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
