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"#
