Chemistry Lesson Part 2

continuation…

Now P(A) is in direct proportion to [A] and the same for B. So we can conclude that the probability of getting m-molecules of A and n-molecules of B in close enough proximity and with high enough energy to cause a reaction is directly related to the product [A]^m*[B]^n and thus the rate of the reaction, a measure of how many times those individual reactions are taking place is also directly related to this product. This very long discussion leads us to a very nice result.

   Suppose mA + nB <--> xC + yD is a balanced equation of a reaction in equilibrium. We have shown that the forward rate, F=k [A]^m*[B]^n, for some constant of proportionality k, and through the same reasoning, the reverse rate, R=c [C]^x*[D]^y, for some constant c. Now, equilibrium was defined as the state when F=R so (drum roll please) we can define a constant called Keq, the so called equilibrium constant for this equation at this temperature, defined as:
Keq = k/c = [C]^x*[D]^y/[A]^m*[B]^n.    This constant, Keq, is singular to this equation at this temperature. If we add various amounts of one or more of the substances A,B,C,D to the mixture in equilibrium, it will reestablish equilibrium in such a way that the relative concentrations still satisfy Keq.  They might all be different but the ratio will hold.

Example: In water at 25 degrees Celsius (room temperature), the water molecules H2O have a certain kinetic energy distribution in which some of the molecules will have sufficiently high energy so that their collisions could free a Hydrogen nucleus from the electron it is sharing with a greedier Oxygen nucleus. (When two H atoms share their respective electrons with the O atom to fill the outer shell of the O, the bond between the H's and O is a polar covalent bond, meaning that there is a slight negative charge on one side of the water molecule and two slight positive charges on the opposite side due to the  greedier pull of the O on the shared electrons.) This freed H-nucleus, H+, essentially a proton, has a positive charge and quickly aligns it self with another water molecule, drawn by the negatively charged O-side of the H2O molecule. The H2O molecule that lost the H+, but held its electron, is now a hydroxide ion OH- and there is a corresponding hydronium ion H3O+. That is we have an ongoing reaction which we can write in two ways (equivalent):
H2O <--> H+  +  OH-    or       2H2O <--> H3O+  +  OH-

Again, once H+ is freed up it is captured by another H2O to become a H3O+. So [H+]=[H3O]+ and they are just two ways of describing the hydronium ion.

In this case one of our reactants, H2O, is a solvent which means its molecules are everywhere, it has a concentration of 1. Thus we get Keq= [H+][OH-] which we can show is Keq=10^-14, and since the concentrations of hydronium ions and hydroxide ions are the same in pure water auto ionization, they are both 10^-7. Defining the function p=-log, we get pH = p([H+]) =7 for pure water.  :) :)