Why does pH + pOH = 14?

Question

Noah Aldrich · Accepted Answer

Because water undergoes a measurable equilibrium at #298*K#.

The equilibrium can be expressed as follows: #2H_2OrightleftharpoonsH_3O^+ + HO^-# And, #K_w# #=# #[H_3O^+][HO^-]#. This equilibrium has been carefully measured (by conductivity experiments) over a range of temperatures. #[H_2O]# does not appear in the equilibrium because its concentration is effectively constant. At #298*K# we know that #K_w = 10^-14.# How do we know this? By careful and repeated measurements of the conductivity of the water solvent. So #K_w = [H_3O^+][HO^-]=10^-14.# This is an arithmetic expression, so we can take the logarithm of both sides to get: We can divide, multiply, add to, and subtract from it as long as we do it to both sides of the equation. #log_10K_w = log_10[H_3O^+]+ log_10[HO^-]=log_10(10^-14)# But, by definition of logarithms, if #a^b=c#, #log_(a)c=b#, then #log_10(10^-14)=-14#, and thus, #log_10K_w = log_10[H_3O^+]+ log_10[HO^-]=-14# But by definition, #-log_10[H_3O^+]=pH#, and #-log_10[HO^-]=pOH# #pK_w = pH+ pOH=14# as required. Given that this is a bond-breaking reaction, how would you expect the equilibrium to evolve at temperatures greater than #298K#? Would #pH# increase or decrease?

Caleb Adams · Answer

undefined pH + pOH = 14 because it is a fundamental property of water at 25°C. In water, the concentration of hydrogen ions (H⁺) multiplied by the concentration of hydroxide ions (OH⁻) always equals 1.0 x 10^(-14) M². Taking the negative logarithm (base 10) of both sides of this equation yields pH + pOH = 14.