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?
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
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.
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.
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
To find the pH of a 0.15 mol L–1 solution of methylamine (CH3NH2), you need to consider the dissociation of methylamine in water and calculate the concentration of H+ ions.
Method 1:
You correctly set up the dissociation reaction: CH3NH2 + H2O → CH3NH3+ + OH-
Then, you created an ICE (Initial, Change, Equilibrium) table and assumed the change in concentration of CH3NH2 to be (0.15 - x) M.
To calculate Ka, you used the expression Ka = ([CH3NH3+][OH-])/[CH3NH2]. Since the solution is neutral at equilibrium (the concentration of [H+] is equal to [OH-]), you can write the expression as Ka = ([H+][OH-])/[CH3NH2].
Substituting the values, you obtained x^2/0.15 = 2.0 x 10–11.
Solving this equation gives you x = sqrt(2.0 x 10–11 * 0.15) ≈ 1.07 x 10–6.
Finally, you calculated the pH using pH = -log[H+] = -log(1.07 x 10–6) ≈ 5.97.
Method 2:
In this method, you used the relationship Ka x Kb = Kw to find Kb, the base dissociation constant.
Kb = Kw/Ka = (1 x 10^-14)/(2 x 10^-11) = 5 x 10^-4.
Using the equation Kb = ([OH-]^2)/[CH3NH2] and substituting values, you obtained (x^2)/0.15 = 5 x 10^-4.
Solving this equation gives you x = sqrt[(5 x 10^-4) x 0.15] ≈ 2.74 x 10^-3.
Again, you calculated the pH using pH = -log[H+] = -log(2.74 x 10^-3) ≈ 2.56.
Comparing the calculated pH values from Method 1 (pH ≈ 5.97) and Method 2 (pH ≈ 2.56) with the answer options, it seems neither is correct. It's possible that there is an error in the answer choices or in the calculations.
Double-check your calculations and ensure you have the correct answer options to resolve the discrepancy.