# Question fb9c4

May 16, 2017

The pH is 11.74.

#### Explanation:

The chemical equation is

$\text{CH"_3"NH"_2 + "H"_2"O" ⇌ "CH"_3"NH"_3^"+" + "OH"^"-}$

Let's re-write this as

$\text{B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-}$

Then we can use an ICE table to do the calculation.

$\textcolor{w h i t e}{m m m m m m m m} \text{B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m l l} 0.200 \textcolor{w h i t e}{m m m m m l l} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1":color(white)(m)"0.200-} x \textcolor{w h i t e}{m m m m m} x \textcolor{w h i t e}{m m m} x$

K_text(b) = (["BH"^"+"]["OH"^"-"])/(["B"]) = x^2/("0.200-"x) = 1.5 × 10^"-4"

Check for negligibility:

0.200/(1.5 × 10^"-4") = 1300 > 400. ∴ x ≪ 0.200

x^2/0.200 = 1.5 × 10^"-4"

x^2 = 0.200 × 1.5 × 10^"-4" = 3.00 × 10^"-5"

x = 5.48 × 10^"-3"

["OH"^"-"] = 5.48 × 10^"-3" color(white)(l)"mol/L"

"pOH" = -log(5.48 × 10^"-3") = 2.26#

$\text{pH = 14.00 - pOH = 14.00 - 2.26} = 11.74$