# What is the pH of a 0.26 mol/L solution of methylamine ("CH"_3"NH"_2, K_text(b) = 4.4 × 10^"-4")?

Apr 21, 2017

pH = 12.03

#### Explanation:

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

The chemical equation is

$\text{CH"_3"NH"_2 + "H"_2"O" ⇌ "CH"_3"NH"_3^"+" + "OH"^"-"; K_text(b) = 4.4 × 10^"-4}$

Let's rewrite this as

$\textcolor{w h i t e}{m m m m m m m m m} \text{B + H"_2"O" ⇌ "BH"^"+" + "OH"^"-}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m m} 0.26 \textcolor{w h i t e}{m m m m m m} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mmm)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mm)"+} x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m l} 0.26 - x \textcolor{w h i t e}{m m m m m l} x \textcolor{w h i t e}{m m l l} x$

K_text(b) = (["BH"^"+"]["OH"^"-"])/(["B"]) = (x × x)/(0.26-x) = x^2/(0.26-x) = 4.4 × 10^"-4"

Check for negligibility:

0.26/(4.4 × 10^"-4") = 5900 ≫ 400.

x ≪ 0.26

Then

x^2/0.26 = 4.4 × 10^"-4"

x^2 = 0.26 × 4.4 × 10^"-4" = 1.14 × 10^"-4"

x = 1.07 × 10^"-2"

["OH"^"-"] = x color(white)(l)"mol/L" = 1.07 × 10^"-2"color(white)(l)"mol/L"

"pOH" = "-log"["OH"^"-"] = "-log"(1.07 × 10^"-2") = 1.97

$\text{pH" = "14.00 - pH" = "14.00 - 1.97} = 12.03$