# Question 1fb6d

May 7, 2017

$\text{pH =5.18; pOH = 8.82}$

#### Explanation:

We can use an ICE table to calculate the concentrations of the ions in solution.

The chemical equation is

$\text{HOCl + H"_2"O" ⇌ "H"_3"O"^"+" + "OCl"^"-"; K_text(a) = 3.5 × 10^"-8}$

Let's rewrite this as

$\textcolor{w h i t e}{m m m m m m m m m} \text{HA" +color(white)(ll) "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-}$
$\text{I/mol·L"^"-1":color(white)(mmll)"0.001 25} \textcolor{w h i t e}{m m m m m m} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mmmll)"-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1":color(white)(mll)"0.001 25 -} x \textcolor{w h i t e}{m m m m m l} x \textcolor{w h i t e}{m m l l} x$

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

${x}^{2} / \text{0.001 25" = 3.5 × 10^"-8}$

${x}^{2} = \text{0.001 25" × 3.5 × 10^"-8" = 4.38 × 10^"-11}$

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#

$\text{pOH" = "14.00 - pH = 14.00 - 5.18} = 8.82$