# What happens to the pH of water as the temperature is raised ?

Aug 16, 2015

This is good question, and can be answered by reflecting on the bond-breaking nature of the acid-base reaction.

#### Explanation:

We write the autoprotolysis of water as:

${H}_{2} O r i g h t \le f t h a r p \infty n s {H}^{+} + O {H}^{-}$

${K}_{w}$ is quoted at 298K and is equal to ${10}^{-} 14$, so as you know the equilibrium should lie strongly to the left. Since the extent of reaction is so small, we would be justified in saying that this is an endothermic reaction.

Another way to look at it is to consider that it is a bond-breaking reaction that requires energy to break a strong $O - H$ bond. Given that it is a bond-breaking reaction, increase of temperature would probably drive the reaction further to the right, and result in greater concentrations of ${H}^{+}$. Since, by definition, $p H = - {\log}_{10} \left\{{H}^{+}\right\}$, $p H$ should indeed decrease at higher temperature.

Aug 16, 2015

The pH decreases because the concentration of hydrogen ions increases.

#### Explanation:

The dissociation of water is given by:

${H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$

${K}_{w} = {10}^{- 14} \text{ } m o {l}^{2.} {\mathrm{dm}}^{- 6}$$\text{ at "25^0"C}$

and $\Delta H$ is positive.

This means that as the temperature increases Le Chatelier's Principle predicts that more water will dissociate as the position of equilibrium shifts to the right.

This means the concentration of ${H}^{+}$ ions increases.

Since $p H = - \log \left[{H}^{+}\right]$ this means the $p H$ will decrease.

At ${100}^{0} \text{C}$ the $p H$ drops to 6.14.

This does not mean that the water has become more acidic since $\left[{H}^{+}\right]$ will always equal $\left[O {H}^{-}\right]$ for pure water.