We know that #"p"K_"a" = 10.20#.
∴ #K_"a" = 10^"-10.20" = 6.31 × 10^"-11"#
Let's set up an ICE table for the calculation of #["H"_3"O"^+]#.
#color(white)(mmmmmmm)"HA"color(white)(l) +color(white)(l) "H"_2"O"color(white)(l) ⇌ color(white)(l) "H"_3"O"^+color(white)(l) +color(white)(l) "A"^"-"#
#"I/mol·L"^"-1": color(white)(m)0.8470color(white)(mmmmmmm)0color(white)(mmmml)0#
#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmmll)+xcolor(white)(mmll)+x#
#"E/mol·L"^"-1":color(white)(ll)"0.8470 -"xcolor(white)(mmmmmll)xcolor(white)(mmmml)x#
The #K_"a"# expression is
#K_"a" = (["H"_3"O"^+]["A"^"-"])/(["HA"]) = 6.31 × 10^"-11"#
#K_a = (x × x)/(0.8470-x) = x^2/(0.8470-x) = 6.31 × 10^"-11"#
Check if #x ≪ 0.8470#:
#0.8470/K_"a" = 0.8470/(6.31 × 10^"-11") = 1.34 × 10^10 ≫ 400#
∴ #x ≪ 0.8470#, and the equation becomes
#x^2/0.8470 = 6.31 × 10^"-11"#
#x^2 = 0.8470 × 6.31 × 10^"-11" = 5.34 × 10^"-11"#
#x = 7.31 × 10^"-6"#
#["H"^+] = 7.31 × 10^-6 "mol/L"#
#"pH" = -log["H"^+] = -log(7.31 × 10^-6) = 5.14#
