We can use an ICE table to calculate the concentrations of the ions in solution.
The chemical equation is
#"HOCl + H"_2"O" ⇌ "H"_3"O"^"+" + "OCl"^"-"; K_text(a) = 3.5 × 10^"-8"#
Let's rewrite this as
#color(white)(mmmmmmmmm)"HA" +color(white)(ll) "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"#
#"I/mol·L"^"-1":color(white)(mmll)"0.001 25"color(white)(mmmmmm)0color(white)(mmm)0#
#"C/mol·L"^"-1":color(white)(mmmll)"-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mm)"+"x#
#"E/mol·L"^"-1":color(white)(mll)"0.001 25 -"xcolor(white)(mmmmml)xcolor(white)(mmll)x#
#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = (x × x)/("0.001 25 -"color(white)(l)x) = x^2/("0.001 25 -"color(white)(l)x) = 3.5 × 10^"-8"#
Check for negligibility:
#"0.001 25"/(3.5 × 10^"-8") = "36 000" ≫ 400#.
∴ #x ≪ "0.001 25"#.
Then
#x^2/"0.001 25" = 3.5 × 10^"-8"#
#x^2 = "0.001 25" × 3.5 × 10^"-8" = 4.38 × 10^"-11"#
#x = 6.61 × 10^"-6"#
#["H"_3"O"^"+"] = x color(white)(l)"mol/L" = 6.61 × 10^"-6"color(white)(l)"mol/L"#
#"pH" = "-log"["H"_3"O"^"+"] = "-log"(6.61 × 10^"-6") = 5.18#
#"pOH" = "14.00 - pH = 14.00 - 5.18" = 8.82#
