#"pH = 2.68"#
#["H"_3"O"^"+"] = 10^"-pH" color(white)(l)"mol/L" = 10^"-2.68"color(white)(l)"mol/L" = "0.002 09 mol/L"#
We can use an ICE table to help us calculate the value of #K_text(a)#.
#color(white)(mmmmmmll)"HA" + "H"_2"O" ⇌ "H"_3"O"^"+" + color(white)(l)"A"^"-"#
#"I/mol·L"^"-1": color(white)(mll)0.21 color(white)(mmmmmm)0color(white)(mmml)0#
#"C/mol·L"^"-1": color(white)(ml)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+"x#
#"E/mol·L"^"-1": color(white)(m)"0.21-"xcolor(white)(mmmmml)xcolor(white)(mmml)x#
At equilibrium,
#["H"_3"O"^"+"] = "0.002 09 mol/L"#
∴ #x = "0.002 09"#
#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = x^2/(0.21-x) = "0.002 09"^2/("0.21-0.002 09") = (4.16 × 10^"-6")/0.208#
#= 2.1 × 10^"-5"#
