a few pitfalls, so to solve these problems you're going to need a chart available to reference the relative strength of the conjugate acid base pairs you can find one of those very easily online or in a textbook. So here's a very simple example here where we have ammonium ion and hydroxide which we know hydroxide is a strong base forming an equilibrium with ammonium and water so in this case if you have a chart handy you'll find out that hydroxide is a stronger base than is water so that means that the tendency here is going to be for this hydroxide to abstract the proton from the ammonium ion to go forward and make the products NH3 and water so since the forward reaction is what's favored because in this case hydroxide is a stronger base then the equilibrium will lie to the right meaning that our products, right, are more favored than our reactants and that's the simplest way to describe equilibrium systems in acid base.