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#
