Let's set up an ICE table for the calculation of #["H"_3"O"^+]#.
#color(white)(mmmmmmm)"HF"color(white)(l) +color(white)(l) "H"_2"O"color(white)(l) ⇌ color(white)(l) "H"_3"O"^+color(white)(l) +color(white)(l) "F"^"-"#
#"I/mol·L"^"-1": color(white)(mll)0.35color(white)(mmmmmmmll)0color(white)(mmmll)0#
#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmmml)"+x"color(white)(mmm)"+x"#
#"E/mol·L"^"-1":color(white)(ll)"0.35 -"xcolor(white)(mmmmmmm)xcolor(white)(mmmll)x#
The #K_"a"# expression is
#K_"a" = (["H"_3"O"^+]["F"^"-"])/(["HF"]) = 6.8 × 10^"-4"#
#K_a = (x × x)/(0.35-x) = x^2/(0.35-x) = 6.8 × 10^"-4"#
#0.35/K_"a" = 0.35/(6.31 × 10^"-4") = 515 > 400#
∴ #x ≪ 0.35#, and the equation becomes
#x^2/0.35 = 6.8 × 10^"-4"#
#x^2 = 0.35 × 6.8 × 10^"-4" = 2.38 × 10^"-4"#
#x = 1.54 × 10^"-2"#
#["H"^+] = 1.54 × 10^"-2"color(white)(l) "mol/L"#
#"pH" = "-log"["H"_3"O"^"+"] = "-log"(1.54 × 10^"-2") = 1.81#
