Question #e67f8

As you know, the pH of pure water is based on water's self-ionization reaction which results in the production of hydronium and hydroxide ions.

#H_2O_((l)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + OH_((aq))^(-)#

The equilibrium constant for this reaction, #K_(eq)#, is equal to

#K_(eq) = ([H_3O^(+)] * [OH^(-)])/([H_2O]^2)#

Since water's concentration is assumed to be constant at any given temperature, you can write this expression as

#underbrace(K_(eq) * [H_2O]^2)_(color(blue)(=K_W)) = [H_3O^(+)] * [OH^(-)]#

The product between #K_(eq)# and the concentration of water is called the ion product of water and is equal to

#K_W = [H_3O^(+)] * [OH^(-)]#

Now, if #K_W# increases with themperature, then the concentrations of the hydronium and hydroxide ions formed by the self-ionization reaction will increase as well.

That happens because the formation of the hydronium and hydroxide ions is actually an endothermic process. That means that the equilibrium will counteract any increase in temperature by favoring the forward reaction, since producing hydronium and hydroxide ions consumes energy (think Le Chatelier's Principle.

At any temperature, the pH of pure water is given by the equation

#pH_"water" = -log([H_3O^(+)])#

SInce you're dealing with the negative log, any increase in the concentration of the hydronium ions will result in a smaller value for the pH.

However, you need to remember that the water will still be neutral, even if the pH is less than 7!

That happens because the concentrations of hydronium and hydroxide ions are equal, so you can't say that the pH is acidic.