At 25°C, the methylaminium ion, CH3NH3+, has a Ka of 2.0 x 10–11. Which of the following is the pH of a 0.15 mol L–1 solution of methylamine?

Question

i tried out two methods because im not getting the right answer:

H20 + Ch3Nh2 -> CH3NH3+ + OH-

I made a table gave me 0.15 - x

Ka = x^2 / 0.15

x^2 = 3x10^-12
x = 1.73x10^-6

plugged that into

pH = - log ( 1.73x10^-6) --> 5.76 but that's incorrect

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then i tried using Ka x Kb = Kw
Kb =Kw/Ka

= 1X10^-14 / 2X10^-11
=5X10^-26

I then used the 0.15 - x (from the table)

giving me x^2 / 0.15 = 5X10^-26

this led me to a closer answer to the correct pH however is still incorrect.

Please help

Answer 1

In your first attempt I saw two errors. There may be more but I quit looking. One error is that x = (OH^-) but you used it as (H^+). The second is I don't know where the 3E-12 came from. In the second trial you made a math error right off the bat and I quit looking. 5E-4 is more like Kw/Ka.

..CH3NH2 + HOH ==> CHNH3^+ + OH^-
I..0.15...............0.......0
C..-x................x.......x
E.0.15-x.............x.......x

Kb for CH3NH2 = (Kw/Ka for CH3NH3^+) = (x)(x)/(0.15-x)
Solve for x = (OH^-)</b) and convert to pH. Make sure you can use 0.15 for 0.15-x. It could be that you would need to use the quadratic for that. Also, that's the methylammonium ion.

Answer 2

Ka = x^2 /( 0.15 -x)
x^2+2E-11x-.15Ka=0

x^2+2E-11 x -.3E-11-0

x=(-2E-11+-sqrt(4E-22 +1.2E-11))/2

x= -1E-11+sqrt(3E-12)
x=1.73E-6
I agree with your pH.

Answer 3

Hey DrBob222 thankyou for getting me on the right track. I got the right answer. But I'm wondering...

So in the last steps using the solved x value which is [OH-]

the next step is pOH = -log[OH-]

is this the pOH of CH3NH2 or of just the hydroxide ions in solutions.

because i needed to further

pH = 14 - pOH