When dealing with very large or very small numbers, it is often convenient to use logarithms, as the unwealdy numbers then become easier to use. |

Logarithms

A logarithm is the number to which 10 (or the base being used) must be raised to obtain the required number.

100 = 10 1000 = 10 |

The base (10 in this case) is sometimes written after the word 'log' to avoid any doubt, as many areas of science use the natural log base 'e' usually written as 'ln' (natural log). However, the assumption is often made that the word 'log' means log(10) by default.

Logarithms are used to turn either very large, or very small numbers, into numbers that are rather more convenient to handle.

Examples Avogadro's number 6.02 x 10 i.e. log The ionic product of water = 1 x 10 Log |

Definition of 'p'

The letter 'p' is used to indicate the negative logarithm (base
10) of a number. This is used when the numbers are likely to be very small,
so that the final value becomes conveniently sized __and__ positive.

The letter 'p' is never a capital in this form, it stands for the word ' potential'.

Example The ionic product of water, Kw = 1 x 10 pKw is the negative log of Kw Log Kw = -14 Therefore pKw = 14 |

Definition of pH

pH is defined as the negative of the logarithm (base 10) of the hydrogen ion concentration

pH = -log [H |

Example At 25ºC, the hydrogen ion concentration of pure water is 1 x 10 The logarithm of 1 x 10 The negative of -7 = +7 Therefore the pH of pure water at 25ºC = 7 |

Definition of pOH

This is basically (no pun intended) the same as pH, but from the point of
view of the OH- ions. Thus, the negative logarithm of the OH^{-} ion
concentration, [OH^{-}], gives the pOH.

Note: at 25ºC, pH + pOH = 14. It is therefore a simple matter to obtain one from the other.

Example: Show that pKw = pH + pOH Kw = [H Kw = 1 x 10 [H pH = 7, pOH = 7 and pKw = 14 Therefore pKw = pH + pOH = 14 (at 25ºC) |

Calculations involving pH and pOH

If we are dealing with a strong acid, or base, then calculations are straightforward. It is a matter of treating the hydrogen ion concentration as the molar concentration of the acid (or base) for monobasic acids (or bases).

Example: Calculate the pH of 0.25 M hydrochloric acid As hydrochloric acid dissociates 100% according to the equation
Then 0.25 M hydrochloric acid gives 0.25 mol dm The pH of this solution is: pH = - log 0.25 = 0.6 |

Example: Calculate
the pH of 0.1 mol dm As potassium hydroxide dissociates 100% according to the equation
Then 0.1 M potassium hydroxide gives 0.1 mol dm The pOH of this solution is: pOH = - log 0.1 = 1 And as: pOH + pH = 14 therefore pH = 14 -1 = 13 |

When dealing with a weak acid (or base) the degree of dissociation into ions must be known, as given by the Ka (or pKa). This is dealt with in the next section.