You know how to calculate the pH of an acid in solution or a base in solution, but calculating the pH of two acids or two bases in solution is a little bit trickier. Using the formula described below, you can estimate the pH for a monoprotic two-chemical mixture of this kind. This equation neglects the autoionization of water, since the value for water will make a negligible contribution to the pH in any case.
Click on Resources link to retrieve a table of dissociation constants (pKa values) for various common acids. Write down the pKa values for the two acids you are using.
Convert pKa values to Ka values using the following equation: Ka = 10^-pKa, so if pKa is 5, Ka = 10^-5.
Write down the following formula: [H+]^2 = Ka1 [A1] + Ka2 [A2], where [H+] is the concentration of hydrogen ions and A1 and A2 are the concentrations of the acids you are using. Remember that in chemistry, brackets like [ ] denote concentrations.
Plug in the concentrations of your acids and their Ka values, then calculate [H+]. For example, suppose acid 1 has a Ka of 1 x 10^-5, while acid 2 has a Ka of 3 x 10^-4, and both are present in 1 molar concentration. Your equation would be as follows: [H+]^2 = 1 x 10^-5 [1] + 3 x 10^-4 [1] [H+]^2 = 1 x 10^-5 + 3 x 10^-4 [H+]^2 = 3.1 x 10^-4 [H+] = 0.018
Take the log of the value you just found on your calculator, then take its negative. The -log of an [H+] concentration is the pH. For example: log 0.018 = -1.75, therefore, pH = 1.75.
Click on the Resources link to retrieve a table of dissociation constants (pKb values) for various common bases. Write down the pKb values for the two bases you are using.
Convert pKb values to Kb values by using the following equation: Kb = 10^-pKb, so if pKb is 5, Kb = 10^-5.
Write down the following formula: [OH-]^2 = Kb1 [B1] + Kb2 [B2], where [OH-] is the concentration of hydroxide ions and B1 and B2 are the concentrations of the bases you are using.
Plug in the concentrations of your bases and their Kb values, then calculate [OH-]. For example, suppose base 1 has a Kb of 1 x 10^-5, while base 2 has a Kb of 3 x 10^-4, and both are present in 1 molar concentration. Your equation would be as follows: [OH-]^2 = 1 x 10^-5 [1] + 3 x 10^-4 [1] [OH-]^2 = 1 x 10^-5 + 3 x 10^-4 [OH-]^2 = 3.1 x 10^-4 [OH-] = 0.018
Take the log of the value you just found on your calculator, then take its negative. The -log of an [OH-] concentration is the pOH. For example, log 0.018 = -1.75, therefore, pOH = 1.75.
Subtract the pOH from 14 to get the pH. For example, 14.0 - 1.75 = pH = 12.3.
