I would argue that the temperature has nothing to do with the value of #K_text(a)#.
In this problem, it is the #"pH"# and the initial concentration of the acid that determine the value of #K_text(a)#.
Let's set up an ICE table to solve this problem.
#color(white)(mmmmmmm)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#
#"I/mol·L"^"-1":color(white)(mml)6color(white)(mmmmmml)0color(white)(mmm)0#
#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)"+"xcolor(white)(mml)"+"x#
#"E/mol·L"^"-1":color(white)(ml)"6 -"color(white)(l)xcolor(white)(mmmmml)xcolor(white)(mmll)x#
We must use the #"pH"# to calculate the value of #x#.
#"pH = 2.5"#
∴ #["H"_3"O"^"+"] = 10^"-2.5"color(white)(l) "mol/L" = "0.0031 mol/L" = x#
The #K_"a"# expression is:
#K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x) = x^2/(0.100-x) = 0.0031^2/(6 - 0.0031) = (1.00 × 10^"-5")/6 = 2 × 10^"-6"#
