Calculations with pH

The pH scale was devised in order to have a convenient measure of acidity in an aqueous solution.  There is a tremendous range in the value of the concentration of the H₃O⁺ ion in ordinary aqueous solutions.  Consider the following solutions, first a 10 M solution of HCl (not exactly ordinary, but certainly possible) and secondly a solution which 10 M in NaOH.  These two solutions would be very close to the upper and lower limit of the concentration of H₃O⁺ ion in an aqueous solution. In the 10 M HCl solution the [H₃O⁺] would be 10 M, while in the 10 M solution of NaOH the [H₃O⁺] would be 1.0 x 10^-15 M.  Notice what a range in magnitude this involves, 10 to 10^-15.  The pH scale compresses this range for us.  In terms of pH for the two solutions above the range would be from -1 to 15.

The symbol pH comes from power of Hydrogen ion concentration, aren't chemists clever?  pH is defined as follows:

pH = -log[H₃O⁺] 

The negative sign is so that the pH of all solutions with a  H₃O⁺ ion concentration of less than 1 M will have a positive value.  This is true most of the time so pH will usually be a positive value.  pH can have a negative value this will be the case if we have a solution with H₃O⁺ ion concentration greater than one molar.  The price we have to pay by having the negative sign in the definition of pH is that we have an inverse relationship between pH and acidity. That is the higher the acidity of the solution the lower the pH.  A pH of 7.0 means that we have a neutral solution, a pH below 7.0 means an acidic solution, while a pH above 7.0 means that we have a basic solution.  A pOH scale analogous to the pH scale can be devised using the negative logarithm of the hydroxide ion concentration of a solution.

pOH = -log[OH^-] 

Next lets look at the relationship between pH and pOH.

K_w =  [H₃O⁺][OH^-] = 1.0 x 10^-14 

Taking the negative logarithm of both sides we obtain

-log[H₃O⁺] -log[OH^-]  = 14.00

From the definitions of pH and pOH we get

pH + pOH = 14.00

  Example 1

The H₃O⁺ ion concentration in a freshly opened bottle of wine was 3.5 x 10^-4 M, the unused portion of the wine after standing open to the air for three weeks was found to have a H₃O⁺ ion concentration of 1.1 x 10^-3 M.  Calculate the pH of the wine in each of these situations.

The fresh wine:

pH = -log[H₃O⁺] = -log(3.5 x 10^-4) = 3.46

After the wine had partially oxidized to vinegar:

pH = -log[ H₃O⁺] = -log(1.1 x 10^-3) = 2.96