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, "2H"_2"O" ⇌ "H"_3"O"^"+" + "OH"^"-"; K_text(w) = 1.00 × 10^"-14"
At 60 °C, "2H"_2"O" ⇌ "H"_3"O"^"+" + "OH"^"-"; K_text(w) = 1 × 10^"-13"
The value of K_text(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_text(w) increases, the autoionization of water must be endothermic.
What are the concentrations of "H"_3"O"^"+" and "OH"^"-" at 60 °C?
"2H"_2"O" ⇌ "H"_3"O"^"+" + "OH"^"-"
color(white)(mmmmmll)xcolor(white)(mmml)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"
