How many mol of HCl would you have to add to 0.25 liter of 0.200 M methylamine solution to make a buffer with a pH of 11.00?

I tried plugging it into the Henderson Hasselbalch Equation:

CH

_{3}NH

_{2} + H

^{+} CH

_{3}NH

_{3}[tex] K_a = \frac {1.0*10^{-14}} {4.4*10^{-4}} = 2.3*10^{-11} [/tex]

[tex]0.25L * 0.200M = 0.050 \ mol [/tex]

[tex]mol \ of\ CH_3NH_2=0.050-x[/tex]

[tex]mol\ of\ CH_3NH_3=x[/tex]

[tex]11.00 = -log(2.3*10^{-11})+log \frac {x} {0.050-x}[/tex]

When I solved the equation, I got 0.035 mol of HCl. However, the answer for this question is supposed to be 0.015 mol. Also, when I tried to enter the equation into an online smart calculator (wolfram), I ended up with no real answers.

Was my method of calculation incorrect? Did I make a mistake in my calculations?

**EDIT:** I realized my stupid mistake. The Henderson Hasselbalch Equation calls for base over acid, and I accidentally did acid over base. So it should be the following:

[tex]11.00 = -log(2.3*10^{-11})+log \frac {0.050-x} {x}[/tex]

After solving the equation, I got 0.015 mol of HCl, which was the right answer.

Sorry about that!