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)(ml)0.100color(white)(mmmmml)0color(white)(mmm)0#
#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)+xcolor(white)(ml)+x#
#"E/mol·L"^"-1":color(white)(m)"0.100-"xcolor(white)(mmmmll)xcolor(white)(mmll)x#
We must use the #"pH"# to calculate the value of #x#.
#"pH = 2.00"#
#["H"_3"O"^"+"] = 10^"-2.00" color(white)(l)"mol/L" = "0.0100 mol/L" = x#
The #K_"a"# expression is:
#K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x) #
∴ #K_"a" = x^2/(0.100-x) = 0.100^2/(0.100-0.010) = 0.0100/0.090 =0.11#
