Question

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

Answer 1

pH = 12.03

Explanation:

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

The chemical equation is

`"CH"_3"NH"_2 + "H"_2"O" ⇌ "CH"_3"NH"_3^"+" + "OH"^"-"; K_text(b) = 4.4 × 10^"-4"`

Let's rewrite this as

`color(white)(mmmmmmmmm)"B + H"_2"O" ⇌ "BH"^"+" + "OH"^"-"`
`"I/mol·L"^"-1":color(white)(mmm)0.26color(white)(mmmmmm)0color(white)(mmm)0`
`"C/mol·L"^"-1":color(white)(mmm)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mm)"+"x`
`"E/mol·L"^"-1":color(white)(ml)0.26-xcolor(white)(mmmmml)xcolor(white)(mmll)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`

`"pH" = "14.00 - pH" = "14.00 - 1.97" = 12.03`