Chemistry 112

pH + pOH = 14.00 at 25 °C

This useful relationship can be derived starting with K_w:

K_w = [H₃O⁺]_e[OH^–]_e = 1.0×10^–14

Taking the log of both sides gives

log([H₃O⁺]_e[OH^–]_e) = log(1.0×10^–14) = –14.00

Using the property of logs that the log of a product is the sum of the logs of the multipliers gives

log([H₃O⁺]_e) + log([OH^–]_e) = –14.00

Multiplying both sides by –1 gives

–log([H₃O⁺]_e) + –log([OH^–]_e) = 14.00

Using the definitions of pH ( = –log[H₃O⁺]_e ) and pOH ( = –log[OH^–]_e ) gives

pH + pOH = 14.00

This relationship is only true at 25 °C because the value of K_w changes at all other temperatures. The 14.00 really should be written as –logK_w = pK_w, so a more general relationship is

pH + pOH = pK_w

which is true at all temperatures.