# How to Calculate Ph And Poh

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As a chemistry student, one of the basic skills you learn is how to calculate the pH and pOH of acids and bases. The concepts and the calculation aren’t difficult if you’re familiar with logarithms and the concentration of solutions.

#### TL;DR (Too Long; Didn't Read)

To calculate pH = - log(H3O+ ion concentration). Calculation for pOH is - log(OH- ion concentration).

## Meaning of pH and pOH

For acids and bases, the solution concentration involves numbers that can change over a wide range of values – over a million to one. Unlike most units, which are linear, pH and pOH are both based on the common (base 10) logarithm, letting you express values in one or two digits that would otherwise span many orders of magnitude. Though this takes getting used to, the compactness of pH and pOH units is convenient and saves time and confusion. The pH unit indicates acidity, where smaller numbers mean greater concentrations of H3O+ (hydronium) ions, and range from over 14 (very alkaline) down to negative numbers (very acid; these negative numbers are used mainly for academic purposes). On this scale, deionized water has a pH of 7. The pOH scale is much like pH, but reversed. It uses the same numbering system as pH, but measures OH- ions. On this scale, water has the same value (7), but you'll find bases at the low end and acids at the high end.

## Calculating pH

To calculate pH from the molar concentration of an acid, take the common log of the H3O+ ion concentration, and then multiply by -1: pH = - log(H3O+). For example, what is the pH of a 0.1 M solution of hydrochloric acid (HCl), assuming the acid is completely disassociated into ions in the solution? The concentration of H3O+ ions is 0.1 moles per liter. pH = - log(.1) = -(-1) = 1.

## Calculating pOH

The calculation for pOH follows the same rules as for pH, but uses the concentration of OH- ions: pOH = - log(OH-). For example, find the pOH of a 0.02 M solution of sodium hydroxide (KOH). The concentration of OH- ions is 0.02 moles per liter. pOH = - log(.02) = -(-1.7) = 1.7.