The chemical equation is
#"CH"_3"NH"_2 + "H"_2"O" ⇌ "CH"_3"NH"_3^"+" + "OH"^"-"#
Let's re-write this as
#"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"#
Then we can use an ICE table to do the calculation.
#color(white)(mmmmmmmm)"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"#
#"I/mol·L"^"-1":color(white)(mll)0.200color(white)(mmmmmll)0color(white)(mmm)0#
#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mm)"+"x#
#"E/mol·L"^"-1":color(white)(m)"0.200-"xcolor(white)(mmmmm)xcolor(white)(mmm)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#
#"pH = 14.00 - pOH = 14.00 - 2.26" = 11.74#
