# For the water solvent, how does pK_w evolve at HIGHER temperatures...? At 298*K, K_w=10^-14...

Nov 13, 2016

$p H + p O H = 14$, given a temperature of $298 \cdot K$

#### Explanation:

$2 {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} + H {O}^{-}$

As you have correctly interpreted, this is a bond-breaking reaction, which requires substantial energy input to break the strong $O - H$ bond, and the small equilibrium constant, ${10}^{-} 14$ (at $298 \cdot K$), reflects this strength. Now this reaction was exhaustively measured at a precise temperature of $298 \cdot K$. At higher temperatures, given a bond breaking reaction, we would expect that the equilibrium should lie farther over to the RHS, the product side, and thus $p H$ should DECREASE at higher temperatures.

And thus at $100$ ""^@C, $p H = 6.14$, as anticipated (this is almost a tenfold increase in the extent of reaction!). See the lower part of here.. Of course under this scenario, $p O H = 6.14$ necessarily.

Note that typically any equilibrium is a function of temperature. A standard temperature of $298 \cdot K$ is assumed for simplicity.