# Question 0318d

Mar 4, 2017

The reaction is endothermic, and ["H"_3"O"^"+"] = ["OH"^"-"] = 3 × 10^"-7"color(white)(l) "mol/L" at 60 °C.

How do we know that the reaction is endothermic?

At 25 °C, $\text{2H"_2"O" ⇌ "H"_3"O"^"+" + "OH"^"-}$; K_text(w) = 1.00 × 10^"-14"

At 60 °C, $\text{2H"_2"O" ⇌ "H"_3"O"^"+" + "OH"^"-}$; K_text(w) = 1 × 10^"-13"

The value of ${K}_{\textrm{w}}$ increases when we increase the temperature, i.e. the position of equilibrium moves to the right.

Per Le Châtelier's Principle, adding heat to a reaction that absorbs heat will cause the system to respond in a way that removes the heat.

Since the ${K}_{\textrm{w}}$ increases, the autoionization of water must be endothermic.

What are the concentrations of $\text{H"_3"O"^"+}$ and $\text{OH"^"-}$ at 60 °C?

$\text{2H"_2"O" ⇌ "H"_3"O"^"+" + "OH"^"-}$
$\textcolor{w h i t e}{m m m m m l l} x \textcolor{w h i t e}{m m m l} x$

K_text(w) = ["H"_3"O"^"+"]["OH"^"-"] = x^2 = 1 × 10^"-13"

x = sqrt(1 × 10^"-13") = 3.2 × 10^"-7"

["H"_3"O"^"+"] = ["OH"^"-"] = x color(white)(l)"mol/L" = 3 × 10^"-7" color(white)(l)"mol/L"#