Question #1fb6d

1 Answer
May 7, 2017

Answer:

#"pH =5.18; pOH = 8.82"#

Explanation:

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#