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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Baksu888 on March 11, 2020, 06:26:54 PM

Title: HCl added to a methylamine solution (buffer)
Post by: Baksu888 on March 11, 2020, 06:26:54 PM

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:

CH₃NH₂ + H⁺  ::equil::  CH₃NH₃

[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!