pH, pOH, pK_{a}, and pK_{b}
To calculate the pH of an aqueous solution you need to know the concentration of the hydronium ion in moles per liter (molarity). The pH is then calculated using the expression:
pH = - log [H_{3}O^{+}].
Example: Find the pH of a 0.0025 M HCl solution. The HCl is a strong acid and is 100% ionized in water. The hydronium ion concentration is 0.0025 M. Thus:
pH = - log (0.0025) = - ( - 2.60) = 2.60
The hydronium ion concentration can be found from the pH by the reverse of the mathematical operation employed to find the pH.
[H_{3}O^{+}] = 10^{-pH} or [H_{3}O^{+}] = antilog (- pH)
Example: What is the hydronium ion concentration in a solution that has a pH of 8.34?
To calculate the pOH of a solution you need to know the concentration of the hydroxide ion in moles per liter (molarity). The pOH is then calculated using the expression:
pOH = - log [OH^{-}]
Example: What is the pOH of a solution that has a hydroxide ion concentration of 4.82 x 10^{-5} M?
pOH = - log [4.82 x 10^{-5}] = - ( - 4.32) = 4.32
Calculating the Hydroxide Ion Concentration from pOH
The hydroxide ion concentration can be found from the pOH by the reverse mathematical operation employed to find the pOH.
[OH^{-}] = 10^{-pOH} or [OH^{-}] = antilog ( - pOH)
Example: What is the hydroxide ion concentration in a solution that has a pOH of 5.70?
5.70 = - log [OH^{-}]
-5.70 = log[OH^{-}]
[OH^{-}] = 10^{-5.70} = 2.00 x 10^{-6} M
On a calculator calculate 10^{-5.70}, or "inverse" log (- 5.70).
Relationship Between pH and pOH
The pH and pOH of a water solution at 25^{o}C are related by the following equation.
pH + pOH = 14
If either the pH or the pOH of a solution is known, the other can be quickly calculated.
Example: A solution has a pOH of 11.76. What is the pH of this solution?
pH = 14 - pOH = 14 - 11.76 = 2.24
The pK_{a} is calculated using the expression:
pK_{a} = - log (K_{a})
where "K_{a}" is the equilibrium constant for the ionization of the acid.
Example: What is the pK_{a} of acetic acid, if K_{a} for acetic acid is 1.78 x 10^{-5}?
pK_{a} = - log (1.78 x 10^{-5}) = - ( - 4.75) = 4.75
Calculating K_{a} from the pK_{a}
The K_{a} for an acid is calculated from the pK_{a} by performing the reverse of the mathematical operation used to find pK_{a}.
K_{a} = 10^{-pKa} or K_{a} = antilog ( - pK_{a})
Example: Calculate the value of the ionization constant for the ammonium ion, K_{a}, if the pK_{a} is 9.74.
9.74 = - log (K_{a})
-9.74 = log (K_{a})
K_{a} = 10^{-9.74} = 1.82 x 10^{-10}
On a calculator calculate 10^{-9.74}, or "inverse" log ( - 9.74).
The pK_{b} is calculated using the expression:
pK_{b} = - log (K_{b})
where K_{b} is the equilibrium constant for the ionization of a base.
Example: What is the pK_{b} for methyl amine, if the value of K_{b} for methyl amine is 4.4 x 10^{-4}?
pK_{b} = - log (4.4 x 10^{-4}) = - ( - 3.36) = 3.36
The K_{b} for an acid is calculated from the pK_{b} by performing the reverse of the mathematical operation used to find pK_{b}.
K_{b} = 10^{-pKb} or K_{b} = antilog ( - pK_{b})
Example: Calculate the value of the ionization constant, K_{b}, for aniline if the pK_{b} is 9.38.
9.38 = - log (K_{b})
-9.38 = log (K_{b})
K_{b} = 10^{-9.38} = 4.17 x 10^{-10}
On a calculator calculate 10^{-9.38}, or "inverse" log ( - 9.38).