What is the concentration of OH- in a 2.0 M solution of the weak base CH3NH2?

Question

What is the concentration of OH- in a 2.0 M solution of the weak base CH3NH2

CH3NH2 + H2O <-> CH3NH3 + OH-

With Kb = 4.27 x 10^-4

And what is the pH?

Answer 1

B + H20 -->HB+ + OH-
(B is methylamine; HB+ is protonated B; and OH- is hydroxide ion)

Kb= [HB+][OH-]/[B]
([H2O] is largely unchanging and usually ignored)

B starts as 2.0 M, but changes by some amount that ionizes (x)
HB+ and OH- are ions formed from B and water in amount (x) each. We use the subscript "eq" for the value at equilibrium.
If you look at nairb's layout, "I" is initial, "C" is change, "E" is at equilibrium, which can lead to a nice mnemonic "ICE".

So by the time equilibrium is reached, Beq is not 2.0, but 2.0-x
HB+, which didn't exist at first, is now x
OH- is made at the same time, so its concentration is also x
Kb is given as 4.27 x 10^-4 (and we can use Kb for the number where convenient)

Therefore, Kb= 4.27 x 10^-4 = x*x/(2.0-x) = x^2/(2.0-x)

x^2 + Kbx - 2Kb = 0

x= -Kb +/-sqrt(Kb^2 + 8Kb)/2

x=2.9 x 10^-2 (0.029)

Let's check: If x = 0.029, then Beq= 2.0-x ~ 1.97M
Does x^2/[B] = 4.27 x 10^-4?
0.029^2 = 0.000841 or 8.41 x 10^-4, and dividing 0.000841 by 1.97 = 0.0004269 or 4.27 x 10^-4

The check is good, and the concentration of OH- is 0.029 M (2.9 x 10^-2 M).

BTW, how bad is ignoring the loss of B at Beq? Not too bad...it makes x=0.0292, only about a 1% error.

BTW#2...Methylamine may be a "weak" base compared to KOH, for example, but it's much stronger than ammonium hydroxide (Kb 1.8 x 10^-5).

Answer 2

Hope this helps:

1) CH3NH2 + H2O <===> CH3NH3+ + OH¯

Methylamine, CH3NH2, is a weak base that reacts according to the equation above. The value of the ionization constant, Kb, is 5.25 x 10¯4. Methylamine forms salts such as methylammonium nitrate, (CH3NH3+) (NO3¯).

CH3NH2 + H2O <===> CH3NH3+ + OH¯

Kb = ([CH3NH3+] [OH¯]) ÷ [CH3NH2] = 5.25 x 10¯4

CH3NH2 CH3NH3+ OH¯
I 0.225 0 0
C -x +x +x
E 0.225 - x x x

5.25 x 10¯4 = [(x) (x)] / (0.225 - x)

neglect the minus x to get 5.25 x 10¯4 = x2 / 0.225

x = [square root]((5.25 x 10¯4) (0.225))

[OH¯] = 1.09 x 10¯2

(Note: quadratic gives 1.06 x 10¯2)

Answer 3

BrO- acts as a base by utilising right here equation: BrO- + H2O <---> HBrO + OH- Kb = [HBrO][OH-] / [HBrO] recognize that [HBrO] = [OH-] = x, and [BrO-] = 0.545-x. anticipate that x is small while in comparison with 0.545, so, Kb = 4X10^-6 = x^2 / 0.545 x = [OH-] = a million.5X10^-3 M

Answer 4

28%

Answer 5

pH=14-pOH

pOH = - log [OH]

[OH]=2.9 x 10^-2, so pOH = 1.54, and pH = 14 - 1.54 = 12.46

pH = 12.46

Answer 6

You just gave me a headache... ouch